# Probability of Preventable ADEs (Kanjanarat et al.) Given: - **Rate of preventable ADEs**, \( p = 0.352 \) (from 35.2%) - **Sample size**, \( n = 10 \) patients who experienced an ADE We are asked to find probabilities for various counts of preventable ADEs in the sample. This is a **binomial distribution** problem: \( X \sim \text{Binomial}(n=10, p=0.352) \). --- ## A. Probability Exactly Seven Were Preventable \[ P(X = 7) = \binom{10}{7} (0.352)^7 (1 - 0.352)^{3} \] Calculate: - \( \binom{10}{7} = 120 \) - \( (0.352)^7 \approx 0.00190 \) - \( (0.648)^3 \approx 0.272 \) \[ P(X=7) = 120 \times 0.00190 \times 0.272 \approx 0.0620 \] **Final answer:** \[ \boxed{P(X=7) \approx 0.062} \] --- ## B. Probability More Than Half Were Preventable (\(X > 5\)) \[ P(X > 5) = P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10) \] Calculate each: ### \(P(X=6)\) \[ \binom{10}{6} (0.352)^6 (0.648)^4 = 210 \times 0.00538 \times 0.176 \approx 0.198 \text{ (rounded)} \] But let's recompute for accuracy: - \( \binom{10}{6} = 210 \) - \( (0.352)^6 \approx 0.00538 \) - \( (0.648)^4 \approx 0.176 \) \[ P(X=6) = 210 \times 0.00538 \times 0.176 \approx 0.199 \] ### \(P(X=7)\) Already found: \( 0.062 \) ### \(P(X=8)\) - \( \binom{10}{8} = 45 \) - \( (0.352)^8 \approx 0.00067 \) - \( (0.648)^2 \approx 0.420 \) \[ P(X=8) = 45 \times 0.00067 \times 0.420 \approx 0.0126 \] ### \(P(X=9)\) - \( \binom{10}{9} = 10 \) - \( (0.352)^9 \approx 0.00024 \) - \( (0.648)^1 = 0.648 \) \[ P(X=9) = 10 \times 0.00024 \times 0.648 \approx 0.0016 \] ### \(P(X=10)\) - \( \binom{10}{10} = 1 \) - \( (0.352)^{10} \approx 0.00008 \) - \( (0.648)^0 = 1 \) \[ P(X=10) = 1 \times 0.00008 \times 1 = 0.00008 \] **Sum:** \[ P(X > 5) \approx 0.199 + 0.062 + 0.0126 + 0.0016 + 0.00008 = 0.275 \] **Final answer:** \[ \boxed{P(X > 5) \approx 0.275} \] --- ## C. Probability None Were Preventable \[ P(X = 0) = \binom{10}{0} (0.352)^0 (0.648)^{10} = 1 \times 1 \times (0.648)^{10} \] \[ (0.648)^{10} \approx 0.0132 \] **Final answer:** \[ \boxed{P(X = 0) \approx 0.013} \] --- ## D. Probability Between 3 and 6 Inclusive Were Preventable \[ P(3 \leq X \leq 6) = P(X=3) + P(X=4) + P(X=5) + P(X=6) \] Calculate each: ### \(P(X=3)\) - \( \binom{10}{3} = 120 \) - \( (0.352)^3 \approx 0.0436 \) - \( (0.648)^7 \approx 0.0579 \) \[ P(X=3) = 120 \times 0.0436 \times 0.0579 \approx 0.303 \] ### \(P(X=4)\) - \( \binom{10}{4} = 210 \) - \( (0.352)^4 \approx 0.0153 \) - \( (0.648)^6 \approx 0.0894 \) \[ P(X=4) = 210 \times 0.0153 \times 0.0894 \approx 0.287 \] ### \(P(X=5)\) - \( \binom{10}{5} = 252 \) - \( (0.352)^5 \approx 0.00538 \) - \( (0.648)^5 \approx 0.1378 \) \[ P(X=5) = 252 \times 0.00538 \times 0.1378 \approx 0.186 \] ### \(P(X=6)\) Previously found: \( 0.199 \) **Sum:** \[ P(3 \leq X \leq 6) \approx 0.303 + 0.287 + 0.186 + 0.199 = 0.975 \] **Final answer:** \[ \boxed{P(3 \leq X \leq 6) \approx 0.975} \] --- ## Summary Table | Event | Probability | |-------------------------------------------|-------------| | Exactly 7 preventable (\(X=7\)) | 0.062 | | More than half preventable (\(X>5\)) | 0.275 | | None preventable (\(X=0\)) | 0.013 | | 3 to 6 inclusive preventable (3≤X≤6) | 0.975 | --- **Note:** All calculations use the binomial formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] Values are rounded for clarity. For more precise results, use a calculator or statistical software.