Example 47

EXAMPLE 47 probability using combinations: Acceptance sampling

Continuing with Example 46, if both cell phones in the sample of size 2 are defective,we will reject the batch and cancel our contract with the supplier.

What is the number of ways that both cell phones will be defective?
What is the number of outcomes in this sample space?
What is the probability that both cell phones will be defective?

Solution

From Example 46, the number of ways that both cell phones will be defective is
$_{3} C_{2} \cdot_{17} C_{0} = 3 \cdot 1 = 3$
The number of outcomes in the sample space is given by the number of ways of selecting 2 cell phones out of a batch of 20; that is,
$N (S) =_{20} C_{2} = \frac{20!}{2! (20 - 2)!} = \frac{20 \cdot 19 \cdot 18!}{2! \cdot 18!} = \frac{380}{2} = 190$
Therefore, the probability that both cell phones will be defective is given by
$P (Both defective) = \frac{number of ways both defective}{number of outcomes in sample space} = \frac{3}{190} \approx 0.01579$