Suppose that 55% of all babies born in a particular hospital are boys. If 7 babies born in the hospital are randomly selected, what is the probability that at least 3 of them are boys?
Carry your intermediate computations to at least four decimal places, and round your answer to two decimal places.
Solution
Given that ,
p = 0.55
q = 1 - p = 1 - 0.55 = 0.45
n = 7
Using binomial probability formula ,
P(X = x) = (n C x) * p x * (1 - p)n - x
P(X 3 ) = 1 - P( x <3)
= 1 - P(X = 0) P(X = 1) P(X = 2)
= 1 - (7 C 0) * 0.55 0 * (0.45)7 (7 C 1) * 0.55 1 * (0.45)6 (7 C 2) * 0.55 2 * (0.45)5
=1-0.1529
probability=0.8471
answer=0.85
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