In a certain country, the true probability of baby being a girl is 0.45. Among the next seven randomly selected births in this country, what is the probability that at least one of them is a boy? Would it be a significant event if none of the next 7 births resulted in a boy? How do you know?

Given that the the true probability of being a girl is 0.45, the probability of being a boy is:

P(boy)= 1- P(girl)

= =1-0.45

= 0.55

This implies that the percentage of being a boy, p = 0.55

Every baby can either be girl or a boy, which are the only two possible outcomes. Therefore, we will use the binomial distribution which gives the probability of exactly x successes on n repeated trials and is given by:

P(X=x) = (n!/(x!(n-x)!)).px.(1-p)n-x

given that 7 babies are selected, it implies that n=7

The probability of selecting at least 1 boy:

where

P(X≥1) = P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) +P(X=7)

By using the formula

P(X=x) = (n!/(x!(n-x)!)).px .(1-p)n-x

P(X=1) = (7!/(1!(7-1)!)).(0.55)1 .(0.45)6 = 0.0320

P(X=2) = (7!/(2!(7-2)!)).(0.55)2 .(0.45)5 = 0.1172

P(X=3) = (7!/(3!(7-3)!)).(0.55)3 .(0.45)4 = 0.2388

P(X=4) = (7!/(4!(7-4)!)).(0.55)4 .(0.45)3 = 0.2918

P(X=5) = (7!/(5!(7-5)!)).(0.55)5 .(0.45)2 = 0.2140

P(X=6) = (7!/(6!(7-6)!)).(0.55)6 .(0.45)1 = 0.0872

P(X=7) = (7!/(7!(7-7)!)).(0.55)7 .(0.45)0 = 0.0152

Thus

P(X≥1) = 0.0320+0.1172+0.2388+0.2918+0.2140+0.0872+0.0152 = 0.9963

Answer: The probability of selecting at least 1 boy= 0.9963

It would be a significant event if none of the next 7 births resulted in a boy if the P(no boy)≤ 0.05

P(no boy) = P(X=0)

where

P(X=0) = (7!/(0!(7-0)!)).(0.55)0 .(0.45)7 = 0.0037

Since the P(no boy) = P(X=0) = 0.0037<0.05, we conclude that it would be a significant event if none of the next 7 births resulted in a boy