At a local company, 15% of the employees are women. every day, 9% of them bring their lunch to work, while only 3% of the men bring lunch. Find the probability that a randomly selected employee a. is a woman goven that the person brings their lunch to work.
b. brings their lunch to work given that person is a woman.
c. is a woman given that the person brings their lunch to work.

a) 0.3462 = 34.62% that a randomly selected employee is a woman given that the person brings their lunch to work.

b) 0.09 = 9% probability that a randomly selected employee brings their lunch to work given that person is a woman.

c) 0.3462 = 34.62% that a randomly selected employee is a woman given that the person brings their lunch to work.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

In which

P(B|A) is the probability of event B happening, given that A happened.

is the probability of both A and B happening.

P(A) is the probability of A happening.

Questions a/c:

Questions a and c are the same, so:

Event A: Brings lunch to work.

Event B: Is a woman.

Probability of a person bringing lunch to work:

9% of 15%(woman)

3% of 100 - 15 = 85%(man). So

Probability of a person bringing lunch to work and being a woman:

9% of 15%, so:

Desired probability:

0.3462 = 34.62% that a randomly selected employee is a woman given that the person brings their lunch to work.

Question b:

Event A: Woman

Event B: Brings lunch

15% of the employees are women.

This means that

Probability of a person bringing lunch to work and being a woman:

9% of 15%, so:

Desired probability:

0.09 = 9% probability that a randomly selected employee brings their lunch to work given that person is a woman.

answer are

step-by-step explanation:

the answer is 13 cm