Question 727336: Suppose that 40 % of the population is male and that 80% own a computer. If 70 % of females own computers, what is the probability that:
(a) A randomly selected person does NOT own a computer?
P( person does not own a computer ) =
(b) A randomly selected person owns a computer?
P( person owns a computer ) =
(c) If the person selected owns a computer, what is the probability that the person was male?
P( Male | owns computer ) =
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! Parts a) and b)
40 % of the population is male, so 60% is female
Suppose that 40 % of the population is male and that 80% own a computer, which means that 80% of the guys, or 0.8*0.4 = 0.32 = 32% of the population owns a computer and is male. This isn't the whole story since we've yet to count females.
70 % of females own computers, so 0.7*0.6 = 0.42 = 42% of the population owns computers and are female. All together, 32+42 = 74% of the population owns a computer. Now we're counting both males and females.
So 100 - 74 = 26% of the population does not own a computer. This is because you either own a computer or you don't own a computer.
So the answer to part a) is 26% or 0.26
The answer to part b) is 74% or 0.74
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Part c)
P( Male | owns computer ) = P(Male AND owns computer)/P(owns computer)
P( Male | owns computer ) = (0.32)/(0.74)
P( Male | owns computer ) = 0.43243243243243
which is roughly 43.24%
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