How do I calculate conditional probability? Specific problem included?

Question

Hi all, can someone help me understand the calculating of conditional probability? I know the formula, but let me pose this problem to you:

1) A bin has 10 bolts

2) 2 bolts are defective

3) Select 2 bolts one after the other without replacement

4) What are the conditional probabilities of both defective and nondefective bolts

What gets me on this problem is that the bolts will decrease with each one taken out of the bin, and its messing with my math. Any help will be awesome. Thanks!

~Sil

Answer 1

The question you have to answer (item 4) is not clearly worded, but the following should give you what you need to understand all the possibilities.

The idea of conditional probability is that the likelihood of a certain outcome depends on (i.e., is "conditioned" on) what has happened before. In this case, the probabilities for the second bolt depend on what happened with the first bolt.

Some numbers:
First bolt:
2/10 probability it's defective, 8/10 probability it's good.

Second bolt:
If first bolt was good:
2/9 prob. defective, 7/9 prob. good
If first bolt was defective:
1/9 prob. defective, 8/9 prob. good

Probability both are defective:
2/10 x 1/9 = 1/45
Probability both are good:
8/10 x 7/9 = 28/45
Probability one good and one defective:
2/10 x 8/9 + 8/10 x 2/9 = 16/45
(Note that the defective one can occur either first or second,
so we have to add the probabilities of each of these events.)

A "conditional" probability is used in each of these calculations. It is the second number in each multiplication, which is dependent on what happened when the first bolt was drawn.

Answer 2

Both Bolts defective:
2/10 for the first bolt, 1/9 for the second ( because one of the defective bolts has already been withdrawn) probability is 2/90=1/10*1/9

Both bolts good : First choice 8/10 second choice 7/9 (as there are only 7 good bolts in the 9 remaining bolts) probability 7*8/90

best of luck - Mike

Answer 3

There are some aspects of your question which are not specified.

Q: Will you know whether the bolts removed are defective or not after you have removed two, and are you therfore interested in the resultant "informed' probability calculation?

If the answer to this question is "no", then the probabilities for each of the selections would remain constant and equal.

However, clearly if after taking the first two bolt out of the bin you have selected no defective bolt the probability of one or two defective bolts in the next selection would be higher. And, if you have selected the two defective bolts during the first selection the chance of one or two in the next selection is zero. And so on . . .

There are lots of variables you have not included in your question.