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

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:

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:

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

In this question:

60 total candies means that

12 are faulty, which means that

Seven are chosen, so

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

This is . So

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

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐