Clarifying whether hypergeometric distribution can be used here 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. 

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


