Question 1083436
Number of samples of size 3 from 9 is 9C3=84, and that is the denominator.
the numerator is 5C3=10/84, because there are 10 ways 5 non-defective printers may be chosen in a sample of 3.
10/84=5/42
probability printers are not defective is 5/9*4/8*3/7 for the first three, because the sample space changes. (60/504), which is 10/84=5/42