Probability: Defective Parts

[teme92]

Homework Statement

A manager must assess the quality of a new batch of 25 components ready for shipping.
Rather than assess each component, a sample of 5 is randomly selected and tested. The
quality control speci fication is that if there are 2 or more defectives in the sample, the
quality manager must reject the batch. Suppose there are actually 3 defective components
in this batch of 25.

What is the probability that the batch is rejected?

Homework Equations

The Attempt at a Solution

So I'm looking for the probability that 2 or 3 defective parts are taken in the sample of 5. I'm having trouble wrapping my head around the problem a bit.

I said the probability of choosing all 3 is:

(3/25).(2/24).(1/23)=1/2300

What's confusing me is that this has only three picks when there's five parts picked in the sample. Am I missing something here? Any help would be much appreciated.

 

[LCKurtz]

Homework Statement

A manager must assess the quality of a new batch of 25 components ready for shipping.
Rather than assess each component, a sample of 5 is randomly selected and tested. The
quality control specification is that if there are 2 or more defectives in the sample, the
quality manager must reject the batch. Suppose there are actually 3 defective components
in this batch of 25.

What is the probability that the batch is rejected?

Homework Equations

The Attempt at a Solution

So I'm looking for the probability that 2 or 3 defective parts are taken in the sample of 5. I'm having trouble wrapping my head around the problem a bit.

I said the probability of choosing all 3 is:

(3/25).(2/24).(1/23)=1/2300

What's confusing me is that this has only three picks when there's five parts picked in the sample. Am I missing something here? Any help would be much appreciated.

You have 3 bad and 22 good in the batch. If x is the number of bad, how many ways can you pick x out of the bad and 5-x out of the good? Have you studied the hypergeometric distribution yet?

 

[teme92]

This is a sample exam question yes so I would've done hypergeometric distribution yes. I just don't quite understand how to use it so I never thought that's what this problem was.

I think I know now when it should be used though. So:

P(X=2)+P(X+3)

P(X=2)=[(mCk).(N-m)C(n-k)]/(25C5)

where N=25, m=3, k=2, n=5 and use the same approach for P(X=3) except k=3.

Is this how the problem is done?

 

[LCKurtz]

That's the idea.