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

Answer:

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

Step-by-step 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:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

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.

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

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

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

$P(X = 3) = h(3,70,7,12) = \frac{C_{12,3}*C_{48,4}}{C_{60,7}} = 0.1108$

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