Clarifying whether hypergeometric distribution can be used here

Question

A batch of 140 widgets is inspected by choosing a sample of 5 widgets. Assume 10 of the widgets are defective. Find the probability of selecting a sample of five widgets with at least one defective widget.

My approach is to this problem is to think about what distribution I should use first and I feel hypergeometric works best.

If that's appropriate then my set up is to find 1-P(sample of five with no defective widget), which would be $1- ({}_{10}C_0 \cdot {}_{130}C_5/{}_{140}C_5).$

Am I using the principle of hypergeometric distribution correctly and if not, what should I do instead? if this is correct then would the above equation be the pmf and 0.31 be the probability?

thanks in advance for your help.

Answer 1

You are correct that the hypergeometric distribution will work here.

$$P(X \ge 0) = 1 - P(X = 0) = 1 - \frac{{10 \choose 0}{130 \choose 5}}{{140 \choose 5}} = 1- 0.6865 = 0.3135.$$

So your method is right, and your computation is correct to two places. If you write it out in terms of factorials, there is quite a bit of cancellation, and the remaining products can be done on a calculator.

In R statistical software it is easy to compute all six probabilities in this distribution:

 x = 0:5;  pdf = dhyper(x, 10, 130, 5)
 cbind(x, pdf)
# x          pdf
# 0 6.864927e-01
# 1 2.724177e-01
# 2 3.861039e-02
# 3 2.413149e-03
# 4 6.547304e-05
# 5 6.043665e-07