# Given Information - Two independent production units: \(U_1\) (60% output and \(U_2\)40% output). - item is either Defective (D) or Non-Defective (N). - Probability an item is non-defective: - \(P(N | U_1) = .8\) - \((N | U_2) = .2\) - A shipment consists of 6 items, all from the same (unknown) unit. - Let \(X\) be the number of non-defective items in the shipment: - \(X | U_1 \sim \text{Binomial}(6, .8)\) - \(X | U_2 \sim \text{Binomial}(6, .2)\) - **Inspection Rule:** - If more than 3 items are non-defective, shipment is accepted. - If exactly 3 items are non-defective, shipment is accepted with probability \(\frac{1}{2}\). - If fewer than 3 items are non-defective, shipment is rejected. - Let \(A\) denote the event that the shipment is accepted. # What Have to Find (a) \(P(A | U_1)\) (b) \(P(A | U_2)\) (c) \(P(A)\) (overall probability a shipment is accepted) (d) \(P(U_1 | A)\) (probability the shipment came from \(U_1\) if accepted) # Definition/Concept Used - **Binomial Probability:** \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\) - **Law of Total Probability:** \(P(A) = P(A|U_1)P(U_1) + P(A|U_2)P(U_2)\) - **Bayes' Theorem:** \(P(U_1|A) = \frac{P(A|U_1)P(U_1)}{P(A)}\) # Step-by-Step Solution ### (a) Find \(P(A | U_1)\) \(X | U_1 \sim \text{Binomial}(6, .8)\) - **More than 3 non-defective:** \(X = 4, 5, 6\) - **Exactly 3 non-defective:** \(X = 3\) (accepted with probability \(\frac{1}{2}\)) \[ P(A | U_1) = \sum_{k=4}^6 P(X = k | U_1) + \frac{1}{2}P(X=3|U_1) \] Calculate binomial probabilities: \[ P(X = k | U_1) = \binom{6}{k} (.8)^k (.2)^{6-k} \] \[ \begin{align*} P(X=6|U_1) &= \binom{6}{6}(.8)^6(.2)^ = 1 \cdot .262144 = .262144 \\ P(X=5|U_1) &= \binom{6}{5}(.8)^5(.2)^1 = 6 \cdot .32768 \cdot .2 = .393216 \\ P(X=4|U_1) &= \binom{6}{4}(.8)^4(.2)^2 = 15 \cdot .4096 \cdot .04 = .24576 \\ P(X=3|U_1) &= \binom{6}{3}(.8)^3(.2)^3 = 20 \cdot .512 \cdot .008 = .08192 \\ \end{align*} \] \[ P(A|U_1) = .262144 + .393216 + .24576 + \frac{1}{2}(.08192) \] \[ P(A|U_1) = .90112 \] --- ### (b) Find \(P(A | U_2)\) \(X | U_2 \sim \text{Binomial}(6, .2)\) - **More than 3 non-defective:** \(X = 4, 5, 6\) - **Exactly 3 non-defective:** \(X = 3\) (accepted with probability \(\frac{1}{2}\)) \[ P(X = k | U_2) = \binom{6}{k} (.2)^k (.8)^{6-k} \] \[ \begin{align*} P(X=6|U_2) &= \binom{6}{6}(.2)^6 = 1 \cdot .000064 = .000064 \\ P(X=5|U_2) &= \binom{6}{5}(.2)^5(.8)^1 = 6 \cdot .00032 \cdot .8 = .001536 \\ P(X=4|U_2) &= \binom{6}{4}(.2)^4(.8)^2 = 15 \cdot .0016 \cdot .64 = .01536 \\ P(X=3|U_2) &= \binom{6}{3}(.2)^3(.8)^3 = 20 \cdot .008 \cdot .512 = .08192 \\ \end{align*} \] \[ P(A|U_2) = .000064 + .001536 + .01536 + \frac{1}{2}(.08192) \] \[ P(A|U_2) = .057184 \] --- ### (c) Find \(P(A)\) (overall probability a shipment is accepted) Using the Law of Total Probability: \[ P(A) = P(A|U_1)P(U_1) + P(A|U_2)P(U_2) \] \[ P(A) = (.90112)(.6) + (.057184)(.4) \] \[ P(A) = .540672 + .022874 \] \[ P(A) = .563546 \] --- ### (d) Find \(P(U_1 | A)\) Using Bayes' Theorem: \[ P(U_1|A) = \frac{P(A|U_1)P(U_1)}{P(A)} \] \[ P(U_1|A) = \frac{.90112 \times .6}{.563546} \] \[ P(U_1|A) = \frac{.540672}{.563546} \] \[ P(U_1|A) = .9594 \] --- # Summary (Final Answers Only) - \(P(A|U_1) = .9011\) - \(P(A|U_2) = .0572\) - \(P(A) = .5635\) - \(P(U_1|A) = .9594\)

# Given Information - Two independent production units: - \(U_1\) produces 60% of total output. - \(U_2\) produces 40% of total output. - Probabilities of non-defective items: - \(P(N | U_1) = 0.8\) - \(P(N | U_2) = 0.2\) - A shipment consists of 6 items from the same unit. - Let \(X\) be the number of non-defective items in the shipment: - \(X | U_1 \sim \text{Binomial}(6, 0.8)\) - \(X | U_2 \sim \text{Binomial}(6, 0.2)\) - **Acceptance Rule:** - Accept if \(X > 3\) - Accept with probability \(\frac{1}{2}\) if \(X = 3\) - Reject if \(X < 3\) - Let \(A\) denote the event that the shipment is accepted. # What to Find (a) \(P(A | U_1)\) (b) \(P(A | U_2)\) (c) \(P(A)\) (overall probability that a shipment is accepted) (d) \(P(U_1 | A)\) (probability the shipment came from \(U_1\) if accepted) # Definitions/Concepts Used - **Binomial Probability:** \[ P(X=k) = \binom{n}{k} p^k (1-p)^{n-k} \] - **Law of Total Probability:** \[ P(A) = P(A|U_1)P(U_1) + P(A|U_2)P(U_2) \] - **Bayes' Theorem:** \[ P(U_1|A) = \frac{P(A|U_1)P(U_1)}{P(A)} \] # Step-by-Step Solution ### (a) Find \(P(A | U_1)\) Using the binomial distribution for \(U_1\): \[ P(A | U_1) = P(X \geq 4 | U_1) + \frac{1}{2}P(X = 3 | U_1) \] Calculate binomial probabilities for \(X | U_1\): \[ P(X = k | U_1) = \binom{6}{k} (0.8)^k (0.2)^{6-k} \] Calculating individual probabilities: - \(P(X=6|U_1) = (0.8)^6 = 0.262144\) - \(P(X=5|U_1) = 6 \cdot (0.8)^5 \cdot (0.2) = 0.393216\) - \(P(X=4|U_1) = 15 \cdot (0.8)^4 \cdot (0.2)^2 = 0.24576\) - \(P(X=3|U_1) = 20 \cdot (0.8)^3 \cdot (0.2)^3 = 0.08192\) Thus, \[ P(A|U_1) = 0.262144 + 0.393216 + 0.24576 + \frac{1}{2}(0.08192 = 0.90112 \] ### (b) Find \(P(A | U_2)\) Using the binomial distribution for \(U_2\): \[ P(A | U_2) = P(X \geq 4 | U_2) + \frac{1}{2}P(X = 3 | U_2) \] Calculating probabilities for \(X | U_2\): - \(P(X=6|U_2) = (0.2)^6 = 0.000064\) - \(P(X=5|U_2) = 6 \cdot (0.2)^5 \cdot (0.8) = 0.001536\) - \(P(X=4|U_2) = 15 \cdot (0.2)^4 \cdot (0.8)^2 = 0.01536\) - \(P(X=3|U_2) = 20 \cdot (0.2)^3 \cdot (0.8)^3 = 0.08192\) Thus, \[ P(A|U_2) = 0.000064 + 0.001536 + 0.01536 + \frac{1}{2}(0.08192) = 0.057184 \] ### (c) Find \(P(A)\) Using the Law of Total Probability: \[ P(A) = P(A|U_1)P(U_1) + P(A|U_2)P(U_2) \] \[ P(A) = (0.90112)(0.6) + (0.057184)(0.4) = 0.540672 + 0.022874 = 0.563546 \] ### (d) Find \(P(U_1 | A)\) Using Bayes' Theorem: \[ P(U_1|A) = \frac{P(A|U_1)P(U_1)}{P(A)} = \frac{(0.90112)(0.6)}{0.563546} = \frac{0.540672}{0.563546} \approx 0.9594 \] # Summary (Final Answers Only) - \(P(A|U_1) \approx 0.9011\) - \(P(A|U_2) \approx 0.0572\) - \(P(A) \approx 0.5635\) - \(P(U_1|A) \approx 0.9594\)

# Given Information - Three independent evaluation models: - Model \(M_1\): \(P(M_1) = 0.4\) - Model \(M_2\): \(P(M_2) = 0.35\) - Model \(M_3\): \(P(M_3) = 0.25\) - **Pass probabilities for Test A (Credit Score Filter)**: - \(P(\text{Pass A} | M_1) = 0.85\) - \(P(\text{Pass A} | M_2) = 0.80\) - \(P(\text{Pass A} | M_3) = 0.75\) - **Pass probabilities for Test B (Income Verification)**: - \(P(\text{Pass B} | M_1) = 0.90\) - \(P(\text{Pass B} | M_2) = 0.70\) - \(P(\text{Pass B} | M_3) = 0.60\) - An application is approved only if it passes both tests A and B. # What to Find - Probability that an application is approved, \(P(A)\). # Definitions/Concepts Used - **Total Probability**: \[ P(A) = P(A | M_1)P(M_1) + P(A | M_2)P(M_2) + P(A | M_3)P(M_3) \] - **Independent Events**: If \(X\) and \(Y\) are independent, then \(P(X \cap Y) = P(X)P(Y)\). ### Calculation of \(P(A | M_i)\) For each model \(M_i\), the application is approved if it passes both tests A and B: \[ P(A | M_i) = P(\text{Pass A} | M_i) \times P(\text{Pass B} | M_i) \] Calculating for each model: - For \(M_1\): \[ P(A | M_1) = P(\text{Pass A} | M_1) \times P(\text{Pass B} | M_1) = 0.85 \times 0.90 = 0.765 \] - For \(M_2\): \[ P(A | M_2) = P(\text{Pass A} | M_2) \times P(\text{Pass B} | M_2) = 0.80 \times 0.70 = 0.56 \] - For \(M_3\): \[ P(A | M_3) = P(\text{Pass A} | M_3) \times P(\text{Pass B} | M_3) = 0.75 \times 0.60 = 0.45 \] ### Overall Probability of Approval \(P(A)\) Using the Total Probability formula: \[ P(A) = P(A | M_1)P(M_1) + P(A | M_2)P(M_2) + P(A | M_3)P(M_3) \] Substituting the values: \[ P(A) = (0.765)(0.4) + (0.56)(0.35) + (0.45)(0.25) \] Calculating each term: - \(0.765 \times 0.4 = 0.306\) - \(0.56 \times 0.35 = 0.196\) - \(0.45 \times 0.25 = 0.1125\) Now summing these: \[ P(A) = 0.306 + 0.196 + 0.1125 = 0.6145 \] # Summary (Final Answer Only) - \(P(A) = 0.6145\)

# Given Information - Two factories producing microprocessors: - Factory \(F_1\): produces 60% of total chips. - Factory \(F_2\): produces 40% of total chips. - Probabilities of working chips: - \(P(\text{Working} | F_1) = 0.7\) - \(P(\text{Working} | F_2) = 0.5\) - Let \(x\) denote the number of working chips among 4: - \(x | F_1 \sim \text{Binomial}(4, 0.7)\) - \(x | F_2 \sim \text{Binomial}(4, 0.5)\) - Let event \(A = \{x = 2\}\). # What to Find - Find \(P(F_1 | x = 2)\), the probability that the shipment came from Factory \(F_1\) given that exactly 2 chips are working. # Definitions/Concepts Used - **Bayes' Theorem**: \[ P(F_1 | A) = \frac{P(A | F_1) P(F_1)}{P(A)} \] - **Binomial Probability**: \[ P(X = k | n, p) = \binom{n}{k} p^k (1-p)^{n-k} \] # Step-by-Step Solution ### Calculate \(P(A | F_1)\) Using the binomial distribution for Factory \(F_1\): \[ P(A | F_1) = P(x = 2 | F_1) = \binom{4}{2} (0.7)^2 (0.3)^2 \] Calculating: \[ \binom{4}{2} = 6 \] \[ (0.7)^2 = 0.49 \quad \text{and} \quad (0.3)^2 = 0.09 \] Thus, \[ P(A | F_1) = 6 \cdot 0.49 \cdot 0.09 = 0.2646 \] ### Calculate \(P(A | F_2)\) Using the binomial distribution for Factory \(F_2\): \[ P(A | F_2) = P(x = 2 | F_2) = \binom{4}{2} (0.5)^2 (0.5)^2 \] Calculating: \[ \binom{4}{2} = 6 \] \[ (0.5)^2 = 0.25 \] Thus, \[ P(A | F_2) = 6 \cdot 0.25 \cdot 0.25 = 0.375 \] ### Calculate \(P(A)\) Using Total Probability Using the Law of Total Probability: \[ P(A) = P(A | F_1) P(F_1) + P(A | F_2) P(F_2) \] Substituting the values: \[ P(A) = (0.2646)(0.6) + (0.375)(0.4) \] Calculating each term: \[ 0.2646 \times 0.6 = 0.15876 \] \[ 0.375 \times 0.4 = 0.15 \] Thus, \[ P(A) = 0.15876 + 0.15 = 0.30876 \] ### Apply Bayes' Theorem Now calculate \(P(F_1 | A)\): \[ P(F_1 | A) = \frac{P(A | F_1) P(F_1)}{P(A)} \] Substituting the values: \[ P(F_1 | A) = \frac{(0.2646)(0.6)}{0.30876} \] Calculating: \[ P(F_1 | A) = \frac{0.15876}{0.30876} \approx 0.513 \] # Summary (Final Answer Only) - \(P(F_1 | x = 2) \approx 0.513\)

# Given Information - Two automated production systems: - System \(S_1\): produces 50% of total devices. - System \(S_2\): produces 50% of total devices. - Probabilities of working devices: - \(P(\text{Working} | S_1) = 0.6\) - \(P(\text{Working} | S_2) = 0.4\) - Let \(x\) denote the number of working devices among the 6 selected: - \(x | S_1 \sim \text{Binomial}(6, 0.6)\) - \(x | S_2 \sim \text{Binomial}(6, 0.4)\) - **Approval Rule:** - Approved if \(x > 3\) - Approved with probability \(\frac{1}{2}\) if \(x = 3\) - Rejected if \(x < 3\) - Let event \(A\) denote that the shipment is approved. # What to Find - Overall probability that the shipment is approved, \(P(A)\). # Definitions/Concepts Used - **Binomial Probability**: \[ P(X = k | n, p) = \binom{n}{k} p^k (1-p)^{n-k} \] - **Law of Total Probability**: \[ P(A) = P(A | S_1)P(S_1) + P(A | S_2)P(S_2) \] # Step-by-Step Solution ### Calculate \(P(A | S_1)\) Using the binomial distribution for \(S_1\): \[ P(A | S_1) = P(x > 3 | S_1) + \frac{1}{2}P(x = 3 | S_1) \] Calculating the probabilities: - For \(x = 4, 5, 6\): \[ P(x = k | S_1) = \binom{6}{k} (0.6)^k (0.4)^{6-k} \] Calculating individual probabilities: - \(P(x=6 | S_1)\): \[ P(x = 6 | S_1) = (0.6)^6 = 0.046656 \] - \(P(x=5 | S_1)\): \[ P(x = 5 | S_1) = \binom{6}{5} (0.6)^5 (0.4)^1 = 6 \cdot 0.07776 \cdot 0.4 = 0.186624 \] - \(P(x=4 | S_1)\): \[ P(x = 4 | S_1) = \binom{6}{4} (0.6)^4 (0.4)^2 = 15 \cdot 0.1296 \cdot 0.16 = 0.31104 \] - \(P(x=3 | S_1)\): \[ P(x = 3 | S_1) = \binom{6}{3} (0.6)^3 (0.4)^3 = 20 \cdot 0.216 \cdot 0.064 = 0.27648 \] Summing the probabilities: \[ P(A | S_1) = P(x > 3 | S_1) + \frac{1}{2} P(x = 3 | S_1) = (0.046656 + 0.186624 + 0.31104) + \frac{1}{2}(0.27648) \] \[ P(A | S_1) = 0.54432 + 0.13824 = 0.68256 \] ### Calculate \(P(A | S_2)\) Using the binomial distribution for \(S_2\): \[ P(A | S_2) = P(x > 3 | S_2) + \frac{1}{2}P(x = 3 | S_2) \] Calculating the probabilities: - For \(x = 4, 5, 6\): \[ P(x = k | S_2) = \binom{6}{k} (0.4)^k (0.6)^{6-k} \] Calculating individual probabilities: - \(P(x=6 | S_2)\): \[ P(x = 6 | S_2) = (0.4)^6 = 0.004096 \] - \(P(x=5 | S_2)\): \[ P(x = 5 | S_2) = \binom{6}{5} (0.4)^5 (0.6)^1 = 6 \cdot 0.01024 \cdot 0.6 = 0.06144 \] - \(P(x=4 | S_2)\): \[ P(x = 4 | S_2) = \binom{6}{4} (0.4)^4 (0.6)^2 = 15 \cdot 0.0256 \cdot 0.36 = 0.13824 \] - \(P(x=3 | S_2)\): \[ P(x = 3 | S_2) = \binom{6}{3} (0.4)^3 (0.6)^3 = 20 \cdot 0.064 \cdot 0.216 = 0.27648 \] Summing the probabilities: \[ P(A | S_2) = P(x > 3 | S_2) + \frac{1}{2} P(x = 3 | S_2) = (0.004096 + 0.06144 + 0.13824) + \frac{1}{2}(0.27648) \] \[ P(A | S_2) = 0.203776 + 0.13824 = 0.342016 \] ### Calculate Overall Probability \(P(A)\) Using the Law of Total Probability: \[ P(A) = P(A | S_1) P(S_1) + P(A | S_2) P(S_2) \] \[ P(A) = (0.68256)(0.5) + (0.342016)(0.5) \] \[ P(A) = 0.34128 + 0.171008 = 0.512288 \] # Summary (Final Answer Only) - \(P(A) \approx 0.5123\)

# Given Information - Three machines in a factory: - **Machine A**: Produces 30% of items with a defect rate of 2% (0.02). - **Machine B**: Produces 45% of items with a defect rate of 3% (0.03). - **Machine C**: Produces 25% of items with a defect rate of 5% (0.05). # What to Find - The probability that a randomly selected defective item came from Machine C, denoted as \(P(C | D)\). # Definitions/Concepts Used - **Bayes' Theorem**: \[ P(C | D) = \frac{P(D | C) P(C)}{P(D)} \] - **Total Probability of Defective Items**: \[ P(D) = P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C) \] # Step-by-Step Solution ### Calculate Probabilities of Defective Items 1. **Probability of selecting an item from each machine**: - \(P(A) = 0.3\) - \(P(B) = 0.45\) - \(P(C) = 0.25\) 2. **Probability of an item being defective from each machine**: - \(P(D | A) = 0.02\) - \(P(D | B) = 0.03\) - \(P(D | C) = 0.05\) ### Calculate Overall Probability of a Defective Item \(P(D)\) Using the total probability formula: \[ P(D) = P(D | A) P(A) + P(D | B) P(B) + P(D | C) P(C) \] Substituting the values: \[ P(D) = (0.02)(0.3) + (0.03)(0.45) + (0.05)(0.25) \] Calculating each term: - From Machine A: \(0.02 \times 0.3 = 0.006\) - From Machine B: \(0.03 \times 0.45 = 0.0135\) - From Machine C: \(0.05 \times 0.25 = 0.0125\) Summing these probabilities: \[ P(D) = 0.006 + 0.0135 + 0.0125 = 0.032 \] ### Apply Bayes' Theorem to Find \(P(C | D)\) Using Bayes' theorem: \[ P(C | D) = \frac{P(D | C) P(C)}{P(D)} \] Substituting the values: \[ P(C | D) = \frac{(0.05)(0.25)}{0.032} \] Calculating: \[ P(C | D) = \frac{0.0125}{0.032} \approx 0.390625 \] # Summary (Final Answer Only) - \(P(C | D) \approx 0.3906\)

# Given Information - Two factories producing microprocessors: - **Factory \(F_1\)**: produces 60% of total chips. - **Factory \(F_2\)**: produces 40% of total chips. - Probabilities of working chips: - \(P(\text{Working} | F_1) = 0.7\) - \(P(\text{Working} | F_2) = 0.5\) - Let \(x\) denote the number of working chips among 4: - \(x | F_1 \sim \text{Binomial}(4, 0.7)\) - \(x | F_2 \sim \text{Binomial}(4, 0.5)\) - Let event \(A = \{x = 2\}\). # What to Find - Find \(P(F_1 | x = 2)\), the probability that the shipment came from Factory \(F_1\) given that exactly 2 chips are working. # Definitions/Concepts Used - **Bayes' Theorem**: \[ P(F_1 | A) = \frac{P(A | F_1) P(F_1)}{P(A)} \] - **Binomial Probability**: \[ P(X = k | n, p) = \binom{n}{k} p^k (1-p)^{n-k} \] # Step-by-Step Solution ### Calculate \(P(A | F_1)\) Using the binomial distribution for Factory \(F_1\): \[ P(A | F_1) = P(x = 2 | F_1) = \binom{4}{2} (0.7)^2 (0.3)^2 \] Calculating: - \(\binom{4}{2} = 6\) - \((0.7)^2 = 0.49\) and \((0.3)^2 = 0.09\) Thus, \[ P(A | F_1) = 6 \cdot 0.49 \cdot 0.09 = 0.2646 \] ### Calculate \(P(A | F_2)\) Using the binomial distribution for Factory \(F_2\): \[ P(A | F_2) = P(x = 2 | F_2) = \binom{4}{2} (0.5)^2 (0.5)^2 \] Calculating: - \(\binom{4}{2} = 6\) - \((0.5)^2 = 0.25\) Thus, \[ P(A | F_2) = 6 \cdot 0.25 \cdot 0.25 = 0.375 \] ### Calculate \(P(A)\) Using Total Probability Using the Law of Total Probability: \[ P(A) = P(A | F_1) P(F_1) + P(A | F_2) P(F_2) \] Substituting the values: \[ P(A) = (0.2646)(0.6) + (0.375)(0.4) \] Calculating each term: \[ 0.2646 \times 0.6 = 0.15876 \] \[ 0.375 \times 0.4 = 0.15 \] Thus, \[ P(A) = 0.15876 + 0.15 = 0.30876 \] ### Apply Bayes' Theorem Now calculate \(P(F_1 | A)\): \[ P(F_1 | A) = \frac{P(A | F_1) P(F_1)}{P(A)} \] Substituting the values: \[ P(F_1 | A) = \frac{(0.2646)(0.6)}{0.30876} \] Calculating: \[ P(F_1 | A) = \frac{0.15876}{0.30876} \approx 0.513 \] # Summary (Final Answer Only) - \(P(F_1 | x = 2) \approx 0.513\)

# Given Information - Two automated production systems: - **System \(S_1\)**: produces 50% of total devices. - **System \(S_2\)**: produces 50% of total devices. - Probabilities of working devices: - \(P(\text{Working} | S_1) = 0.6\) - \(P(\text{Working} | S_2) = 0.4\) - Let \(x\) denote the number of working devices among the 6 selected: - \(x | S_1 \sim \text{Binomial}(6, 0.6)\) - \(x | S_2 \sim \text{Binomial}(6, 0.4)\) - **Approval Rule:** - Approved if \(x > 3\) - Approved with probability \(\frac{1}{2}\) if \(x = 3\) - Rejected if \(x < 3\) - Let event \(A\) denote that the shipment is approved. # What to Find - Overall probability that the shipment is approved, \(P(A)\). # Definitions/Concepts Used - **Binomial Probability**: \[ P(X = k | n, p) = \binom{n}{k} p^k (1-p)^{n-k} \] - **Law of Total Probability**: \[ P(A) = P(A | S_1)P(S_1) + P(A | S_2)P(S_2) \] # Step-by-Step Solution ### Calculate \(P(A | S_1)\) Using the binomial distribution for \(S_1\): \[ P(A | S_1) = P(x > 3 | S_1) + \frac{1}{2} P(x = 3 | S_1) \] Calculating the probabilities: - For \(x = 4, 5, 6\): \[ P(x = k | S_1) = \binom{6}{k} (0.6)^k (0.4)^{6-k} \] Calculating individual probabilities: - \(P(x=6 | S_1)\): \[ P(x = 6 | S_1) = (0.6)^6 = 0.046656 \] - \(P(x=5 | S_1)\): \[ P(x = 5 | S_1) = \binom{6}{5} (0.6)^5 (0.4)^1 = 6 \cdot 0.07776 \cdot 0.4 = 0.186624 \] - \(P(x=4 | S_1)\): \[ P(x = 4 | S_1) = \binom{6}{4} (0.6)^4 (0.4)^2 = 15 \cdot 0.1296 \cdot 0.16 = 0.31104 \] - \(P(x=3 | S_1)\): \[ P(x = 3 | S_1) = \binom{6}{3} (0.6)^3 (0.4)^3 = 20 \cdot 0.216 \cdot 0.064 = 0.27648 \] Summing the probabilities: \[ P(A | S_1) = P(x > 3 | S_1) + \frac{1}{2} P(x = 3 | S_1) = (0.046656 + 0.186624 + 0.31104) + \frac{1}{2}(0.27648) \] \[ P(A | S_1) = 0.54432 + 0.13824 = 0.68256 \] ### Calculate \(P(A | S_2)\) Using the binomial distribution for \(S_2\): \[ P(A | S_2) = P(x > 3 | S_2) + \frac{1}{2} P(x = 3 | S_2) \] Calculating the probabilities: - For \(x = 4, 5, 6\): \[ P(x = k | S_2) = \binom{6}{k} (0.4)^k (0.6)^{6-k} \] Calculating individual probabilities: - \(P(x=6 | S_2)\): \[ P(x = 6 | S_2) = (0.4)^6 = 0.004096 \] - \(P(x=5 | S_2)\): \[ P(x = 5 | S_2) = \binom{6}{5} (0.4)^5 (0.6)^1 = 6 \cdot 0.01024 \cdot 0.6 = 0.06144 \] - \(P(x=4 | S_2)\): \[ P(x = 4 | S_2) = \binom{6}{4} (0.4)^4 (0.6)^2 = 15 \cdot 0.0256 \cdot 0.36 = 0.13824 \] - \(P(x=3 | S_2)\): \[ P(x = 3 | S_2) = \binom{6}{3} (0.4)^3 (0.6)^3 = 20 \cdot 0.064 \cdot 0.216 = 0.27648 \] Summing the probabilities: \[ P(A | S_2) = P(x > 3 | S_2) + \frac{1}{2} P(x = 3 | S_2) = (0.004096 + 0.06144 + 0.13824) + \frac{1}{2}(0.27648) \] \[ P(A | S_2) = 0.203776 + 0.13824 = 0.342016 \] ### Calculate Overall Probability \(P(A)\) Using the Law of Total Probability: \[ P(A) = P(A | S_1) P(S_1) + P(A | S_2) P(S_2) \] \[ P(A) = (0.68256)(0.5) + (0.342016)(0.5) \] \[ P(A) = 0.34128 + 0.171008 = 0.512288 \] # Summary (Final Answer Only) - \(P(A) \approx 0.5123\)

# Given Information - Let \(x_1, x_2, \ldots, x_n\) be a random sample from the Normal distribution: \(x_i \sim N(\mu, \sigma^2)\) where both parameters \(\mu \in \mathbb{R}\) and \(\sigma^2 > 0\) are unknown. - The joint density of the sample is given by: \[ f(x; \mu, \sigma^2) = \left(2\pi\sigma^2\right)^{-n/2} \exp\left(-\frac{1}{2\sigma^2} \sum_{i=1}^n (x_i - \mu)^2\right) \] - Additional information provided: \(\sum (x_i - \mu)^2 = \sum (x_i - \bar{x})^2 + n(\bar{x} - \mu)^2\) - Fisher Information matrix for one observation: \[ I(\mu, \sigma^2) = \begin{pmatrix} \frac{1}{\sigma^2} & 0 \\ 0 & \frac{1}{2\sigma^4} \end{pmatrix} \] - Cramer-Rao Lower Bound for unbiased estimators is given by the inverse of \(nI(\mu, \sigma^2)\). # Questions ## (a) Show that the statistics \(T_1 = \sum x_i\) and \(T_2 = \sum x_i^2\) form a jointly sufficient statistic for \((\mu, \sigma^2)\). ### Solution 1. **Factorization Theorem**: The joint density can be factored into a product of two functions: one that depends on the data only through \(T_1\) and \(T_2\), and another that does not depend on \(\mu\) and \(\sigma^2\). 2. **Joint Density**: \[ f(x; \mu, \sigma^2) = \left(2\pi\sigma^2\right)^{-n/2} \exp\left(-\frac{1}{2\sigma^2} \left(\sum_{i=1}^n x_i^2 - 2\mu \sum_{i=1}^n x_i + n\mu^2\right)\right) \] 3. **Factorization**: - Let \(T_1 = \sum x_i\) - Let \(T_2 = \sum x_i^2\) The joint density can be expressed as: \[ f(x; \mu, \sigma^2) = g(T_1, T_2; \mu, \sigma^2) \cdot h(x) \] where \(g(T_1, T_2; \mu, \sigma^2)\) depends on the data only through \(T_1\) and \(T_2\). 4. **Conclusion**: Thus, \(T_1\) and \(T_2\) are jointly sufficient statistics for \((\mu, \sigma^2)\). ## (b) Find the Maximum Likelihood Estimators (MLEs) of \(\mu\) and \(\sigma^2\). ### Solution 1. **Log-Likelihood Function**: \[ \mathcal{L}(\mu, \sigma^2) = -\frac{n}{2} \ln(2\pi) - \frac{n}{2} \ln(\sigma^2) - \frac{1}{2\sigma^2} \sum_{i=1}^n (x_i - \mu)^2 \] 2. **Partial Derivative with respect to \(\mu\)**: \[ \frac{\partial \mathcal{L}}{\partial \mu} = \frac{1}{\sigma^2} \sum_{i=1}^n (x_i - \mu) = 0 \implies \hat{\mu} = \bar{x} \] 3. **Partial Derivative with respect to \(\sigma^2\)**: \[ \frac{\partial \mathcal{L}}{\partial \sigma^2} = -\frac{n}{2\sigma^2} + \frac{1}{2\sigma^4} \sum_{i=1}^n (x_i - \mu)^2 = 0 \] Rearranging gives: \[ n\sigma^2 = \sum_{i=1}^n (x_i - \hat{\mu})^2 \implies \hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2 \] 4. **Conclusion**: The MLEs are: - \(\hat{\mu} = \bar{x}\) - \(\hat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (x_i - \bar{x})^2\) ## (c) Compute the Fisher Information for the sample. ### Solution 1. **Fisher Information for \(n\) observations**: \[ I_n(\mu, \sigma^2) = n I(\mu, \sigma^2) = n \begin{pmatrix} \frac{1}{\sigma^2} & 0 \\ 0 & \frac{1}{2\sigma^4} \end{pmatrix} = \begin{pmatrix} \frac{n}{\sigma^2} & 0 \\ 0 & \frac{n}{2\sigma^4} \end{pmatrix} \] ## (d) Find the Cramer-Rao Lower Bound for unbiased estimators of \(\mu\). ### Solution 1. **Cramer-Rao Lower Bound**: The Cramer-Rao Lower Bound for an unbiased estimator of \(\mu\) is given by: \[ \text{CRLB}(\mu) = \left(I_n(\mu, \sigma^2)\right)^{-1}_{11} = \frac{\sigma^2}{n} \] # Summary - **(a)** \(T_1 = \sum x_i\) and \(T_2 = \sum x_i^2\) are jointly sufficient statistics for \((\mu, \sigma^2)\). - **(b)** MLEs are \(\hat{\mu} = \bar{x}\) and \(\hat{\sigma}^2 = \frac{1}{n} \sum (x_i - \bar{x})^2\). - **(c)** Fisher Information for the sample is \(I_n(\mu, \sigma^2) = \begin{pmatrix} \frac{n}{\sigma^2} & 0 \\ 0 & \frac{n}{2\sigma^4} \end{pmatrix}\). - **(d)** Cramer-Rao Lower Bound for \(\mu\) is \(\frac{\sigma^2}{n}\).

# Given Information - **Probability of radiation spike**: \(P(\text{Spike}) = 0.3\) \(P(\text{No Spike}) = 0.7\) - **Detection probabilities given a spike**: - \(P(D | S_1, \text{Spike}) = 0.9\) - \(P(D | S_2, \text{Spike}) = 0.8\) - \(P(D | S_3, \text{Spike}) = 0.7\) - **False alarm probabilities given no spike**: - \(P(D | S_1, \text{No Spike}) = 0.1\) - \(P(D | S_2, \text{No Spike}) = 0.2\) - \(P(D | S_3, \text{No Spike}) = 0.3\) - **Probabilities of selecting sensors**: - \(P(S_1) = 0.5\) - \(P(S_2) = 0.3\) - \(P(S_3) = 0.2\) # What to Find - Overall probability of detection, \(P(D)\). # Definitions/Concepts Used - **Law of Total Probability**: \[ P(D) = P(D | \text{Spike})P(\text{Spike}) + P(D | \text{No Spike})P(\text{No Spike}) \] - **Total Detection Probability for each Sensor**: \[ P(D | \text{Spike}) = \sum_{i=1}^{3} P(D | S_i, \text{Spike}) P(S_i) \] \[ P(D | \text{No Spike}) = \sum_{i=1}^{3} P(D | S_i, \text{No Spike}) P(S_i) \] # Step-by-Step Solution ### Calculate \(P(D | \text{Spike})\) \[ P(D | \text{Spike}) = P(D | S_1, \text{Spike}) P(S_1) + P(D | S_2, \text{Spike}) P(S_2) + P(D | S_3, \text{Spike}) P(S_3) \] Substituting the values: \[ P(D | \text{Spike}) = (0.9)(0.5) + (0.8)(0.3) + (0.7)(0.2) \] Calculating each term: - \(0.9 \times 0.5 = 0.45\) - \(0.8 \times 0.3 = 0.24\) - \(0.7 \times 0.2 = 0.14\) Now summing these: \[ P(D | \text{Spike}) = 0.45 + 0.24 + 0.14 = 0.83 \] ### Calculate \(P(D | \text{No Spike})\) \[ P(D | \text{No Spike}) = P(D | S_1, \text{No Spike}) P(S_1) + P(D | S_2, \text{No Spike}) P(S_2) + P(D | S_3, \text{No Spike}) P(S_3) \] Substituting the values: \[ P(D | \text{No Spike}) = (0.1)(0.5) + (0.2)(0.3) + (0.3)(0.2) \] Calculating each term: - \(0.1 \times 0.5 = 0.05\) - \(0.2 \times 0.3 = 0.06\) - \(0.3 \times 0.2 = 0.06\) Now summing these: \[ P(D | \text{No Spike}) = 0.05 + 0.06 + 0.06 = 0.17 \] ### Calculate Overall Probability \(P(D)\) Using the Law of Total Probability: \[ P(D) = P(D | \text{Spike})P(\text{Spike}) + P(D | \text{No Spike})P(\text{No Spike}) \] Substituting the values: \[ P(D) = (0.83)(0.3) + (0.17)(0.7) \] Calculating each term: - \(0.83 \times 0.3 = 0.249\) - \(0.17 \times 0.7 = 0.119\) Now summing these: \[ P(D) = 0.249 + 0.119 = 0.368 \] # Summary (Final Answer Only) - \(P(D) = 0.368\)

# Given Information - Probability that Emma answers correctly: \(P(E) = 0.62\) - Probability that Liam answers correctly: \(P(L) = 0.80\) # Questions ## A. What is the probability that both Emma and Liam answer correctly? ### Solution The events of Emma's and Liam's answers are independent. Therefore, the probability that both answer correctly is given by: \[ P(E \text{ and } L) = P(E) \times P(L) \] Substituting the values: \[ P(E \text{ and } L) = 0.62 \times 0.80 = 0.496 \] ### Summary - The probability that both Emma and Liam answer correctly is **0.496**. --- ## B. What is the probability that at least one of them answers correctly? ### Solution The probability that at least one answers correctly can be found using the complement rule: \[ P(\text{at least one correct}) = 1 - P(\text{neither correct}) \] The probability that neither answers correctly is: \[ P(\text{neither correct}) = P(E') \times P(L') \] Where \(P(E')\) and \(P(L')\) are the probabilities that Emma and Liam answer incorrectly: \[ P(E') = 1 - P(E) = 1 - 0.62 = 0.38 \] \[ P(L') = 1 - P(L) = 1 - 0.80 = 0.20 \] Now calculate \(P(\text{neither correct})\): \[ P(\text{neither correct}) = 0.38 \times 0.20 = 0.076 \] Thus, the probability that at least one answers correctly: \[ P(\text{at least one correct}) = 1 - 0.076 = 0.924 \] ### Summary - The probability that at least one of them answers correctly is **0.924**. --- ## C. Assuming the question is answered correctly by Liam, what is the probability that Emma answered correctly? ### Solution We need to find \(P(E | L)\), the probability that Emma answers correctly given that Liam answered correctly. Using the definition of conditional probability: \[ P(E | L) = \frac{P(E \text{ and } L)}{P(L)} \] From part A, we have \(P(E \text{ and } L) = 0.496\) and \(P(L) = 0.80\): \[ P(E | L) = \frac{0.496}{0.80} = 0.62 \] ### Summary - The probability that Emma answered correctly given that Liam answered correctly is **0.62**. --- ## D. Are the events of Liam and Emma getting the answer correct mutually exclusive? ### Solution Two events are mutually exclusive if they cannot happen at the same time. In this case, both Emma and Liam can answer correctly simultaneously. To illustrate: \[ P(E \text{ and } L) = 0.496 > 0 \] Since \(P(E \text{ and } L) > 0\), they are not mutually exclusive events. ### Summary - The events of Liam and Emma getting the answer correct are **not mutually exclusive** because they can both answer correctly at the same time.

# Given Information - **Probabilities of answering correctly**: - Probability that Emma answers correctly: \(P(E) = 0.62\) - Probability that Liam answers correctly: \(P(L) = 0.80\) # What to Find A. The probability that both Emma and Liam answer correctly. B. The probability that at least one of them answers correctly. C. Assuming Liam answers correctly, the probability that Emma answered correctly. D. Whether the events of Liam and Emma getting the answer correct are mutually exclusive. # Definitions/Concepts Used - **Independent Events**: Two events are independent if the occurrence of one does not affect the occurrence of the other. - **Complement Rule**: The probability of at least one event occurring can be found using the complement of the probability of neither event occurring. - **Conditional Probability**: The probability of event \(A\) given event \(B\) is given by: \[ P(A | B) = \frac{P(A \cap B)}{P(B)} \] # Step-by-Step Solution ### A. Probability that both Emma and Liam answer correctly The events are independent, so: \[ P(E \text{ and } L) = P(E) \times P(L) \] Substituting the values: \[ P(E \text{ and } L) = 0.62 \times 0.80 = 0.496 \] ### B. Probability that at least one of them answers correctly Using the complement rule: \[ P(\text{at least one correct}) = 1 - P(\text{neither correct}) \] First, calculate the probabilities of neither answering correctly: \[ P(E') = 1 - P(E) = 1 - 0.62 = 0.38 \] \[ P(L') = 1 - P(L) = 1 - 0.80 = 0.20 \] Now calculate \(P(\text{neither correct})\): \[ P(\text{neither correct}) = P(E') \times P(L') = 0.38 \times 0.20 = 0.076 \] Thus, the probability that at least one answers correctly is: \[ P(\text{at least one correct}) = 1 - 0.076 = 0.924 \] ### C. Probability that Emma answered correctly given Liam answered correctly Using conditional probability: \[ P(E | L) = \frac{P(E \text{ and } L)}{P(L)} \] From part A, we have \(P(E \text{ and } L) = 0.496\) and \(P(L) = 0.80\): \[ P(E | L) = \frac{0.496}{0.80} = 0.62 \] ### D. Are the events of Liam and Emma getting the answer correct mutually exclusive? To determine if they are mutually exclusive, check if both can happen at the same time: \[ P(E \text{ and } L) = 0.496 > 0 \] Since \(P(E \text{ and } L) > 0\), the events are not mutually exclusive. # Summary (Final Answers Only) - A. The probability that both Emma and Liam answer correctly is **0.496**. - B. The probability that at least one of them answers correctly is **0.924**. - C. The probability that Emma answered correctly given that Liam answered correctly is **0.62**. - D. The events of Liam and Emma getting the answer correct are **not mutually exclusive**.

# Given Information - The amount of insurance claims (in thousands of rupees) follows a Gamma distribution: - Shape parameter: \(\alpha = 2\) - Scale parameter: \(\theta\) (unknown) - A random sample of 5 claim amounts is observed: - \(x_1 = 4\), \(x_2 = 6\), \(x_3 = 5\), \(x_4 = 3\), \(x_5 = 7\) # What to Find (a) The Maximum Likelihood Estimator (MLE) of \(\theta\). (b) The mean claim amount. (c) The variance of the claim amount. (d) A 95% confidence interval for \(\theta\). # Definitions/Concepts Used - **Gamma Distribution**: The probability density function of a Gamma distribution is given by: \[ f(x; \alpha, \theta) = \frac{x^{\alpha - 1} e^{-x/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] - **Maximum Likelihood Estimation (MLE)**: The MLE is found by maximizing the likelihood function. - **Mean of Gamma Distribution**: The mean \(E(X)\) of a Gamma distribution is given by: \[ E(X) = \alpha \theta \] - **Variance of Gamma Distribution**: The variance \(\text{Var}(X)\) is given by: \[ \text{Var}(X) = \alpha \theta^2 \] - **Confidence Interval for \(\theta\)**: A confidence interval can be constructed using the properties of the Chi-squared distribution. # Step-by-Step Solution ### (a) Find the MLE of \(\theta\) The likelihood function for a Gamma distribution with parameters \(\alpha\) and \(\theta\) is given by: \[ L(\theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{n} \frac{x_i^{\alpha - 1} e^{-x_i/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] For our sample: \[ L(\theta) = \frac{1}{\theta^{2n} \Gamma(2)} \prod_{i=1}^{n} x_i^{1} e^{-\sum x_i / \theta} \] Taking the natural logarithm to find the log-likelihood: \[ \log L(\theta) = -2n \log \theta + \sum_{i=1}^{n} \log x_i - \frac{\sum_{i=1}^{n} x_i}{\theta} - n \log \Gamma(2) \] Differentiating with respect to \(\theta\) and setting it to zero for maximization: \[ \frac{d}{d\theta} \log L(\theta) = -\frac{2n}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \] This leads to: \[ \sum_{i=1}^{n} x_i = 2n \theta \] Thus, \[ \hat{\theta} = \frac{\sum_{i=1}^{n} x_i}{2n} \] Calculating \(\sum_{i=1}^{5} x_i = 4 + 6 + 5 + 3 + 7 = 25\) and \(n = 5\): \[ \hat{\theta} = \frac{25}{2 \times 5} = \frac{25}{10} = 2.5 \] ### (b) Estimate the mean claim amount Using the MLE found: \[ E(X) = \alpha \hat{\theta} = 2 \times 2.5 = 5 \] ### (c) Estimate the variance of the claim amount Using the MLE found: \[ \text{Var}(X) = \alpha \hat{\theta}^2 = 2 \times (2.5)^2 = 2 \times 6.25 = 12.5 \] ### (d) Construct a 95% confidence interval for \(\theta\) To construct a confidence interval for \(\theta\), we can use the Chi-squared distribution. The estimator \(\frac{2n\hat{\theta}}{\sigma^2}\) follows a Chi-squared distribution with \(2n\) degrees of freedom. The confidence interval is given by: \[ \left(\frac{2n\hat{\theta}}{\chi^2_{\alpha/2, 2n}}, \frac{2n\hat{\theta}}{\chi^2_{1-\alpha/2, 2n}}\right) \] Using \(n = 5\), we have \(2n = 10\). The critical values for \( \chi^2 \) with 10 degrees of freedom at \(\alpha = 0.05\) are approximately: - \(\chi^2_{0.025, 10} \approx 19.675\) - \(\chi^2_{0.975, 10} \approx 3.247\) Then: \[ \text{Lower Bound} = \frac{2 \times 5 \times 2.5}{19.675} \approx \frac{25}{19.675} \approx 1.273 \] \[ \text{Upper Bound} = \frac{2 \times 5 \times 2.5}{3.247} \approx \frac{25}{3.247} \approx 7.699 \] Thus, the 95% confidence interval for \(\theta\) is approximately: \[ (1.273, 7.699) \] # Summary (Final Answers Only) - (a) MLE of \(\theta\) is **2.5**. - (b) Estimated mean claim amount is **5**. - (c) Estimated variance of the claim amount is **12.5**. - (d) 95% confidence interval for \(\theta\) is approximately **(1.273, 7.699)**.

# Given Information - Prevalence of the disease in the population: \(P(D) = 0.02\) (2%) - Test sensitivity (True Positive Rate): \(P(T^+ | D) = 0.99\) (99%) - Test specificity (True Negative Rate): \(P(T^- | D') = 0.95\) (95%) Where: - \(D\): Event that a person has the disease. - \(D'\): Event that a person does not have the disease. - \(T^+\): Event that a person tests positive. - \(T^-\): Event that a person tests negative. # What to Find - The probability that a randomly selected person who tests positive actually has the disease, denoted as \(P(D | T^+)\). # Definitions/Concepts Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability** for \(P(T^+)\): \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D') P(D') \] Where \(P(T^+ | D') = 1 - P(T^- | D') = 1 - 0.95 = 0.05\). # Step-by-Step Solution ### Calculate \(P(T^+)\) Using the total probability formula: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D') P(D') \] Calculating \(P(D')\): \[ P(D') = 1 - P(D) = 1 - 0.02 = 0.98 \] Now substituting the values: \[ P(T^+) = (0.99)(0.02) + (0.05)(0.98) \] Calculating each term: - \(0.99 \times 0.02 = 0.0198\) - \(0.05 \times 0.98 = 0.049\) Summing these: \[ P(T^+) = 0.0198 + 0.049 = 0.0688 \] ### Calculate \(P(D | T^+)\) using Bayes' Theorem Now applying Bayes' theorem: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] Substituting the values: \[ P(D | T^+) = \frac{(0.99)(0.02)}{0.0688} \] Calculating: \[ P(D | T^+) = \frac{0.0198}{0.0688} \approx 0.288 \] # Summary (Final Answer Only) - The probability that a randomly selected person who tests positive actually has the disease is approximately **0.288** (or **28.8%**).

# Given Information - The lifetimes of components follow a Gamma distribution: - Shape parameter: \(\alpha = 3\) - Scale parameter: \(\theta\) (unknown) - A random sample of 5 lifetimes (in hours) is observed: - \(x_1 = 12\), \(x_2 = 9\), \(x_3 = 15\), \(x_4 = 18\), \(x_5 = 6\) # What to Find (a) The Maximum Likelihood Estimator (MLE) of \(\theta\). (b) The mean lifetime of the components. (c) The estimated variance of the lifetime. # Definitions/Concepts Used - **Gamma Distribution**: The probability density function of a Gamma distribution is: \[ f(x; \alpha, \theta) = \frac{x^{\alpha - 1} e^{-x/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] - **Maximum Likelihood Estimation (MLE)**: The MLE is found by maximizing the likelihood function. - **Mean of Gamma Distribution**: The mean \(E(X)\) of a Gamma distribution is given by: \[ E(X) = \alpha \theta \] - **Variance of Gamma Distribution**: The variance \(\text{Var}(X)\) is given by: \[ \text{Var}(X) = \alpha \theta^2 \] # Step-by-Step Solution ### (a) Find the MLE of \(\theta\) The likelihood function for a Gamma distribution with parameters \(\alpha\) and \(\theta\) is given by: \[ L(\theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{5} \frac{x_i^{\alpha - 1} e^{-x_i/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] For our sample: \[ L(\theta) = \frac{1}{\theta^{3n} \Gamma(3)} \prod_{i=1}^{5} x_i^{2} e^{-\sum_{i=1}^{5} x_i / \theta} \] Taking the logarithm to find the log-likelihood: \[ \log L(\theta) = -3n \log \theta + 2 \sum_{i=1}^{5} \log x_i - \frac{\sum_{i=1}^{5} x_i}{\theta} - n \log \Gamma(3) \] Differentiating with respect to \(\theta\) and setting it to zero for maximization: \[ \frac{d}{d\theta} \log L(\theta) = -\frac{3n}{\theta} + \frac{\sum_{i=1}^{5} x_i}{\theta^2} = 0 \] This leads to: \[ \sum_{i=1}^{5} x_i = 3n \theta \] Thus, \[ \hat{\theta} = \frac{\sum_{i=1}^{5} x_i}{3n} \] Calculating \(\sum_{i=1}^{5} x_i = 12 + 9 + 15 + 18 + 6 = 60\) and \(n = 5\): \[ \hat{\theta} = \frac{60}{3 \times 5} = \frac{60}{15} = 4 \] ### (b) Estimate the mean lifetime of the components Using the MLE found: \[ E(X) = \alpha \hat{\theta} = 3 \times 4 = 12 \] ### (c) Find the estimated variance of the lifetime Using the MLE found: \[ \text{Var}(X) = \alpha \hat{\theta}^2 = 3 \times (4)^2 = 3 \times 16 = 48 \] # Summary (Final Answers Only) - (a) MLE of \(\theta\) is **4**. - (b) Estimated mean lifetime of the components is **12** hours. - (c) Estimated variance of the lifetime is **48** hours².

# Given Information - The load at which a component fails follows a Gamma distribution: - Shape parameter: \(\alpha = 4\) - Scale parameter: \(\theta\) (unknown) - A sample of 5 failure loads (in tons) is observed: - \(x_1 = 20\), \(x_2 = 28\), \(x_3 = 24\), \(x_4 = 32\), \(x_5 = 16\) # What to Find (a) The Maximum Likelihood Estimator (MLE) of \(\theta\). (b) The mean failure load. (c) The variance of the failure load. (d) A 95% confidence interval for \(\theta\). # Definitions/Concepts Used - **Gamma Distribution**: The probability density function of a Gamma distribution is: \[ f(x; \alpha, \theta) = \frac{x^{\alpha - 1} e^{-x/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] - **Maximum Likelihood Estimation (MLE)**: The MLE is found by maximizing the likelihood function. - **Mean of Gamma Distribution**: The mean \(E(X)\) is given by: \[ E(X) = \alpha \theta \] - **Variance of Gamma Distribution**: The variance \(\text{Var}(X)\) is given by: \[ \text{Var}(X) = \alpha \theta^2 \] # Step-by-Step Solution ### (a) Find the MLE of \(\theta\) The likelihood function for a Gamma distribution is given by: \[ L(\theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{5} \frac{x_i^{\alpha - 1} e^{-x_i/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] For our sample: \[ L(\theta) = \frac{1}{\theta^{4n} \Gamma(4)} \prod_{i=1}^{5} x_i^{3} e^{-\sum_{i=1}^{5} x_i / \theta} \] Taking the logarithm to find the log-likelihood: \[ \log L(\theta) = -4n \log \theta + 3 \sum_{i=1}^{5} \log x_i - \frac{\sum_{i=1}^{5} x_i}{\theta} - n \log \Gamma(4) \] Differentiating with respect to \(\theta\) and setting it to zero for maximization: \[ \frac{d}{d\theta} \log L(\theta) = -\frac{4n}{\theta} + \frac{\sum_{i=1}^{5} x_i}{\theta^2} = 0 \] This leads to: \[ \sum_{i=1}^{5} x_i = 4n \theta \] Thus, \[ \hat{\theta} = \frac{\sum_{i=1}^{5} x_i}{4n} \] Calculating \(\sum_{i=1}^{5} x_i = 20 + 28 + 24 + 32 + 16 = 120\) and \(n = 5\): \[ \hat{\theta} = \frac{120}{4 \times 5} = \frac{120}{20} = 6 \] ### (b) Estimate the mean failure load Using the MLE found: \[ E(X) = \alpha \hat{\theta} = 4 \times 6 = 24 \] ### (c) Estimate the variance of the failure load Using the MLE found: \[ \text{Var}(X) = \alpha \hat{\theta}^2 = 4 \times (6)^2 = 4 \times 36 = 144 \] ### (d) Construct a 95% confidence interval for \(\theta\) To construct a confidence interval for \(\theta\), we can use the Chi-squared distribution. The estimator \(\frac{4n\hat{\theta}}{\sigma^2}\) follows a Chi-squared distribution with \(4n\) degrees of freedom. The confidence interval is given by: \[ \left(\frac{4n\hat{\theta}}{\chi^2_{\alpha/2, 4n}}, \frac{4n\hat{\theta}}{\chi^2_{1-\alpha/2, 4n}}\right) \] Using \(n = 5\), we have \(4n = 20\). The critical values for \( \chi^2 \) with 20 degrees of freedom at \(\alpha = 0.05\) are approximately: - \(\chi^2_{0.025, 20} \approx 32.671\) - \(\chi^2_{0.975, 20} \approx 10.851\) Then: \[ \text{Lower Bound} = \frac{4 \times 5 \times 6}{32.671} \approx \frac{120}{32.671} \approx 3.674 \] \[ \text{Upper Bound} = \frac{4 \times 5 \times 6}{10.851} \approx \frac{120}{10.851} \approx 11.058 \] Thus, the 95% confidence interval for \(\theta\) is approximately: \[ (3.674, 11.058) \] # Summary (Final Answers Only) - (a) MLE of \(\theta\) is **6**. - (b) Estimated mean failure load is **24** tons. - (c) Estimated variance of the failure load is **144** tons². - (d) 95% confidence interval for \(\theta\) is approximately **(3.674, 11.058)**.

# Given Information - The amount of insurance claims (in thousands of rupees) follows a Gamma distribution: - Shape parameter: \(\alpha = 2\) - Scale parameter: \(\theta\) (unknown) - A random sample of 5 claim amounts is observed: - \(x_1 = 4\), \(x_2 = 6\), \(x_3 = 5\), \(x_4 = 3\), \(x_5 = 7\) # What to Find (a) The Maximum Likelihood Estimator (MLE) of \(\theta\). (b) The mean claim amount. (c) The variance of the claim amount. (d) A 95% confidence interval for \(\theta\). # Definitions/Concepts Used - **Gamma Distribution**: The probability density function of a Gamma distribution is given by: \[ f(x; \alpha, \theta) = \frac{x^{\alpha - 1} e^{-x/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] - **Maximum Likelihood Estimation (MLE)**: The MLE is found by maximizing the likelihood function. - **Mean of Gamma Distribution**: The mean \(E(X)\) is given by: \[ E(X) = \alpha \theta \] - **Variance of Gamma Distribution**: The variance \(\text{Var}(X)\) is given by: \[ \text{Var}(X) = \alpha \theta^2 \] # Step-by-Step Solution ### (a) Find the MLE of \(\theta\) The likelihood function for a Gamma distribution is given by: \[ L(\theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{5} \frac{x_i^{\alpha - 1} e^{-x_i/\theta}}{\theta^{\alpha} \Gamma(\alpha)} \] For our sample: \[ L(\theta) = \frac{1}{\theta^{2n} \Gamma(2)} \prod_{i=1}^{5} x_i^{1} e^{-\sum_{i=1}^{5} x_i / \theta} \] Taking the logarithm to find the log-likelihood: \[ \log L(\theta) = -2n \log \theta + \sum_{i=1}^{5} \log x_i - \frac{\sum_{i=1}^{5} x_i}{\theta} - n \log \Gamma(2) \] Differentiating with respect to \(\theta\) and setting it to zero for maximization: \[ \frac{d}{d\theta} \log L(\theta) = -\frac{2n}{\theta} + \frac{\sum_{i=1}^{5} x_i}{\theta^2} = 0 \] This leads to: \[ \sum_{i=1}^{5} x_i = 2n \theta \] Thus, \[ \hat{\theta} = \frac{\sum_{i=1}^{5} x_i}{2n} \] Calculating \(\sum_{i=1}^{5} x_i = 4 + 6 + 5 + 3 + 7 = 25\) and \(n = 5\): \[ \hat{\theta} = \frac{25}{2 \times 5} = \frac{25}{10} = 2.5 \] ### (b) Estimate the mean claim amount Using the MLE found: \[ E(X) = \alpha \hat{\theta} = 2 \times 2.5 = 5 \] ### (c) Estimate the variance of the claim amount Using the MLE found: \[ \text{Var}(X) = \alpha \hat{\theta}^2 = 2 \times (2.5)^2 = 2 \times 6.25 = 12.5 \] ### (d) Construct a 95% confidence interval for \(\theta\) To construct a confidence interval for \(\theta\), we can use the Chi-squared distribution. The estimator \(\frac{2n\hat{\theta}}{\sigma^2}\) follows a Chi-squared distribution with \(2n\) degrees of freedom. The confidence interval is given by: \[ \left(\frac{2n\hat{\theta}}{\chi^2_{\alpha/2, 2n}}, \frac{2n\hat{\theta}}{\chi^2_{1-\alpha/2, 2n}}\right) \] Using \(n = 5\), we have \(2n = 10\). The critical values for \(\chi^2\) with 10 degrees of freedom at \(\alpha = 0.05\) are approximately: - \(\chi^2_{0.025, 10} \approx 19.675\) - \(\chi^2_{0.975, 10} \approx 3.247\) Then: \[ \text{Lower Bound} = \frac{2 \times 5 \times 2.5}{19.675} \approx \frac{25}{19.675} \approx 1.273 \] \[ \text{Upper Bound} = \frac{2 \times 5 \times 2.5}{3.247} \approx \frac{25}{3.247} \approx 7.699 \] Thus, the 95% confidence interval for \(\theta\) is approximately: \[ (1.273, 7.699) \] # Summary (Final Answers Only) - (a) MLE of \(\theta\) is **2.5**. - (b) Estimated mean claim amount is **5**. - (c) Estimated variance of the claim amount is **12.5**. - (d) 95% confidence interval for \(\theta\) is approximately **(1.273, 7.699)**.

# Given Information - A random sample \(x_1, x_2, \ldots, x_n\) is drawn from a Gamma distribution with: - Unknown shape parameter \(\alpha > 0\) - Unknown scale parameter \(\theta > 0\) - The probability density function (PDF) is given by: \[ f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x^{\alpha - 1} e^{-x/\theta}, \quad x > 0 \] # What to Find (a) The likelihood function \(L(\alpha, \theta)\). (b) The likelihood equations for the Maximum Likelihood Estimators (MLEs) of \(\alpha\) and \(\theta\). (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). (d) Find the Fisher Information matrix for one observation. (e) State whether closed-form solutions for the MLEs exist. Justify your answer briefly. # Definitions/Concepts Used - **Likelihood Function**: The likelihood function is the joint probability of the observed data as a function of the parameters. - **Sufficient Statistics**: A statistic is sufficient for a parameter if the conditional distribution of the sample given the statistic does not depend on the parameter. - **Fisher Information**: It measures the amount of information that an observable random variable carries about an unknown parameter. # Step-by-Step Solution ### (a) Write down the likelihood function \(L(\alpha, \theta)\) The likelihood function is given by the product of the densities for the observed sample: \[ L(\alpha, \theta) = \prod_{i=1}^n f(x_i; \alpha, \theta) = \prod_{i=1}^n \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x_i^{\alpha - 1} e^{-x_i/\theta} \right) \] This simplifies to: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^n x_i^{\alpha - 1} e^{-\sum_{i=1}^n x_i/\theta} \] ### (b) Obtain the likelihood equations for the MLEs of \(\alpha\) and \(\theta\) Taking the natural logarithm of the likelihood function gives us the log-likelihood: \[ \log L(\alpha, \theta) = -n \log \Gamma(\alpha) - n \alpha \log \theta + (\alpha - 1) \sum_{i=1}^n \log x_i - \frac{\sum_{i=1}^n x_i}{\theta} \] Differentiating with respect to \(\alpha\) and \(\theta\) and setting the equations to zero for maximization: \[ \frac{\partial \log L}{\partial \alpha} = -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} - n \log \theta + \sum_{i=1}^n \log x_i = 0 \] \[ \frac{\partial \log L}{\partial \theta} = -\frac{n \alpha}{\theta} + \frac{\sum_{i=1}^n x_i}{\theta^2} = 0 \] ### (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\) According to the factorization theorem, we can express the likelihood function in a form showing dependence only on \(T_1\) and \(T_2\): \[ L(\alpha, \theta) = g(T_1, T_2; \alpha, \theta) \cdot h(x) \] The form of the likelihood shows that \(T_1\) and \(T_2\) contain all necessary information about \(\alpha\) and \(\theta\) from the observed data, confirming they are jointly sufficient statistics. ### (d) Find the Fisher Information matrix for one observation The Fisher Information matrix \(I(\alpha, \theta)\) contains elements defined as: \[ I_{ij} = -E\left[\frac{\partial^2 \log L}{\partial \theta_i \partial \theta_j}\right] \] Calculating the expected values from the second derivatives of the log-likelihood, we can derive the Fisher Information matrix for parameters \(\alpha\) and \(\theta\): 1. For \(\alpha\): \[ I_{\alpha\alpha} = \frac{n \Gamma'(\alpha)^2}{\Gamma(\alpha)^2} - n \frac{\Gamma''(\alpha)}{\Gamma(\alpha)} \] 2. For \(\theta\): \[ I_{\theta\theta} = \frac{n \alpha}{\theta^2} \] 3. For \(\alpha\) and \(\theta\): \[ I_{\alpha\theta} = 0 \] Thus, the Fisher Information matrix is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] ### (e) State whether closed-form solutions for the MLEs exist Closed-form solutions for the MLEs of \(\alpha\) and \(\theta\) do not exist in general for Gamma distributions. The equations derived for the MLEs involve the digamma function \(\psi(\alpha)\) and are typically solved numerically. The presence of the logarithm and the gamma function complicates the equations, indicating that numerical approaches, such as Newton-Raphson, are often necessary to find MLEs. # Summary - (a) The likelihood function is given by: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] - (b) The likelihood equations involve solving: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} - n \log \theta + \sum_{i=1}^{n} \log x_i = 0 \] \[ -\frac{n \alpha}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \] - (c) The statistics \(T_1 = \sum x_i\) and \(T_2 = \sum \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). - (d) The Fisher Information matrix for one observation is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] - (e) Closed-form solutions for the MLEs do not exist; numerical methods are required.

# Given Information - A random sample \(x_1, x_2, \ldots, x_n\) is drawn from a Gamma distribution with: - Unknown shape parameter \(\alpha > 0\) - Unknown scale parameter \(\theta > 0\) - The probability density function (PDF) is given by: \[ f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x^{\alpha - 1} e^{-x/\theta}, \quad x > 0 \] # What to Find (a) The likelihood function \(L(\alpha, \theta)\). (b) The likelihood equations for the Maximum Likelihood Estimators (MLEs) of \(\alpha\) and \(\theta\). (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). (d) Find the Fisher Information matrix for one observation. (e) State whether closed-form solutions for the MLEs exist and justify your answer briefly. # Definitions/Concepts Used - **Likelihood Function**: The likelihood function is the joint probability of the observed data as a function of the parameters. - **Sufficient Statistics**: A statistic is sufficient for a parameter if the conditional distribution of the sample given the statistic does not depend on the parameter. - **Fisher Information**: It measures the amount of information that an observable random variable carries about an unknown parameter. # Step-by-Step Solution ### (a) Write down the likelihood function \(L(\alpha, \theta)\) The likelihood function for \(n\) observations is given by the product of the densities for the observed sample: \[ L(\alpha, \theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{n} \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x_i^{\alpha - 1} e^{-x_i/\theta} \right) \] This simplifies to: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] ### (b) Obtain the likelihood equations for the MLEs of \(\alpha\) and \(\theta\) Taking the natural logarithm of the likelihood function gives us the log-likelihood: \[ \log L(\alpha, \theta) = -n \log \Gamma(\alpha) - n \alpha \log \theta + (\alpha - 1) \sum_{i=1}^{n} \log x_i - \frac{\sum_{i=1}^{n} x_i}{\theta} \] Differentiating with respect to \(\alpha\) and \(\theta\) and setting the derivatives to zero: 1. **For \(\alpha\)**: \[ \frac{\partial \log L}{\partial \alpha} = -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} - n \log \theta + \sum_{i=1}^{n} \log x_i = 0 \] 2. **For \(\theta\)**: \[ \frac{\partial \log L}{\partial \theta} = -\frac{n \alpha}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \] Thus, the likelihood equations are: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^{n} \log x_i = n \log \theta \] \[ \sum_{i=1}^{n} x_i = n \alpha \theta \] ### (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\) Using the factorization theorem, we can express the likelihood function in a form that shows dependence only on \(T_1\) and \(T_2\): \[ L(\alpha, \theta) = g(T_1, T_2; \alpha, \theta) \cdot h(x) \] The form of the likelihood shows that \(T_1\) and \(T_2\) contain all necessary information about \(\alpha\) and \(\theta\) from the observed data, confirming they are jointly sufficient statistics. ### (d) Find the Fisher Information matrix for one observation The Fisher Information matrix \(I(\alpha, \theta)\) contains elements defined as: \[ I_{ij} = -E\left[\frac{\partial^2 \log L}{\partial \theta_i \partial \theta_j}\right] \] Calculating the second derivatives of the log-likelihood: 1. For \(\alpha\): \[ I_{\alpha\alpha} = n \left( \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} \right)^2 - n \frac{\Gamma''(\alpha)}{\Gamma(\alpha)} \] 2. For \(\theta\): \[ I_{\theta\theta} = \frac{n \alpha}{\theta^2} \] 3. For \(\alpha\) and \(\theta\): \[ I_{\alpha\theta} = 0 \] Thus, the Fisher Information matrix is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] ### (e) State whether closed-form solutions for the MLEs exist Closed-form solutions for the MLEs of \(\alpha\) and \(\theta\) do not exist in general for Gamma distributions. The equations derived for the MLEs involve the digamma function \(\psi(\alpha)\) and are typically solved numerically. The presence of the logarithm and the gamma function complicates the equations, indicating that numerical approaches, such as Newton-Raphson, are often necessary to find MLEs. # Summary - (a) The likelihood function is given by: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] - (b) The likelihood equations involve solving: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^{n} \log x_i = n \log \theta \] \[ \sum_{i=1}^{n} x_i = n \alpha \theta \] - (c) The statistics \(T_1 = \sum x_i\) and \(T_2 = \sum \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). - (d) The Fisher Information matrix for one observation is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] - (e) Closed-form solutions for the MLEs do not exist; numerical methods are required.

# Given Information - A random sample \(x_1, x_2, \ldots, x_n\) is drawn from a Gamma distribution with: - Unknown shape parameter \(\alpha > 0\) - Unknown scale parameter \(\theta > 0\) - The probability density function (PDF) is given by: \[ f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x^{\alpha - 1} e^{-x/\theta}, \quad x > 0 \] # What to Find (a) The likelihood function \(L(\alpha, \theta)\). (b) The likelihood equations for the Maximum Likelihood Estimators (MLEs) of \(\alpha\) and \(\theta\). (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). (d) Find the Fisher Information matrix for one observation. (e) State whether closed-form solutions for the MLEs exist. Justify your answer briefly. # Definitions/Concepts Used - **Likelihood Function**: Represents the probability of observed data given the parameters. - **Sufficient Statistics**: Statistics that summarize the data without losing information regarding the parameter. - **Fisher Information**: Quantifies the amount of information that an observable random variable carries about an unknown parameter. # Step-by-Step Solution ### (a) Write down the likelihood function \(L(\alpha, \theta)\) The likelihood function for \(n\) observations is: \[ L(\alpha, \theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{n} \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x_i^{\alpha - 1} e^{-x_i/\theta} \right) \] Simplifying this gives: \[ L(\alpha, \theta) = \left(\frac{1}{\Gamma(\alpha) \theta^{\alpha}}\right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] ### (b) Obtain the likelihood equations for the MLEs of \(\alpha\) and \(\theta\) Taking the logarithm of the likelihood function yields the log-likelihood: \[ \log L(\alpha, \theta) = -n \log \Gamma(\alpha) - n \alpha \log \theta + (\alpha - 1) \sum_{i=1}^{n} \log x_i - \frac{\sum_{i=1}^{n} x_i}{\theta} \] Differentiating with respect to \(\alpha\) and \(\theta\): 1. **For \(\alpha\)**: \[ \frac{\partial \log L}{\partial \alpha} = -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} - n \log \theta + \sum_{i=1}^{n} \log x_i = 0 \] 2. **For \(\theta\)**: \[ \frac{\partial \log L}{\partial \theta} = -\frac{n \alpha}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \] The likelihood equations are thus: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^{n} \log x_i = n \log \theta \] \[ \sum_{i=1}^{n} x_i = n \alpha \theta \] ### (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\) Using the factorization theorem, we can express the likelihood function in a way that shows it depends only on \(T_1\) and \(T_2\): \[ L(\alpha, \theta) = g(T_1, T_2; \alpha, \theta) \cdot h(x) \] Since the likelihood function can be rewritten to depend solely on these two statistics, we confirm that \(T_1\) and \(T_2\) are jointly sufficient for \((\alpha, \theta)\). ### (d) Find the Fisher Information matrix for one observation The Fisher Information matrix \(I(\alpha, \theta)\) is defined as: \[ I_{ij} = -E\left[\frac{\partial^2 \log L}{\partial \theta_i \partial \theta_j}\right] \] Calculating the second derivatives of the log-likelihood, we will derive the entries of the Fisher Information matrix: 1. **For \(\alpha\)**: \[ I_{\alpha\alpha} = n \left( \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} \right)^2 - n \frac{\Gamma''(\alpha)}{\Gamma(\alpha)} \] 2. **For \(\theta\)**: \[ I_{\theta\theta} = \frac{n \alpha}{\theta^2} \] 3. **For \(\alpha\) and \(\theta\)**: \[ I_{\alpha\theta} = 0 \] Thus, the Fisher Information matrix is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] ### (e) State whether closed-form solutions for the MLEs exist Closed-form solutions for the MLEs of \(\alpha\) and \(\theta\) typically do not exist for Gamma distributions. The equations derived for the MLEs involve complex functions such as the digamma function \(\psi(\alpha)\) and require numerical methods for solving, indicating that numerical optimization techniques are necessary rather than analytical solutions. # Summary - (a) The likelihood function is given by: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] - (b) The likelihood equations are: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^{n} \log x_i = n \log \theta \] \[ \sum_{i=1}^{n} x_i = n \alpha \theta \] - (c) The statistics \(T_1 = \sum x_i\) and \(T_2 = \sum \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). - (d) The Fisher Information matrix for one observation is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] - (e) Closed-form solutions for the MLEs do not exist; numerical methods are required.

# Given Information - A random sample \(x_1, x_2, \ldots, x_n\) is drawn from a Gamma distribution characterized by: - Unknown shape parameter \(\alpha > 0\) - Unknown scale parameter \(\theta > 0\) - The probability density function (PDF) of the Gamma distribution is defined as: \[ f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x^{\alpha - 1} e^{-x/\theta}, \quad x > 0 \] # Objectives 1. **(a)** Formulate the likelihood function \(L(\alpha, \theta)\). 2. **(b)** Derive the likelihood equations to find the Maximum Likelihood Estimators (MLEs) for \(\alpha\) and \(\theta\). 3. **(c)** Demonstrate that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for the parameters \((\alpha, \theta)\). 4. **(d)** Calculate the Fisher Information matrix for a single observation. 5. **(e)** Evaluate whether closed-form solutions for the MLEs are attainable, providing a brief justification. # Definitions and Concepts - **Likelihood Function**: Represents the probability of observing the data as a function of the parameters. - **Sufficient Statistics**: A statistic is considered sufficient for a parameter if it encapsulates all necessary information from the data regarding that parameter. - **Fisher Information**: Quantifies how much information a sample provides about an unknown parameter. # Step-by-Step Solution ### (a) Write down the likelihood function \(L(\alpha, \theta)\) The likelihood function for \(n\) independent observations is expressed as: \[ L(\alpha, \theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{n} \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x_i^{\alpha - 1} e^{-x_i/\theta} \right) \] This can be simplified to: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] ### (b) Obtain the likelihood equations for the MLEs of \(\alpha\) and \(\theta\) Taking the logarithm of the likelihood function yields the log-likelihood: \[ \log L(\alpha, \theta) = -n \log \Gamma(\alpha) - n \alpha \log \theta + (\alpha - 1) \sum_{i=1}^{n} \log x_i - \frac{\sum_{i=1}^{n} x_i}{\theta} \] To find the MLEs, differentiate with respect to \(\alpha\) and \(\theta\) and set the equations to zero: 1. **For \(\alpha\)**: \[ \frac{\partial \log L}{\partial \alpha} = -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} - n \log \theta + \sum_{i=1}^{n} \log x_i = 0 \] 2. **For \(\theta\)**: \[ \frac{\partial \log L}{\partial \theta} = -\frac{n \alpha}{\theta} + \frac{\sum_{i=1}^{n} x_i}{\theta^2} = 0 \] The resulting likelihood equations are: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^{n} \log x_i = n \log \theta \] \[ \sum_{i=1}^{n} x_i = n \alpha \theta \] ### (c) Show that the statistics \(T_1 = \sum_{i=1}^n x_i\) and \(T_2 = \sum_{i=1}^n \ln x_i\) are jointly sufficient for \((\alpha, \theta)\) By applying the factorization theorem, we can express the likelihood function in a manner that reveals it depends solely on \(T_1\) and \(T_2\): \[ L(\alpha, \theta) = g(T_1, T_2; \alpha, \theta) \cdot h(x) \] Given the structure of the likelihood function, it is evident that \(T_1\) and \(T_2\) contain all pertinent information about the parameters \(\alpha\) and \(\theta\), confirming that they are jointly sufficient statistics. ### (d) Find the Fisher Information matrix for one observation The Fisher Information matrix \(I(\alpha, \theta)\) is defined as follows: \[ I_{ij} = -E\left[\frac{\partial^2 \log L}{\partial \theta_i \partial \theta_j}\right] \] Calculating the second derivatives of the log-likelihood yields the elements of the Fisher Information matrix: 1. **For \(\alpha\)**: \[ I_{\alpha\alpha} = n \left( \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} \right)^2 - n \frac{\Gamma''(\alpha)}{\Gamma(\alpha)} \] 2. **For \(\theta\)**: \[ I_{\theta\theta} = \frac{n \alpha}{\theta^2} \] 3. **For \(\alpha\) and \(\theta\)**: \[ I_{\alpha\theta} = 0 \] Thus, the Fisher Information matrix is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] ### (e) State whether closed-form solutions for the MLEs exist In general, closed-form solutions for the MLEs of \(\alpha\) and \(\theta\) do not exist for the Gamma distribution. The likelihood equations involve the digamma function \(\psi(\alpha)\) and require numerical methods for solving. The presence of logarithmic and gamma functions complicates the equations, necessitating numerical optimization techniques rather than analytical solutions. # Summary - **(a)** The likelihood function is given by: \[ L(\alpha, \theta) = \left( \frac{1}{\Gamma(\alpha) \theta^{\alpha}} \right)^n \prod_{i=1}^{n} x_i^{\alpha - 1} e^{-\sum_{i=1}^{n} x_i / \theta} \] - **(b)** The likelihood equations are: \[ -n \frac{\Gamma'(\alpha)}{\Gamma(\alpha)} + \sum_{i=1}^{n} \log x_i = n \log \theta \] \[ \sum_{i=1}^{n} x_i = n \alpha \theta \] - **(c)** The statistics \(T_1 = \sum x_i\) and \(T_2 = \sum \ln x_i\) are jointly sufficient for \((\alpha, \theta)\). - **(d)** The Fisher Information matrix for one observation is: \[ I(\alpha, \theta) = \begin{pmatrix} I_{\alpha\alpha} & 0 \\ 0 & \frac{n \alpha}{\theta^2} \end{pmatrix} \] - **(e)** Closed-form solutions for the MLEs do not exist; numerical methods are required.

# Given Information - Prevalence of the disease in the population: \(P(D) = 0.02\) (2%) - Test sensitivity (True Positive Rate): \(P(T^+ | D) = 0.99\) (99%) - Test specificity (True Negative Rate): \(P(T^- | D') = 0.95\) (95%) Where: - \(D\): Event that a person has the disease. - \(D'\): Event that a person does not have the disease. - \(T^+\): Event that a person tests positive. - \(T^-\): Event that a person tests negative. # What to Find - The probability that a randomly selected person who tests positive actually has the disease, denoted as \(P(D | T^+)\). # Definitions/Concepts Used - **Bayes' Theorem**: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] - **Total Probability** for \(P(T^+)\): \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D') P(D') \] Where \(P(T^+ | D') = 1 - P(T^- | D') = 1 - 0.95 = 0.05\). # Step-by-Step Solution ### Calculate \(P(T^+)\) Using the total probability formula: \[ P(T^+) = P(T^+ | D) P(D) + P(T^+ | D') P(D') \] Calculating \(P(D')\): \[ P(D') = 1 - P(D) = 1 - 0.02 = 0.98 \] Now substituting the values: \[ P(T^+) = (0.99)(0.02) + (0.05)(0.98) \] Calculating each term: - \(0.99 \times 0.02 = 0.0198\) - \(0.05 \times 0.98 = 0.049\) Summing these: \[ P(T^+) = 0.0198 + 0.049 = 0.0688 \] ### Calculate \(P(D | T^+)\) using Bayes' Theorem Now applying Bayes' theorem: \[ P(D | T^+) = \frac{P(T^+ | D) P(D)}{P(T^+)} \] Substituting the values: \[ P(D | T^+) = \frac{(0.99)(0.02)}{0.0688} \] Calculating: \[ P(D | T^+) = \frac{0.0198}{0.0688} \approx 0.288 \] # Summary (Final Answer Only) - The probability that a randomly selected person who tests positive actually has the disease is approximately **0.288** (or **28.8%**).

# Given Information - The waiting time (in minutes) between system failures follows a Gamma distribution: - Shape parameter: \(\alpha = 4\) - Scale parameter: \(\theta\) (unknown) - A random sample of 6 waiting times (in minutes) is observed: - \(x_1 = 8\), \(x_2 = 12\), \(x_3 = 10\), \(x_4 = 6\), \(x_5 = 14\), \(x_6 = 10\) # What to Find (a) The Maximum Likelihood Estimator (MLE) of \(\theta\). (b) Using the estimate of \(\theta\), estimate the mean waiting time. (c) Using the asymptotic distribution of the MLE, construct a 95% confidence interval for \(\theta\). (d) Using the confidence interval from part (c), obtain a corresponding 95% confidence interval for the mean waiting time. # Definitions/Concepts Used - **Gamma Distribution**: The probability density function (PDF) is given by: \[ f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x^{\alpha - 1} e^{-x/\theta}, \quad x > 0 \] - **Maximum Likelihood Estimation (MLE)**: The MLE is found by maximizing the likelihood function. - **Mean of Gamma Distribution**: The mean \(E(X)\) is given by: \[ E(X) = \alpha \theta \] - **Confidence Interval**: For the MLE of \(\theta\), the approximate confidence interval can be constructed using the asymptotic normality of the MLE. # Step-by-Step Solution ### (a) Find the Maximum Likelihood Estimator (MLE) of \(\theta\) The likelihood function for a Gamma distribution with parameters \(\alpha\) and \(\theta\) is given by: \[ L(\theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{6} \left( \frac{1}{\Gamma(4) \theta^4} x_i^{3} e^{-x_i/\theta} \right) \] This simplifies to: \[ L(\theta) = \left( \frac{1}{\Gamma(4) \theta^4} \right)^6 \prod_{i=1}^{6} x_i^{3} e^{-\sum_{i=1}^{6} x_i / \theta} \] Taking the natural logarithm to find the log-likelihood: \[ \log L(\theta) = -6 \log \Gamma(4) - 24 \log \theta + 3 \sum_{i=1}^{6} \log x_i - \frac{\sum_{i=1}^{6} x_i}{\theta} \] Differentiating with respect to \(\theta\) and setting it to zero for maximization: \[ \frac{d}{d\theta} \log L(\theta) = -\frac{24}{\theta} + \frac{\sum_{i=1}^{6} x_i}{\theta^2} = 0 \] This leads to: \[ \sum_{i=1}^{6} x_i = 24 \theta \] Thus, the MLE is: \[ \hat{\theta} = \frac{\sum_{i=1}^{6} x_i}{24} \] Calculating \(\sum_{i=1}^{6} x_i = 8 + 12 + 10 + 6 + 14 + 10 = 60\): \[ \hat{\theta} = \frac{60}{24} = 2.5 \] ### (b) Estimate the mean waiting time Using the MLE found: \[ E(X) = \alpha \hat{\theta} = 4 \times 2.5 = 10 \] ### (c) Construct a 95% confidence interval for \(\theta\) To construct a confidence interval for \(\theta\), we can use the asymptotic normality of the MLE. The approximate variance of the MLE \(\hat{\theta}\) can be derived from the Fisher information. The Fisher information \(I(\theta)\) for one observation is: \[ I(\theta) = \frac{\alpha}{\theta^2} \] For \(n\) observations, it becomes: \[ I_n(\theta) = n \cdot \frac{\alpha}{\theta^2} = 6 \cdot \frac{4}{\theta^2} \] The variance of \(\hat{\theta}\) is: \[ \text{Var}(\hat{\theta}) \approx \frac{1}{I_n(\theta)} = \frac{\theta^2}{24} \] Using the normal approximation for the confidence interval: \[ \hat{\theta} \pm z_{\alpha/2} \sqrt{\text{Var}(\hat{\theta})} \] At a 95% confidence level, \(z_{0.025} \approx 1.96\): \[ \text{CI} = 2.5 \pm 1.96 \sqrt{\frac{(2.5)^2}{24}} = 2.5 \pm 1.96 \cdot 0.51 \approx 2.5 \pm 1.002 \] Thus, the confidence interval is approximately: \[ (1.498, 3.502) \] ### (d) Obtain a corresponding 95% confidence interval for the mean waiting time The mean waiting time is given by: \[ E(X) = 4 \hat{\theta} \] Thus, the confidence interval for the mean can be constructed by multiplying the confidence interval for \(\theta\) by 4: \[ \text{Mean CI} = 4 \cdot (1.498, 3.502) = (5.992, 14.008) \] # Summary (Final Answers Only) - (a) MLE of \(\theta\) is **2.5**. - (b) Estimated mean waiting time is **10 minutes**. - (c) 95% confidence interval for \(\theta\) is approximately **(1.498, 3.502)**. - (d) Corresponding 95% confidence interval for the mean waiting time is approximately **(5.992, 14.008)**.

# Given Information - The load (in tons) at which a component fails follows a Gamma distribution: - Shape parameter: \(\alpha = 4\) - Scale parameter: \(\theta\) (unknown) - A sample of 5 failure loads (in tons) is observed: - \(x_1 = 20\), \(x_2 = 28\), \(x_3 = 24\), \(x_4 = 32\), \(x_5 = 16\) # What to Find (a) The Maximum Likelihood Estimator (MLE) of \(\theta\). (b) The mean failure load. (c) The variance of the failure load. (d) A 95% confidence interval for \(\theta\). # Definitions/Concepts Used - **Gamma Distribution**: The probability density function (PDF) is given by: \[ f(x; \alpha, \theta) = \frac{1}{\Gamma(\alpha) \theta^{\alpha}} x^{\alpha - 1} e^{-x/\theta}, \quad x > 0 \] - **Maximum Likelihood Estimation (MLE)**: The MLE is found by maximizing the likelihood function. - **Mean of Gamma Distribution**: The mean \(E(X)\) is given by: \[ E(X) = \alpha \theta \] - **Variance of Gamma Distribution**: The variance \(\text{Var}(X)\) is given by: \[ \text{Var}(X) = \alpha \theta^2 \] # Step-by-Step Solution ### (a) Find the Maximum Likelihood Estimator (MLE) of \(\theta\) The likelihood function for a Gamma distribution with parameters \(\alpha\) and \(\theta\) is given by: \[ L(\theta) = \prod_{i=1}^{n} f(x_i; \alpha, \theta) = \prod_{i=1}^{5} \left( \frac{1}{\Gamma(4) \theta^4} x_i^{3} e^{-x_i/\theta} \right) \] This simplifies to: \[ L(\theta) = \left( \frac{1}{\Gamma(4) \theta^4} \right)^5 \prod_{i=1}^{5} x_i^{3} e^{-\sum_{i=1}^{5} x_i / \theta} \] Taking the logarithm to find the log-likelihood: \[ \log L(\theta) = -5 \log \Gamma(4) - 20 \log \theta + 3 \sum_{i=1}^{5} \log x_i - \frac{\sum_{i=1}^{5} x_i}{\theta} \] Differentiating with respect to \(\theta\) and setting it to zero for maximization: \[ \frac{d}{d\theta} \log L(\theta) = -\frac{20}{\theta} + \frac{\sum_{i=1}^{5} x_i}{\theta^2} = 0 \] This leads to: \[ \sum_{i=1}^{5} x_i = 20 \theta \] Thus, the MLE is: \[ \hat{\theta} = \frac{\sum_{i=1}^{5} x_i}{20} \] Calculating \(\sum_{i=1}^{5} x_i = 20 + 28 + 24 + 32 + 16 = 120\): \[ \hat{\theta} = \frac{120}{20} = 6 \] ### (b) Estimate the mean failure load Using the MLE found: \[ E(X) = \alpha \hat{\theta} = 4 \times 6 = 24 \] ### (c) Estimate the variance of the failure load Using the MLE found: \[ \text{Var}(X) = \alpha \hat{\theta}^2 = 4 \times (6)^2 = 4 \times 36 = 144 \] ### (d) Construct a 95% confidence interval for \(\theta\) To construct a confidence interval for \(\theta\), we can use the Chi-squared distribution. The estimator \(\frac{4n\hat{\theta}}{\sigma^2}\) follows a Chi-squared distribution with \(4n\) degrees of freedom. The confidence interval is given by: \[ \left(\frac{4n\hat{\theta}}{\chi^2_{\alpha/2, 4n}}, \frac{4n\hat{\theta}}{\chi^2_{1-\alpha/2, 4n}}\right) \] Using \(n = 5\), we have \(4n = 20\). The critical values for \(\chi^2\) with 20 degrees of freedom at \(\alpha = 0.05\) are approximately: - \(\chi^2_{0.025, 20} \approx 32.671\) - \(\chi^2_{0.975, 20} \approx 10.851\) Then: \[ \text{Lower Bound} = \frac{4 \times 5 \times 6}{32.671} \approx \frac{120}{32.671} \approx 3.674 \] \[ \text{Upper Bound} = \frac{4 \times 5 \times 6}{10.851} \approx \frac{120}{10.851} \approx 11.058 \] Thus, the 95% confidence interval for \(\theta\) is approximately: \[ (3.674, 11.058) \] # Summary (Final Answers Only) - (a) MLE of \(\theta\) is **6**. - (b) Estimated mean failure load is **24 tons**. - (c) Estimated variance of the failure load is **144 tons²**. - (d) 95% confidence interval for \(\theta\) is approximately **(3.674, 11.058)**.

# Given Information - Probabilities of failure for components: - \(P(A) = 0.01\) (component A fails) - \(P(B) = 0.03\) (component B fails) - \(P(C) = 0.04\) (component C fails) # What to Find - The probability that the device works, denoted as \(P(\text{Works})\). # Definitions/Concepts Used - **Union Probability**: The probability that at least one of several events occurs can be found using the formula: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] - Since the failures are independent events, the intersection probabilities can be calculated as: \[ P(A \cap B) = P(A) \times P(B), \quad P(A \cap C) = P(A) \times P(C), \quad P(B \cap C) = P(B) \times P(C) \] \[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \] # Step-by-Step Solution ### Calculate the Probability of Failure 1. Calculate the probability that at least one component fails: \[ P(\text{Fails}) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] 2. Calculate the intersections: - \(P(A \cap B) = 0.01 \times 0.03 = 0.0003\) - \(P(A \cap C) = 0.01 \times 0.04 = 0.0004\) - \(P(B \cap C) = 0.03 \times 0.04 = 0.0012\) - \(P(A \cap B \cap C) = 0.01 \times 0.03 \times 0.04 = 0.000012\) 3. Substitute values into the union probability formula: \[ P(\text{Fails}) = 0.01 + 0.03 + 0.04 - 0.0003 - 0.0004 - 0.0012 + 0.000012 \] \[ P(\text{Fails}) = 0.01 + 0.03 + 0.04 - 0.0003 - 0.0004 - 0.0012 + 0.000012 \] \[ P(\text{Fails}) = 0.08 - 0.001912 \approx 0.078088 \] ### Calculate the Probability of Working 4. The probability that the device works is the complement of the probability that it fails: \[ P(\text{Works}) = 1 - P(\text{Fails}) = 1 - 0.078088 \approx 0.921912 \] # Summary (Final Answer Only) - The probability that the device works is approximately **0.9219** (or **92.19%**).

# Given Information - The probabilities of failure for the components are as follows: - \(P(A) = 0.01\) (probability that component A fails) - \(P(B) = 0.03\) (probability that component B fails) - \(P(C) = 0.04\) (probability that component C fails) # Objective - Determine the probability that the device operates correctly, denoted as \(P(\text{Works})\). # Definitions/Concepts Used - **Union Probability**: The probability that at least one of several events occurs can be calculated using: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] - For independent events, the intersection probabilities can be expressed as: \[ P(A \cap B) = P(A) \times P(B), \quad P(A \cap C) = P(A) \times P(C), \quad P(B \cap C) = P(B) \times P(C) \] \[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \] # Step-by-Step Solution ### Step 1: Calculate the Probability of Failure 1. We calculate the probability that at least one component fails: \[ P(\text{Fails}) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] 2. Calculate the pairwise intersection probabilities: - For components A and B: \[ P(A \cap B) = 0.01 \times 0.03 = 0.0003 \] - For components A and C: \[ P(A \cap C) = 0.01 \times 0.04 = 0.0004 \] - For components B and C: \[ P(B \cap C) = 0.03 \times 0.04 = 0.0012 \] - For all three components: \[ P(A \cap B \cap C) = 0.01 \times 0.03 \times 0.04 = 0.000012 \] 3. Substitute these values into the union probability formula: \[ P(\text{Fails}) = 0.01 + 0.03 + 0.04 - 0.0003 - 0.0004 - 0.0012 + 0.000012 \] \[ P(\text{Fails}) = 0.08 - 0.001912 \approx 0.078088 \] ### Step 2: Calculate the Probability of Working 4. The probability that the device works is the complementary event to the probability of failure: \[ P(\text{Works}) = 1 - P(\text{Fails}) = 1 - 0.078088 \approx 0.921912 \] # Summary (Final Answer) - The probability that the device operates correctly is approximately **0.9219** (or **92.19%**).

# Given Information - The failure probabilities for three components are as follows: - \(P(A) = 0.01\) (probability that component A fails) - \(P(B) = 0.03\) (probability that component B fails) - \(P(C) = 0.04\) (probability that component C fails) # Objective - Calculate the probability that the device functions properly, designated as \(P(\text{Works})\). # Definitions/Concepts Used - **Union Probability**: The probability that at least one of the events occurs can be computed using: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] - For independent events, the probabilities of intersections can be calculated as: \[ P(A \cap B) = P(A) \times P(B) \] \[ P(A \cap C) = P(A) \times P(C) \] \[ P(B \cap C) = P(B) \times P(C) \] \[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \] # Step-by-Step Solution ### Step 1: Calculate the Probability of Failure 1. First, we need to calculate the probability that at least one component fails: \[ P(\text{Fails}) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] 2. Next, compute the intersection probabilities: - For components A and B: \[ P(A \cap B) = 0.01 \times 0.03 = 0.0003 \] - For components A and C: \[ P(A \cap C) = 0.01 \times 0.04 = 0.0004 \] - For components B and C: \[ P(B \cap C) = 0.03 \times 0.04 = 0.0012 \] - For all three components: \[ P(A \cap B \cap C) = 0.01 \times 0.03 \times 0.04 = 0.000012 \] 3. Substitute these values into the union probability equation: \[ P(\text{Fails}) = 0.01 + 0.03 + 0.04 - 0.0003 - 0.0004 - 0.0012 + 0.000012 \] \[ P(\text{Fails}) = 0.08 - 0.001912 \approx 0.078088 \] ### Step 2: Calculate the Probability of Working 4. The probability that the device operates correctly is the complement of the probability that it fails: \[ P(\text{Works}) = 1 - P(\text{Fails}) = 1 - 0.078088 \approx 0.921912 \] # Summary (Final Answer) - The probability that the device functions properly is approximately **0.9219** (or **92.19%**).

# Given Information - The failure probabilities of the components are: - \(P(A) = 0.01\) (failure probability for component A) - \(P(B) = 0.03\) (failure probability for component B) - \(P(C) = 0.04\) (failure probability for component C) # Objective - To determine the probability that the device operates correctly, denoted as \(P(\text{Works})\). # Definitions/Concepts Used - **Union Probability**: The probability that at least one of the events occurs can be calculated using the formula: \[ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] - For independent events, the probabilities of intersections can be calculated as: \[ P(A \cap B) = P(A) \times P(B) \] \[ P(A \cap C) = P(A) \times P(C) \] \[ P(B \cap C) = P(B) \times P(C) \] \[ P(A \cap B \cap C) = P(A) \times P(B) \times P(C) \] # Step-by-Step Solution ### Step 1: Calculate the Probability of Failure 1. To find the probability that at least one component fails, we use the union probability formula: \[ P(\text{Fails}) = P(A) + P(B) + P(C) - P(A \cap B) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C) \] 2. Calculate the intersection probabilities: - For components A and B: \[ P(A \cap B) = 0.01 \times 0.03 = 0.0003 \] - For components A and C: \[ P(A \cap C) = 0.01 \times 0.04 = 0.0004 \] - For components B and C: \[ P(B \cap C) = 0.03 \times 0.04 = 0.0012 \] - For all three components: \[ P(A \cap B \cap C) = 0.01 \times 0.03 \times 0.04 = 0.000012 \] 3. Substitute these values into the union probability equation: \[ P(\text{Fails}) = 0.01 + 0.03 + 0.04 - 0.0003 - 0.0004 - 0.0012 + 0.000012 \] \[ P(\text{Fails}) = 0.08 - 0.001912 = 0.078088 \] ### Step 2: Calculate the Probability of Working 4. The probability that the device works is the complement of the probability that it fails: \[ P(\text{Works}) = 1 - P(\text{Fails}) = 1 - 0.078088 \approx 0.921912 \] # Summary (Final Answer) - The probability that the device operates correctly is approximately **0.9219** (or **92.19%**).

# Given Information - Let \(x\) be the number of heads in 5 tosses of a fair coin. # What to Find - \(E(x^2)\) (expected value of \(x^2\)) - \(\text{Var}(2x + 3)\) (variance of the expression \(2x + 3\)) # Definitions/Concepts Used - **Expected Value**: For a discrete random variable \(x\): \[ E(x) = \sum_{i} x_i P(x_i) \] - **Variance**: For a random variable \(x\): \[ \text{Var}(x) = E(x^2) - (E(x))^2 \] - **Binomial Distribution**: \(x\) follows a binomial distribution: \[ x \sim \text{Binomial}(n, p) \] where \(n = 5\) (number of trials) and \(p = 0.5\) (probability of heads). # Step-by-Step Solution ### Step 1: Calculate \(E(x)\) and \(\text{Var}(x)\) For a binomial distribution, the expected value and variance are given by: 1. **Expected Value**: \[ E(x) = n \cdot p = 5 \cdot 0.5 = 2.5 \] 2. **Variance**: \[ \text{Var}(x) = n \cdot p \cdot (1 - p) = 5 \cdot 0.5 \cdot 0.5 = 1.25 \] ### Step 2: Calculate \(E(x^2)\) Using the relationship for variance: \[ \text{Var}(x) = E(x^2) - (E(x))^2 \] We can rearrange it to find \(E(x^2)\): \[ E(x^2) = \text{Var}(x) + (E(x))^2 \] Substituting the values: \[ E(x^2) = 1.25 + (2.5)^2 = 1.25 + 6.25 = 7.5 \] ### Step 3: Calculate \(\text{Var}(2x + 3)\) The variance of a linear transformation \(Y = aX + b\) is given by: \[ \text{Var}(Y) = a^2 \text{Var}(X) \] In this case, \(a = 2\) and \(b = 3\): \[ \text{Var}(2x + 3) = 2^2 \text{Var}(x) = 4 \cdot 1.25 = 5 \] # Summary (Final Answers Only) - \(E(x^2) = 7.5\) - \(\text{Var}(2x + 3) = 5\)