Question 1072110: Apple claims that their iPhones have about 40% of the American smartphone market. If I randomly choose people with a smartphone and ask whether they have an iPhone, what is the probability that I will survey no more than 4 people before finding someone who has an iPhone?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! p = .4 = probability that a person chosen at random has an iphone.
q = .6 = probability that a person chosen at random doesn't have an iphone.
in order to survey no more than 4 people, then you have to survey up to 4 people only.
if none of them have an iphone, then the probability is .6 * .6 * .6 * .6 = .,6^4 = .1296 that you won't find anybody with an iphone if you survey no more than 4.
this means the probability that you will find somebody in the first 4 that has an iphone equal to 1 - .1296 = .8704.
p(0 out of 4 have it) = .1296
p(1 out of 4 have it) = .3456
p(2 out of 4 have it) = .3456
p(3 out of 4 have it) = .1536
p(4 out of 4 have it) = .0256
probability that at least 1 out of the 4 has it would be equal to:
.3456 + .3456 + .1536 + .0256 = .8704
that's the same as the probability of 1 minus the probability that 0 have it.
this essentially becomes a binomial probability problem where n = 4.
p = .4
q = .6
n = 4
x = 0, 1, 2, 3, 4
p(x) = c(n,x) * p^x * q^(n-x)
c(n,x) = n! / (x! * (n-x)!)
the probability is .8704 that you will survey no more than 4 people and find someone who has an iphone.
that's my take.
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