VIPSolutionsKanjanarat et al. (A-12) estimate the rate of preventable adverse drug events (ADEs) in hospitals to be 35.2 percent. Preventable ADEs typically result from inappropriate care of medication errors, which include errors of commission and errors of omission. Suppose that 10 hospital patients experienced an ADE are chosen at random. Let p 1/4:35 and calculate the probability that A Exactly seven of those drug events were preventable B More than half of those drug events were preventable c. None of those drug events were preventable d. Between three and six inclusive were preventable
Question:
Kanjanarat et al. (A-12) estimate the rate of preventable adverse drug events (ADEs) in hospitals to be 35.2 percent. Preventable ADEs typically result from inappropriate care of medication errors, which include errors of commission and errors of omission. Suppose that 10 hospital patients experienced an ADE are chosen at random. Let p 1/4:35 and calculate the probability that A Exactly seven of those drug events were preventable B More than half of those drug events were preventable c. None of those drug events were preventable d. Between three and six inclusive were preventable
This is AI generated Response.
Asked by: suni
Created at: 2025-09-16 08:56:17
Asked by: suni
Created at: 2025-09-16 08:56:17
Final Answer:
Rendered
Raw Text
Did you like this answer?
# Probability Calculations for Preventable ADEs
Given:
- Probability an ADE is preventable, \( p = 0.352 \) (i.e., 35.2%)
- Number of patients, \( n = 10 \)
- Let \( X \) = number of preventable ADEs among 10 patients
- \( X \sim \text{Binomial}(n=10, p=0.352) \)
---
## Binomial Probability Formula
For \( k \) preventable ADEs:
\[
P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}
\]
---
## A. Exactly Seven Preventable ADEs
\[
P(X = 7) = \binom{10}{7} (0.352)^7 (0.648)^3
\]
Calculate:
- \(\binom{10}{7} = 120\)
- \(0.352^7 \approx 0.001527\)
- \(0.648^3 \approx 0.272\)
\[
P(X = 7) \approx 120 \times 0.001527 \times 0.272 \approx 0.0498
\]
**Probability ≈ 4.98%**
---
## B. More Than Half Preventable (i.e., \(X \geq 6\))
Calculate \(P(X \geq 6) = P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)\)
Calculate each term:
- \(P(X=6) = \binom{10}{6} (0.352)^6 (0.648)^4\)
- \(P(X=7) = \binom{10}{7} (0.352)^7 (0.648)^3\)
- \(P(X=8) = \binom{10}{8} (0.352)^8 (0.648)^2\)
- \(P(X=9) = \binom{10}{9} (0.352)^9 (0.648)^1\)
- \(P(X=10) = \binom{10}{10} (0.352)^{10}\)
Let's estimate:
- \( P(X=6) \approx 210 \times 0.00434 \times 0.273 \approx 0.249 \)
- \( P(X=7) \approx 120 \times 0.00153 \times 0.272 \approx 0.0498 \) (from above)
- \( P(X=8) \approx 45 \times 0.00054 \times 0.420 \approx 0.0102 \)
- \( P(X=9) \approx 10 \times 0.00019 \times 0.648 \approx 0.00123 \)
- \( P(X=10) \approx 1 \times 0.00007 \approx 0.00007 \)
Sum:
\[
P(X \geq 6) \approx 0.0249 + 0.0498 + 0.0102 + 0.00123 + 0.00007 \approx 0.0862
\]
**Probability ≈ 8.6%**
---
## C. None Preventable (\(X=0\))
\[
P(X=0) = (0.648)^{10} \approx 0.0124
\]
**Probability ≈ 1.24%**
---
## D. Between Three and Six Inclusive (\(3 \leq X \leq 6\))
\[
P(3 \leq X \leq 6) = P(X=3)+P(X=4)+P(X=5)+P(X=6)
\]
Calculate each:
- \(P(X=3) = \binom{10}{3} (0.352)^3 (0.648)^7 \approx 120 \times 0.0436 \times 0.0553 \approx 0.289\)
- \(P(X=4) = \binom{10}{4} (0.352)^4 (0.648)^6 \approx 210 \times 0.0153 \times 0.0853 \approx 0.274\)
- \(P(X=5) = \binom{10}{5} (0.352)^5 (0.648)^5 \approx 252 \times 0.0054 \times 0.131 \approx 0.179\)
- \(P(X=6) \approx 0.0849\) (from earlier)
Sum:
\[
P(3 \leq X \leq 6) \approx 0.289 + 0.274 + 0.179 + 0.0849 = 0.827
\]
**Probability ≈ 82.7%**
---
## Summary Table
| Scenario | Probability |
|---------------------------------------------|-------------|
| Exactly 7 preventable | 4.98% |
| More than half (6 or more) preventable | 8.62% |
| None preventable | 1.24% |
| 3 to 6 (inclusive) preventable | 82.7% |
---
**Note:** All results rounded to two decimal places. Calculations use the binomial probability formula with \( p = 0.352 \), \( n = 10 \).
📢 Instructions:
- Ask anything relatd to above question only.
- For new Question goto :