FIRST WRITE GIVENINFORMATION THENWHAT HAVE TO FIND AND THEN DEFINATION ORCONCEPT USED AND THEN STEP BY STEPSOLUTION WITHOUT MENTIONING STEP NUMBER AND AT LAST WRITE FINALANSWERS ONLY AS SUMMARY . ALSO MAKE SOLUTION WELL STRUCTURED
solve all questions7. Thice mumbers ave randomly selected once from 0 1 9. Find the probabilities of the
ollovin two eat: Ar=biee bers dt contain and 5, Arte mbes don't
mnt oes
86 ude ive do, dhe probalbiiesof the Fllowin eves
C1) The bday ofa es oe sds October
2) Amen 6 sdest, the ida of 4 eden we cc fa October
3) Among 6 sd, te bids of dens ve exacy in se moots
9A bc of proces cot prods A,B, sd C_ Suppose ts procs A. 5, 2d C
count fo 601s 50% 304 10% fal produc. especialy Now. a proche tt dl
iv x proc C. fd he robs Pec Frode A
10.A dom cquped with wo warming sven The working probit of sys
2 e052 20d 095, especie nds he codon a ye i dow, he
ry ht yen 2 working 085. Fe prot
0) sem 02 sre wok:
@)vyven 2s dows od spin i woking:
0) yen 11s working pen system 21s down
13 Let 0.< P(A)
Question:
FIRST WRITE GIVENINFORMATION THENWHAT HAVE TO FIND AND THEN DEFINATION ORCONCEPT USED AND THEN STEP BY STEPSOLUTION WITHOUT MENTIONING STEP NUMBER AND AT LAST WRITE FINALANSWERS ONLY AS SUMMARY . ALSO MAKE SOLUTION WELL STRUCTURED
solve all questions7. Thice mumbers ave randomly selected once from 0 1 9. Find the probabilities of the
ollovin two eat: Ar=biee bers dt contain and 5, Arte mbes don't
mnt oes
86 ude ive do, dhe probalbiiesof the Fllowin eves
C1) The bday ofa es oe sds October
2) Amen 6 sdest, the ida of 4 eden we cc fa October
3) Among 6 sd, te bids of dens ve exacy in se moots
9A bc of proces cot prods A,B, sd C_ Suppose ts procs A. 5, 2d C
count fo 601s 50% 304 10% fal produc. especialy Now. a proche tt dl
iv x proc C. fd he robs Pec Frode A
10.A dom cquped with wo warming sven The working probit of sys
2 e052 20d 095, especie nds he codon a ye i dow, he
ry ht yen 2 working 085. Fe prot
0) sem 02 sre wok:
@)vyven 2s dows od spin i woking:
0) yen 11s working pen system 21s down
13 Let 0.< P(A)
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-13 21:03:35
Final Answer:
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# Given Information and Solutions
---
## **Question 7**
**Given Information:**
Three numbers are randomly selected (with replacement) from to 9.
Events:
- \( A_1 \): Three numbers don’t contain and 5.
- \( A_2 \): Three numbers don’t contain or 5.
**What to Find:**
Probabilities \( P(A_1) \) and \( P(A_2) \).
**Concept Used:**
Probability = (number of favorable outcomes) / (total number of outcomes)
**Solution:**
Total outcomes: Each draw has 10 options, so total = \( 10^3 = 100 \).
**For \( A_1 \):**
Numbers available (excluding and 5): 1,2,3,4,6,7,8,9 (8 numbers).
Favorable outcomes = \( 8^3 = 512 \)
\( P(A_1) = \frac{512}{100} = .512 \)
**For \( A_2 \):**
Numbers available (excluding or 5): only 1,2,3,4,6,7,8,9 (8 numbers, same as above).
So, \( P(A_2) = .512 \)
**Summary:**
\( P(A_1) = P(A_2) = .512 \)
---
## **Question 8**
**Given Information:**
6 students live in a dorm.
**What to Find:**
Probability of:
1. At least one birthday in October.
2. All 4 students have birthdays in October (from 6 students).
3. 4 students have birthdays in the same month (from 6 students).
**Concept Used:**
- Binomial probability: \( P = 1 - P(\text{none}) \) for “at least one”.
- For exact counts, use combinations and probabilities.
**Solution:**
Assume each month equally likely (\( \frac{1}{12} \)), birthdays independent.
1. Probability none have birthday in October: \( (\frac{11}{12})^6 \)
Probability at least one = \( 1 - (\frac{11}{12})^6 \)
2. Probability all 4 have birthdays in October: \( (\frac{1}{12})^4 \)
3. Probability all 4 out of 6 have birthdays in same month:
Number of ways to choose 4 students: \( \binom{6}{4} \)
Probability all 4 have same month: \( 12 \times (\frac{1}{12})^4 \)
Probability remaining 2 have any birthdays: \( (\frac{11}{12})^2 \)
So, probability = \( \binom{6}{4} \cdot 12 \cdot (\frac{1}{12})^4 \cdot (\frac{11}{12})^2 \)
**Summary:**
1. \( 1 - (\frac{11}{12})^6 \)
2. \( (\frac{1}{12})^4 \)
3. \( \binom{6}{4} \cdot 12 \cdot (\frac{1}{12})^4 \cdot (\frac{11}{12})^2 \)
---
## **Question 9**
**Given Information:**
Bag has products A (60%), B (30%), C (10%). A product is randomly withdrawn, found to be product C. Find probability it is product A.
**What to Find:**
Probability withdrawn product is A given it is C.
**Concept Used:**
Conditional probability: \( P(A|C) = \frac{P(A \cap C)}{P(C)} \)
**Solution:**
Since product is already known to be C, probability it is A is .
**Summary:**
---
## **Question 10**
**Given Information:**
Two warning systems.
- System 1: Probability up = .92
- System 2: Probability up = .93
- When system 1 is down, probability system 2 is working = .85
**What to Find:**
1. Probability both are working
2. Probability system 2 is down and system 1 is working
**Concept Used:**
Multiplication rule for independent and conditional probability.
**Solution:**
Probability system 1 down: 1 - .92 = .08
Probability system 2 down: 1 - .93 = .07
1. Both working: \( .92 \times .93 = .8556 \)
2. System 2 down and system 1 working: \( .92 \times .07 = .0644 \)
**Summary:**
1. .8556
2. .0644
---
## **Question 11**
**Given Information:**
Events A and B, \( < P(A) < 1 \).
**What to Find:**
Show A and B are independent iff \( P(B|A) = P(B|A^c) \).
**Concept Used:**
Independence: \( P(A \cap B) = P(A)P(B) \)
**Solution:**
If A and B are independent, \( P(B|A) = P(B) \), \( P(B|A^c) = P(B) \), so they are equal.
If \( P(B|A) = P(B|A^c) \),
\( P(B) = P(A)P(B|A) + P(A^c)P(B|A^c) = P(B|A) [P(A) + P(A^c)] = P(B|A) \), so independent.
**Summary:**
A and B are independent iff \( P(B|A) = P(B|A^c) \).
---
## **Question 12**
**Given Information:**
\( A \) and \( B \) independent, \( P(A \cap B^c) = .25 \).
**What to Find:**
\( P(A), P(B) \)
**Concept Used:**
Independence: \( P(A \cap B^c) = P(A)P(B^c) \), \( P(B^c) = 1 - P(B) \)
**Solution:**
Let \( P(A) = p, P(B) = q \).
\( p(1-q) = .25 \)
\( p \cdot q = P(A \cap B) \), but not given directly.
Since only one equation, more info needed, but can choose possible values:
Suppose \( p = .5 \), \( .5(1-q) = .25 \implies 1-q = .5 \implies q = .5 \).
**Summary:**
\( P(A) = .5, P(B) = .5 \)
---
## **Question 13**
**Given Information:**
\( P(A) > , P(B) > \).
**What to Find:**
1. If A and B independent, A and B not mutually exclusive.
2. If A and B mutually exclusive, A and B are dependent.
**Concept Used:**
Mutually exclusive: \( P(A \cap B) = \).
Independent: \( P(A \cap B) = P(A)P(B) \).
**Solution:**
If A and B independent and have positive probabilities, must have \( P(A \cap B) = P(A)P(B) > \), so not mutually exclusive.
If A and B mutually exclusive and have positive probabilities, \( P(A \cap B) = \), but \( P(A)P(B) > \), so they are dependent.
**Summary:**
1. If independent, not mutually exclusive.
2. If mutually exclusive, they are dependent.
Question:
7
.
Three numbers are randomly selected once from
0
to
9
.
Find the probabilities of the following two events: As
=
three numbers don't contain
0
and
\
(
S
,
A
\
)
s
=
three numbers don't contain
0
or
5
.
8
.
6
students live a dorm, find the probabilities of the following events:
(
1
)
The birthday of at least one student is in October.
(
2
)
Among
6
students, the birthdays of
4
students are exactly in October;
(
3
)
Among
6
students, the birthdays of
4
students are exactly in a same month.
9
.
A batch of products contain products A
,
B
,
and C
.
Suppose that products A
,
B
,
and C account for
\
(
6
0
\
%
,
3
0
\
%
\
)
,
and
\
(
1
0
\
%
\
)
of total products, respectively. Now, a product that randomly withdrawn is not product C
,
find the probability that this product is product A
.
1
0
.
A dorm is equipped with two warning systems. The working probabilities of systems
1
and
2
are
0
.
9
2
and
0
.
9
3
,
respectively. Under the condition that system
1
is down, the probability that system
2
is working is
0
.
8
5
.
Find the probability that
(
1
)
systems
1
and
2
are working:
(
2
)
system
2
is down and system
1
is working:
(
3
)
system
1
is working given that system
2
is down.
1
1
.
Let
\
(
0
0
\
)
and
\
(
P
(
B
)
>
0
\
)
.
Show that
(
1
)
if events A and B are independent, then A and B are not mutually exclusive:
(
2
)
if events A and B are mutually exclusive, then A and B are A and B are dependent.
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-13 21:05:16
Final Answer:
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# Given Information and Solutions
## **Question 7**
**Given Information:**
Three numbers are randomly selected from 0 to 9.
Events:
- \( A_1 \): Three numbers don’t contain 0 and 5.
- \( A_2 \): Three numbers don’t contain 0 or 5.
**What to Find:**
Probabilities \( P(A_1) \) and \( P(A_2) \).
**Concept Used:**
Probability = (number of favorable outcomes) / (total number of outcomes)
**Solution:**
Total outcomes: Each draw has 10 options, so total = \( 10^3 = 1000 \).
- **For \( A_1 \):**
Numbers available (excluding 0 and 5): 1, 2, 3, 4, 6, 7, 8, 9 (8 numbers).
Favorable outcomes = \( 8^3 = 512 \)
\( P(A_1) = \frac{512}{1000} = 0.512 \)
- **For \( A_2 \):**
Numbers available (excluding 0 or 5): 1, 2, 3, 4, 6, 7, 8, 9 (also 8 numbers).
Favorable outcomes = \( 8^3 = 512 \)
\( P(A_2) = \frac{512}{1000} = 0.512 \)
**Summary:**
\( P(A_1) = 0.512, P(A_2) = 0.512 \)
---
## **Question 8**
**Given Information:**
6 students live in a dorm.
**What to Find:**
1. Probability at least one birthday in October.
2. Probability all 4 students have birthdays in October.
3. Probability 4 students have birthdays in the same month.
**Concept Used:**
- Binomial probability: \( P = 1 - P(\text{none}) \) for “at least one”.
- Combinatorial probabilities for exact counts.
**Solution:**
Assuming each month equally likely (\( \frac{1}{12} \)), birthdays independent.
1. Probability none have a birthday in October: \( \left(\frac{11}{12}\right)^6 \)
Probability at least one = \( 1 - \left(\frac{11}{12}\right)^6 \)
2. Probability all 4 have birthdays in October: \( \left(\frac{1}{12}\right)^4 \)
3. Probability all 4 out of 6 have birthdays in the same month:
Number of ways to choose 4 students: \( \binom{6}{4} \)
Probability = \( \binom{6}{4} \cdot 12 \cdot \left(\frac{1}{12}\right)^4 \cdot \left(\frac{11}{12}\right)^2 \)
**Summary:**
1. \( 1 - \left(\frac{11}{12}\right)^6 \)
2. \( \left(\frac{1}{12}\right)^4 \)
3. \( \binom{6}{4} \cdot 12 \cdot \left(\frac{1}{12}\right)^4 \cdot \left(\frac{11}{12}\right)^2 \)
---
## **Question 9**
**Given Information:**
Batch contains products A (60%), B (30%), C (10%). A product is randomly withdrawn, found to be product C.
**What to Find:**
Probability the withdrawn product is A given it is C.
**Concept Used:**
Conditional probability: \( P(A|C) = \frac{P(A \cap C)}{P(C)} \)
**Solution:**
Since product is known to be C, probability it is A is irrelevant.
Thus, \( P(A|C) = 0 \) as A and C cannot occur together.
**Summary:**
\( P(A|C) = 0 \)
---
## **Question 10**
**Given Information:**
Two warning systems:
- System 1: Probability up = 0.92
- System 2: Probability up = 0.93
- If system 1 is down, probability system 2 is working = 0.85.
**What to Find:**
1. Probability both are working.
2. Probability system 2 is down and system 1 is working.
3. Probability system 1 is working given that system 2 is down.
**Concept Used:**
Multiplication rule for independent and conditional probability.
**Solution:**
Probability system 1 down: \( 1 - 0.92 = 0.08 \)
Probability system 2 down: \( 1 - 0.93 = 0.07 \)
1. Both working: \( 0.92 \times 0.93 = 0.8556 \)
2. System 2 down and system 1 working:
\( 0.92 \times 0.07 = 0.0644 \)
3. Probability system 1 working given system 2 is down:
System 1 still has probability \( 0.92 \) while system 2 is down.
**Summary:**
1. \( 0.8556 \)
2. \( 0.0644 \)
3. \( 0.92 \)
---
## **Question 11**
**Given Information:**
Events A and B, \( 0 < P(A) < 1 \).
**What to Find:**
1. Show if A and B are independent, then A and B are not mutually exclusive.
2. Show if A and B are mutually exclusive, then A and B are dependent.
**Concept Used:**
Mutually exclusive: \( P(A \cap B) = 0 \).
Independent: \( P(A \cap B) = P(A)P(B) \).
**Solution:**
1. If A and B are independent, \( P(A \cap B) = P(A)P(B) > 0 \), thus not mutually exclusive.
2. If A and B are mutually exclusive, \( P(A \cap B) = 0 \), but \( P(A)P(B) > 0 \), hence they are dependent.
**Summary:**
1. If independent, not mutually exclusive.
2. If mutually exclusive, they are dependent.
---
## **Question 12**
**Given Information:**
Events A and B are independent with \( P(A \cap B^c) = 0.25 \).
**What to Find:**
Find \( P(A) \) and \( P(B) \).
**Concept Used:**
Independence: \( P(A \cap B^c) = P(A)(1 - P(B)) \).
**Solution:**
Let \( P(A) = p, P(B) = q \).
So, \( p(1 - q) = 0.25 \).
Without additional information, we cannot uniquely determine \( p \) and \( q \).
Assuming \( p = 0.5 \):
Then \( 0.5(1 - q) = 0.25 \) ⇒ \( 1 - q = 0.5 \) ⇒ \( q = 0.5 \).
**Summary:**
\( P(A) = 0.5, P(B) = 0.5 \)
---
## **Question 13**
**Given Information:**
\( 0 < P(A) < 1 \), \( 0 < P(B) < 1 \).
**What to Find:**
1. If A and B are independent, then A and B are not mutually exclusive.
2. If A and B are mutually exclusive, then A and B are dependent.
**Concept Used:**
Mutually exclusive: \( P(A \cap B) = 0 \).
Independent: \( P(A \cap B) = P(A)P(B) \).
**Solution:**
1. If A and B are independent and have positive probabilities, then \( P(A \cap B) > 0 \), hence they are not mutually exclusive.
2. If A and B are mutually exclusive with positive probabilities, \( P(A \cap B) = 0 \) implies they are dependent.
**Summary:**
1. If independent, not mutually exclusive.
2. If mutually exclusive, they are dependent.
Question:
give final answer in decimal places and write final solutions as a summary at last
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-13 21:08:07
Final Answer:
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# Final Answers Summary
## **Question 7**
- \( P(A_1) = 0.512 \)
- \( P(A_2) = 0.512 \)
## **Question 8**
1. \( P(\text{at least one in October}) = 1 - \left(\frac{11}{12}\right)^6 \approx 0.494 \)
2. \( P(\text{all 4 in October}) = \left(\frac{1}{12}\right)^4 \approx 0.000579 \)
3. \( P(\text{4 in same month}) = \binom{6}{4} \cdot 12 \cdot \left(\frac{1}{12}\right)^4 \cdot \left(\frac{11}{12}\right)^2 \approx 0.045 \)
## **Question 9**
- \( P(A|C) = 0 \)
## **Question 10**
1. Probability both working: \( 0.8556 \)
2. Probability system 2 down and system 1 working: \( 0.0644 \)
3. Probability system 1 given system 2 is down: \( 0.92 \)
## **Question 11**
- If independent, not mutually exclusive.
- If mutually exclusive, they are dependent.
## **Question 12**
- \( P(A) = 0.5, P(B) = 0.5 \)
## **Question 13**
- If independent, not mutually exclusive.
- If mutually exclusive, they are dependent.
Question:
the correct solutions are like this : Question 7
Three numbers are randomly selected from 0–9.
Total outcomes (with replacement):
10
3
=
1000
10
3
=1000
(A) Three numbers don’t contain 0 and 5
This means neither 0 nor 5 appears.
Allowed digits: 1,2,3,4,6,7,8,9 → 8 digits
8
3
=
512
8
3
=512
?
=
512
1000
=
0.512
P=
1000
512
=0.512
(B) Three numbers don’t contain 0 or 5
Logically, “don’t contain 0 and 5” means neither appears.
“Don’t contain 0 or 5” also means neither appears.
So answer is the same:
0.512
0.512
(If the intention was something different like “not both together”, that would change — but mathematically as written, they are identical.)
Question 8
6 students, 12 equally likely months.
(1) At least one born in October
?
(
none in Oct
)
=
(
11
12
)
6
P(none in Oct)=(
12
11
)
6
(
11
12
)
6
=
0.5937
(
12
11
)
6
=0.5937
?
(
at least one
)
=
1
−
0.5937
P(at least one)=1−0.5937
0.4063
0.4063
(2) Exactly 4 born in October
Binomial:
(
6
4
)
(
1
12
)
4
(
11
12
)
2
(
4
6
)(
12
1
)
4
(
12
11
)
2
=
15
×
1
20736
×
121
144
=15×
20736
1
×
144
121
=
15
×
0.0000482
×
0.8403
=15×0.0000482×0.8403
0.000607
0.000607
(Previous solution overstated it — this is the correct small value.)
(3) Exactly 4 born in same month
Choose the month first:
12 choices
12
×
(
6
4
)
(
1
12
)
4
(
11
12
)
2
12×(
4
6
)(
12
1
)
4
(
12
11
)
2
=
12
×
0.000607
=12×0.000607
0.00729
0.00729
(NOT 0.073 — earlier decimal error.)
Question 9
Given:
P(A)=0.60
P(B)=0.30
P(C)=0.10
But defect rates are missing in the image.
Without defect probabilities for A, B, C, this question cannot be numerically solved.
General Bayes formula:
?
(
?
∣
?
?
)
=
?
(
?
?
∣
?
)
?
(
?
)
?
(
?
?
∣
?
)
?
(
?
)
+
?
(
?
?
∣
?
)
?
(
?
)
+
?
(
?
?
∣
?
)
?
(
?
)
P(A∣ND)=
P(ND∣A)P(A)+P(ND∣B)P(B)+P(ND∣C)P(C)
P(ND∣A)P(A)
A numeric answer requires defect rates.
Question 10
Given:
P(S1 works) = 0.92
P(S2 works) = 0.93
So:
P(S1 down)=0.08
P(S2 down)=0.07
Given:
?
(
?
1
works
∣
?
2
down
)
=
0.85
P(S1 works ∣S2 down)=0.85
First find
?
(
?
1
∣
?
2
)
P(S1∣S2)
Use total probability:
0.92
=
?
(
?
1
∣
?
2
)
(
0.93
)
+
0.85
(
0.07
)
0.92=P(S1∣S2)(0.93)+0.85(0.07)
0.92
=
0.93
?
+
0.0595
0.92=0.93P+0.0595
0.8605
=
0.93
?
0.8605=0.93P
?
(
?
1
∣
?
2
)
=
0.9253
P(S1∣S2)=0.9253
(1) Both working
?
(
?
1
∩
?
2
)
=
?
(
?
1
∣
?
2
)
?
(
?
2
)
=
0.9253
×
0.93
=
0.8605
P(S1∩S2)=P(S1∣S2)P(S2)=0.9253×0.93=
0.8605
(2) S2 down and S1 working
=
0.85
×
0.07
=
0.0595
=0.85×0.07=
0.0595
(3) S1 working given S2 down
Given directly:
0.85
0.85
Question 11
If independent:
?
(
?
∣
?
)
=
?
(
?
)
P(B∣A)=P(B)
Also:
?
(
?
∣
?
ˉ
)
=
?
(
?
)
P(B∣
A
ˉ
)=P(B)
So
?
(
?
∣
?
)
=
?
(
?
∣
?
ˉ
)
P(B∣A)=P(B∣
A
ˉ
)
Conversely, if
?
(
?
∣
?
)
=
?
(
?
∣
?
ˉ
)
P(B∣A)=P(B∣
A
ˉ
)
Then from total probability:
?
(
?
)
=
?
(
?
∣
?
)
?
(
?
)
+
?
(
?
∣
?
ˉ
)
?
(
?
ˉ
)
P(B)=P(B∣A)P(A)+P(B∣
A
ˉ
)P(
A
ˉ
)
If both equal k:
?
(
?
)
=
?
[
?
(
?
)
+
?
(
?
ˉ
)
]
=
?
P(B)=k[P(A)+P(
A
ˉ
)]=k
So:
?
(
?
∣
?
)
=
?
(
?
)
P(B∣A)=P(B)
Hence independent.
✔ Proven.
Question 12
Independent:
?
(
?
)
?
(
?
)
=
0.25
P(A)P(B)=0.25
So any pair satisfying:
?
(
?
)
=
0.25
?
(
?
)
P(B)=
P(A)
0.25
with 0 < P(A),P(B) ≤ 1
Examples:
0.5 and 0.5
1 and 0.25
0.25 and 1
Question 13
If independent:
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∩
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)
=
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>
0
P(A∩B)=P(A)P(B)>0
So cannot be mutually exclusive.
If mutually exclusive:
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∩
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=
0
P(A∩B)=0
But independence requires:
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=
0
P(A)P(B)=0
Which contradicts P(A)>0 and P(B)>0.
Hence mutually exclusive events (with positive probabilities) are dependent.
please consider it and edit calculations
This is AI generated Response.
Asked by: nayan
Created at: 2026-02-13 21:14:07
Final Answer:
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# Revised Solutions with Detailed Calculations
## **Question 7**
**Given Information:**
Three numbers are randomly selected from 0 to 9.
**Total outcomes (with replacement):**
\[
10^3 = 1000
\]
**(A) Three numbers don’t contain 0 and 5**
This means neither 0 nor 5 appears.
Allowed digits: 1, 2, 3, 4, 6, 7, 8, 9 → 8 digits.
Favorable outcomes:
\[
8^3 = 512
\]
\[
P(A_1) = \frac{512}{1000} = 0.512
\]
**(B) Three numbers don’t contain 0 or 5**
This logically means neither 0 nor 5 appears.
Thus, the answer remains the same:
\[
P(A_2) = 0.512
\]
**Summary:**
- \( P(A_1) = 0.512 \)
- \( P(A_2) = 0.512 \)
---
## **Question 8**
**Given Information:**
6 students live in a dorm.
**(1) Probability at least one student is born in October:**
\[
P(\text{none in October}) = \left(\frac{11}{12}\right)^6 \approx 0.5937
\]
\[
P(\text{at least one}) = 1 - P(\text{none}) \approx 1 - 0.5937 = 0.4063
\]
**(2) Probability exactly 4 students have birthdays in October:**
Using the binomial formula:
\[
P(\text{exactly 4 in October}) = \binom{6}{4} \left(\frac{1}{12}\right)^4 \left(\frac{11}{12}\right)^2
\]
Calculating:
\[
\binom{6}{4} = 15
\]
\[
P \approx 15 \times \left(\frac{1}{12}\right)^4 \times \left(\frac{11}{12}\right)^2 \approx 15 \times 0.0005787 \times 0.8403 \approx 0.00729
\]
**(3) Probability that 4 students have birthdays in the same month:**
Choosing the month first, we have 12 choices:
\[
P(\text{4 in same month}) = 12 \times \binom{6}{4} \left(\frac{1}{12}\right)^4 \left(\frac{11}{12}\right)^2 \approx 12 \times 0.000607 \approx 0.00729
\]
**Summary:**
1. \( P(\text{at least one in October}) \approx 0.4063 \)
2. \( P(\text{all 4 in October}) \approx 0.000607 \)
3. \( P(\text{4 in same month}) \approx 0.00729 \)
---
## **Question 9**
**Given Information:**
Products A (60%), B (30%), C (10%).
Defect rates are not provided in the image.
**What to Find:**
To find \( P(A|C) \), we need defect probabilities for A, B, and C.
Using Bayes' theorem:
\[
P(A|C) = \frac{P(C|A)P(A)}{P(C|A)P(A) + P(C|B)P(B) + P(C|C)P(C)}
\]
**Conclusion:**
Numeric answer requires defect rates.
**Summary:**
Insufficient data for a numeric solution.
---
## **Question 10**
**Given Information:**
- \( P(S_1 \text{ works}) = 0.92 \)
- \( P(S_2 \text{ works}) = 0.93 \)
- \( P(S_1 \text{ down}) = 0.08 \)
- \( P(S_2 \text{ down}) = 0.07 \)
- Given \( P(S_1 \text{ works}|S_2 \text{ down}) = 0.85 \)
**(1) Probability both are working:**
\[
P(S_1 \cap S_2) = P(S_1) \cdot P(S_2) = 0.92 \times 0.93 = 0.8556
\]
**(2) Probability system 2 is down and system 1 is working:**
\[
P(S_2 \text{ down} \cap S_1 \text{ working}) = P(S_1 \text{ works}) \cdot P(S_2 \text{ down}) = 0.92 \times 0.07 = 0.0644
\]
**(3) Probability system 1 is working given that system 2 is down:**
Given directly, it is \( 0.85 \).
**Summary:**
1. \( P(S_1 \cap S_2) = 0.8556 \)
2. \( P(S_2 \text{ down and } S_1 \text{ working}) = 0.0644 \)
3. \( P(S_1 \text{ works}|S_2 \text{ down}) = 0.85 \)
---
## **Question 11**
**Given Information:**
Events \( A \) and \( B \), \( 0 < P(A) < 1 \).
**What to Show:**
1. If A and B are independent, then \( P(B|A) = P(B) \) and \( P(B|A^c) = P(B) \), hence not mutually exclusive.
2. If A and B are mutually exclusive, then \( P(A \cap B) = 0 \) but independence requires \( P(A)P(B) = 0 \), which contradicts \( P(A), P(B) > 0 \).
**Summary:**
1. If independent, not mutually exclusive.
2. If mutually exclusive, then dependent.
---
## **Question 12**
**Given Information:**
\( P(A)P(B^c) = 0.25 \).
**What to Find:**
Find \( P(A) \) and \( P(B) \).
**Solution:**
Assuming \( P(B) = q \):
\[
P(B^c) = 1 - q
\]
\[
P(A)(1 - q) = 0.25
\]
**Examples of valid pairs:**
1. \( P(A) = 0.5, P(B) = 0.5 \)
2. \( P(A) = 0.25, P(B) = 1 \)
**Summary:**
Multiple valid pairs exist satisfying \( P(A)P(B^c) = 0.25 \).
---
## **Question 13**
**Given Information:**
\( 0 < P(A) < 1, 0 < P(B) < 1 \).
**What to Show:**
1. If independent, \( P(A \cap B) = P(A)P(B) > 0 \); thus, not mutually exclusive.
2. If mutually exclusive, \( P(A \cap B) = 0 \) contradicts independence if both \( P(A) \) and \( P(B) \) are greater than 0.
**Summary:**
1. If independent, not mutually exclusive.
2. If mutually exclusive, they are dependent.
---
# Final Answers Summary
### **Question 7**
- \( P(A_1) = 0.512 \)
- \( P(A_2) = 0.512 \)
### **Question 8**
1. \( P(\text{at least one in October}) \approx 0.4063 \)
2. \( P(\text{exactly 4 in October}) \approx 0.000607 \)
3. \( P(\text{4 in same month}) \approx 0.00729 \)
### **Question 9**
- Insufficient data for a numeric solution.
### **Question 10**
1. \( P(S_1 \cap S_2) = 0.8556 \)
2. \( P(S_2 \text{ down and } S_1 \text{ working}) = 0.0644 \)
3. \( P(S_1 \text{ works}|S_2 \text{ down}) = 0.85 \)
### **Question 11**
- If independent, not mutually exclusive.
- If mutually exclusive, they are dependent.
### **Question 12**
- Multiple valid pairs exist satisfying \( P(A)P(B^c) = 0.25 \).
### **Question 13**
- If independent, not mutually exclusive.
- If mutually exclusive, they are dependent.
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