Question

A pharmacist receives a shipment of 22 bottles of a drug and has 3 of the bottles tested. If 5 of the 22 bottles are contaminated, what is the probability that exactly 1 of the tested bottles is contaminated?

Answer 1

Answer:

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.

Step-by-step explanation:

$C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this problem, we have that:

Total number of combinations:

$C_{22,3} = \frac{22!}{3!(18)!} = 1540$

Desired combinations:

It is 1 one 5(contamined) and 2 of 17(non contamined). So:

$C_{5,1}*C_{17,2} = 5*17*8 = 680$

What is the probability that exactly 1 of the tested bottles is contaminated?

$P = \frac{680}{1540} = 0.4416$

There is a 44.16% probability that exactly 1 of the tested bottles is contaminated.