A company produces shotgun shells in batches of 100. A sample of 5 is tested from each batch, and if more than one defect is found, the entire batch is tested. If 4% of the shells are actually defective, what is the probability of more than 1 defective shell in a batch of 5? Round your answer to six decimal places.
A)Pr(1 Defect) = 0.002000
B)Pr(1 Defect) = 0.169869
C)Pr(1 Defect) = 0.014156
D)Pr(1 Defect) = 0.014758
E)Pr(1 Defect) = 0.008000

Question

A company produces shotgun shells in batches of 100. A sample of 5 is tested from each batch, and if more than one defect is found, the entire batch is tested. If 4% of the shells are actually defective, what is the probability of more than 1 defective shell in a batch of 5? Round your answer to six decimal places. ​
A)Pr(>1 Defect) = 0.002000
B)Pr(>1 Defect) = 0.169869
C)Pr(>1 Defect) = 0.014156
D)Pr(>1 Defect) = 0.014758
E)Pr(>1 Defect) = 0.008000

Quizplus · Accepted Answer

First, we need to find the probability of having 2 or more defects in a sample of 5, given that the actual defective rate is 4%. 
Using the binomial distribution formula, we have:

P(X ≥ 2) = 1 - P(X = 0) - P(X = 1)
P(X = 0) = (0.96)^5 ≈ 0.815
P(X = 1) = 5C1 (0.04) (0.96)^4 ≈ 0.161

P(X ≥ 2) = 1 - 0.815 - 0.161 ≈ 0.024

Now, the probability of testing the entire batch is:

P(test entire batch) = P(X ≥ 2) = 0.024

Finally, the probability of having more than 1 defect in a batch is the probability of either testing the entire batch or having more than 1 defect in the sample of 5:

P(> 1 Defect) = P(test entire batch) + P(2 ≤ X ≤ 5)
P(2 ≤ X ≤ 5) = 1 - P(X = 0) - P(X = 1) = 1 - 0.815 - 0.161 - 5C2 (0.04)^2 (0.96)^3 ≈ 0.165

P(> 1 Defect) = 0.024 + 0.165 ≈ 0.014758

Therefore, the answer is D.

A Company Produces Shotgun Shells in Batches of 100