Electronic baseball games manufactured by Tempco Electronics are shipped in lots of 25. Before shipping, a quality-control inspector randomly selects a sample of 7 from each lot for testing. If the sample contains any defective games, the entire lot is rejected. What is the probability that a lot containing exactly 2 defective games will still be shipped? (Round your answer to three decimal places.)

Question

anonymous · Answer

For the lot, the probability that a sample is defective is p=2/25.
The probability that such a lot will be missed is the product of
(23/25)((22/24)(21/23)(20/22)(19/21)(18/20)(17/19), because you are sampling without replacement  and must miss the bad lots each time. This is only approximately (23/25)^7, the sampling with replacement value.

anonymous · Answer

0.557? @douglaswinslowcooper

anonymous · Answer

0.578 = (23/25)^7, but this is not the answer for sampling without replacement, that long series of numbers multiplied by each other. Why don't you do that? I will also.

anonymous · Answer

I got 0.0509

anonymous · Answer

I got 0.51. Did you?

anonymous · Answer

I think I may have rounded some of the numbers wrong after dividing..

anonymous · Answer

Easy to make a mistake. I got 0.51 twice. Your value seems too small, as it is much less than 0.578=23/25)^7 our rough estimate.
Multiply the denominators, store. Multiply the numerators, then recall stored denominator and divide by it.

anonymous · Answer

I was just using a basic calculator, this is where I most likely messed up. Thanks! :)

anonymous · Answer

Maybe you meant to write "0.509" rather than "0.0509"?

anonymous · Answer

You are quit welcome. I do think 0.51 is it.

anonymous · Answer

I left for awhile. After multiplying all those numbers, I noticed there was a lot of cancellations I could have done, leaving 
(18*17)/(25*24) = 0.51.