Solving Probability Problem: 5th Good Item on 9th Test

[eku_girl83]

Here's the question:
A box contains 6 good and 8 defective light bulbs. The bulbs are drawn out one at a time, without replacement, and tested. What is the probability that the fifth good item is found on the ninth test?

Could someone explain how I would go about solving this problem? Thanks!

 

[Hurkyl]

If I told you the probability that exactly 4 good items have been found within 8 tests was 0.68, could you solve the problem?

(p.s. 0.68 is probably wrong)

 

[Damned charming :)]

Consider the first 9 balls, this can be done 14C9 ( from 14 choose 9 its on your calculator). If the 9th ball is the 5th good then the first 9 balls must consist of 5 good and 4 bad balls.
The probability of this happening is 6C5*8C4/14C9. If this is true you need the 9th ball to be good. This has probably 5/9.
So the probability is 6C5*8C4/14C9 * 5/9.

 

[matt grime]

Not sure about that answer, Damned.

You want 4 good and 4 bad on the first 8, then to draw a bad on the 9th, which is to draw on of the 4 remaining bad ones from the 6 that are left.

$$\frac{\frac{4}{6}\binom{6}{4}\binom{8}{4}}{\binom{14}{8}}$$but they might well be the same after simplifying

 

[Damned charming :)]

The fifth good item has to found on the 9th test. So you should replace the 4/6 with a 2/6 and this can be rearranged to give my answer. You solution is slighty better and more consistent with student examples of negative binomial etc.

 

[matt grime]

Sorry for switching things over, and yes I agree with your answer entirely now I've thought about it for a second. I also agree that such conditional probabilities would be beyond the scope of the course I imagine the OP is doing.