What is the probability that the second one selected is defective given that the first one was defective?

Question

I have this exercise.

A batch of 500 containers for frozen orange juice contains ten that are defective. Two are selected, at random, without replacement from the batch.

(a) What is the probability that the second one selected is defective given that the first one was defective?

Since it ask us to calculate what the probability is due to the given condition. Well..

The problem seem somehow similar to this exercise

In which the Conditional Approach makes sense, but why can't this approach be used to my problem. ??

Answer 1

I misread the question. The answer the problems is 9/499. Since it's already been pulled a defective one from the batch, the batch decreases,and the the amount of defective decreases aswell.. I misread the question.

Answer 2

The example you provided calculates something different. It calculates $P(A \cap B)$, while your exercise is caluclating $P(A|B)$. The example does, however, calculate a conditional probability of the kind you need (drawing a king, given that the first card drawn was a king).

Answer 3

I agree with the above responses regarding part a. The correct answer to part (a) is 9/499.

We want to find the probability of choosing a second defective item, after selecting a defective item.

There starts with 500 items total, and we have 10 defective items. The probability of grabbing a single defective item is 10/500. Once you attempt to grab another defective item, assuming that you do not replace, your sample size will change from 500 items to 499. Also, you now have 9 defective items, so your probability of grabbing a second defective item is 9/499.

Remember, conditional probability changes the sample size.