A lot of size n = 30 contains three nonconforming units. what is the probability that a sample of five units selected at random contains exactly one nonconforming unit? what is the probability that it contains one or more nonconformances?

Question

Mathematics

Baron Raymond

11 March, 06:05

A lot of size n = 30 contains three nonconforming units. what is the probability that a sample of five units selected at random contains exactly one nonconforming unit? what is the probability that it contains one or more nonconformances?

+1

Answers (1)

Know the Answer?

Kimora · Answer 1

Exactly 1 nonconforming unit = 0.369458128

1 or more nonconforming units = 0.433497537

Due to formatting issues, I'll use the notion C (n, x) for N choose X. Which would be n! / (x! (n-x) !).

Now for the case of exactly 1, the probability will be:

P = C (3,1) * C (30-3,4) / C (30,5)

To explain it, you choose exactly 1 nonconforming item out of the 3 possible, then fill the sample with 4 more conforming items to complete the sample size of 5. Finally, you divide by the number of different ways you can select 5 items out of the entire group of 30. So doing the math, you get P = C (3,1) * C (30-3,4) / C (30,5) = 3*17550/142506 = 0.369458128 = 36.9458128%

For the case of 1 or more nonconforming units you do the sum of x ranging from 1 to 3 with the formula

C (3, x) * C (30-3,5-x) / C (30,5)

or you can set x to 0 and evaluate

1 - C (3,0) * C (30-3,5-0) / C (30,5)

Let's do it both ways.

C (3,1) * C (30-3,5-1) / C (30,5) +

C (3,2) * C (30-3,5-2) / C (30,5) +

C (3,3) * C (30-3,5-3) / C (30,5)

= 3*17550/142506 + 3*2925/142506 + 1*351/142506

= 0.369458128 + 0.061576355 + 0.002463054

= 0.433497537

And doing it the "simple" way

1 - C (3,0) * C (30-3,5-0) / C (30,5)

= 1 - 1*80730/142506 = 1 - 0.566502463 = 0.433497537