Question 268668: I need help with this probability problem.
Suppose that 19% of the employees of a given corporation utilize the company fitness center during the lunch hour. Moreover, assume that 58% of all employees are male, and 13% of all employees are males who utilize the company fitness center during the lunch hour.
A) If we choose an employee at random from this corporation, what is the probability that this person is a female?
B) If we choose an employee at random from this corporation, what is the probability that this person is a male or someone who utilizes the company fitness center during the lunch hour?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Suppose that 19% of the employees of a given corporation utilize the company fitness center during the lunch hour.
Moreover, assume that 58% of all employees are male, and 13% of all employees are males who utilize the company fitness center during the lunch hour.
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A) If we choose an employee at random from this corporation, what is the probability that this person is a female?
P(female) = 0.42
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B) If we choose an employee at random from this corporation, what is the probability that this person is a male or someone who utilizes the company fitness center during the lunch hour?
P(male or uses fitness center)
= P(male) + P(fitness center) - P(male AND uses fitness center)
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= 0.58 + 0.19 - 0.13
= 0.64
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Cheers,
Stan H.
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