Question

asked Nov 4, 2022 68.6k views

1 Answer

Pinturikkio · Answer 1 · 2022-11-08T14:17:50+0000

Answer:

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.

Explanation:

Conditional Probability

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

$P(B|A) = (P(A \cap B))/(P(A))$

In which

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

$P(A \cap B)$ 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

P(A) = 0.09*0.15 + 0.03*0.85 = 0.039

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

9% of 15%, so:

$P(A \cap B) = 0.09*0.15$

Desired probability:

$P(B|A) = (P(A \cap B))/(P(A)) = (0.09*0.15)/(0.039) = 0.3462$

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
P(A) = 0.15

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

9% of 15%, so:

$P(A \cap B) = 0.09*0.15$

Desired probability:

$P(B|A) = (P(A \cap B))/(P(A)) = (0.09*0.15)/(0.15) = 0.09$

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

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