Respuesta :
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.
Step-by-step explanation:
Conditional Probability
We use the conditional probability formula to solve this question. It is
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)}[/tex]
In which
P(B|A) is the probability of event B happening, given that A happened.
[tex]P(A \cap B)[/tex] 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
[tex]P(A) = 0.09*0.15 + 0.03*0.85 = 0.039[/tex]
Probability of a person bringing lunch to work and being a woman:
9% of 15%, so:
[tex]P(A \cap B) = 0.09*0.15[/tex]
Desired probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09*0.15}{0.039} = 0.3462[/tex]
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 [tex]P(A) = 0.15[/tex]
Probability of a person bringing lunch to work and being a woman:
9% of 15%, so:
[tex]P(A \cap B) = 0.09*0.15[/tex]
Desired probability:
[tex]P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.09*0.15}{0.15} = 0.09[/tex]
0.09 = 9% probability that a randomly selected employee brings their lunch to work given that person is a woman.
