College Algebra & Trigonometry, 2018a

334 CHAPTER 7. COMBINATORICS
Example
Given 100 computer components, it is known that 10 of the 100 are defective. If
someone were to choose 6 of these components at random, what is the probability
that:
a) two of them are defective?
b) at least 1 of them is defective?
There are 10 C 2 ways to choose 2 out of the 10 defective components and 90 C 4 ways
to choose 4 non-defective components. The size of the sample space (the number
of ways to choose 6 components from the 100) is 100 C 6 .
So, the probability for part (a) would be:
10C 2 ∗ 90 C 4
100C 6
≈ 0.096
The probability in part (b) is most easily computed by finding the probability that
no defective components are selected and then subtracting that value from 1.
1 − 90 C 6
100C 6
≈ 1 − 0.5223 ≈ 0.4777