Question

At a snack food manufacturing facility, the quality control engineer must ensure that all products feature the appropriate expiration date. Suppose that a box of 60 candy bars includes 12 which do not have the proper printed expiration date. The quality control engineer, in inspecting the box, grabs a handful of seven candy bars. What is the probability that there are exactly 3 faulty candy bars among the seven

Answer 1

0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.

Explanation:

The bars are chosen without replacement, which means that the hypergeometric distribution is used to solve this question.

Hypergeometric distribution:

The probability of x successes is given by the following formula:

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

In which:

x is the number of successes.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

In this question:

60 total candies means that
`N = 70`

12 are faulty, which means that
`k = 12`

Seven are chosen, so
`n = 7`

What is the probability that there are exactly 3 faulty candy bars among the seven?

This is
`P(X = 3)`. So

`P(X = x) = h(x,N,n,k) = (C_(k,x)*C_(N-k,n-x))/(C_(N,n))`

`P(X = 3) = h(3,70,7,12) = (C_(12,3)*C_(48,4))/(C_(60,7)) = 0.1108`

0.1108 = 11.08% probability that there are exactly 3 faulty candy bars among the seven.