a) I read that as "exactly 3 of them ..."
prob(wear hat) = .41
prob(not wear hat) = .59
prob(exactly 3 of 12) = C(12,3)(.41)^3 (.59)^9
= 220(....)(...)
= .13135
b) since 41% are said to wear hats, for 12 people
I would expect 12(.41) or appr 9 people
A taxi driver has observed that, on any given day, the probability that a customer wears a hat is 41%.
a) What is the probability that of his next 12 customers, that 3 of them will have hats?
b) How many of those 12 customers can he expect to be wearing hats?
Note: I have to use distribution formulas but I am not sure which one to use
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