Probability that the object in the box is defective?

Question

A manufactory produces items in a box of two. On the history of the manufacture We observe that

• 92% of the boxes contain two perfect objects,

• 5% of the boxes contain a perfect object and a defective one,

• 3% of the boxes contain two defective objects.

A box is chosen at random and one of the two objects of the box is chosen at random, it turns out defective. What is the probability that the other object in the box is so defective?

Here is my approach:

1. Calling P the event that a box contains a perfect object and D the event that a box contains a defective object

2. We do have: P (PP) = 0.92; P(PD) = 0.05; P(DD)= 0.03. And we have to find P(D|D) = 0.03 / (0.05+0.03) = 3/8? Isn't it?

Please correct me if anything is wrong in my solution. Thank you

Answer 1

You must have: $$P(DD|D)=\frac{P(DD,D)}{P(D)}=\frac{P(DD,D)}{P(DD,D)+P(PD,D)}=\\ =\frac{0.03}{0.03+0.05\cdot 0.5}=\frac{0.03}{0.055}=\frac{6}{11}.$$

Answer 2

You have eight equally likely possibilities -- five boxes with one defective object and three boxes with two defective objects. You are equally likely to have selected any of the eleven defective objects. Six of those times, the other object also will be defective, whereas the other five of those times, it won't be. So the probability that the other object in the same box also is defective is $\frac{6}{11}$.

Given that the object you selected is defective, you are not equally likely to have selected a $PD$ box and a $DD$ box. Your solution assumes those events are equally likely given the information you have, and that is not correct.