Question

A restaurant is considering extending their opening hours, either by extending their hours on weeknights or weekends. 85% of the customers surveyed said that they preferred extended hours on weekends. 60% of the customers were male. 80% of the males preferred extended weekend hours. Given that a randomly selected customer is a female, what is the probability the individual prefers extended weekend hours?

Answer 1

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P(prefer ext) = .85

P(cust is male) = .60

p(cust is female) = .40

p(given male, prefers ext) = .80

Let p be p(given female, prefers ext).

So, p(prefer ext) = .85 = p(cust is male)*p(given male, prefers ext) + p(cust is female)*p(given female, prefers ext)

.85 = .6*.8 + .4*p

p = .925

So, the Given its a female, probability the individual prefers extended weekend hours = 0.925