# Binomial Conditional Probability of a an event

## Determining Binomial Condition Probability of a Random Sample

I have a question about binomial probability involving a conditional event. This problem keeps tripping me up, because while I know how to calculate the binomial probability that a random variable is a failure, i don't know how to calculate the conditional probability of that variable.

My question is as follows:

70% of the total shipments come from factory A, of which 10% are defective.

30% of the total shipments come from factory B, of which 5% are defective.

A random shipment comes in, and a sample of 20 pints is taken, and 1 of the pints is defective.

What is the probability that this shipment came from the Factory A?

## 1 Answer

Edit 1: didn't pay attention to the question. Will edit answer later today.

Edit 2: I've attempted to provide an answer below, however I might be mistaken. Feel free to correct me if I am in error.

$P(1 defective|A) \approx 0.270 \wedge P(1 defective|B) \approx 0.377 \\ P(A) = 0.7 \wedge P(B) = 0.3 \\ P(1D) = 0.7*0.270+0.3*0.377 = 0.189+0.113 = 0.302 \\ P(A|1D) = P(1D|A)*P(A)/P(1D)=0.270*0.7/0.302 \approx 0.626$

• Just one pint out of 20 is defective. – Michael M Oct 11 '13 at 7:53
• Oh, sorry, didn't notice that. Will have to update the answer when I get home. Thanks for noticing. – abaumann Oct 11 '13 at 8:02