Answer to Question #304787 in Statistics and Probability for haylie

Question

1. In a certain country, the true probability of baby being a girl is 0.45.  
2. 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?

Answer 1

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)!)).p^x.(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)!)).p^x .(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