Let X be the number of boys in n selected births. Let p be the probability of getting baby boy on selected birth.
Here n=10. Also the male and female births are equally likely it means chance of baby boy or girl is 1/2
P(Boy) = P(girl) =0.5
p =0.5
From given information we have n =10 fixed number of trials, p is probability of success which is constant for each trial . And each trial is independent of each other.
So X follows Binomial distribution with n=10 and p=0.5
The probability function of Binomial distribution for k number of success, x=k is given as
P(X=k) = [tex] (10Ck) 0.5^{k} (1-0.5)^{10-k} [/tex]
We have to find probability of getting 8 boys in n=10 births
P(X=8) = [tex] (10C8) 0.5^{8} (1-0.5)^{10-8} [/tex]
= 45 * 0.0039 * 0.25
P(X = 8) = 0.0438
The probability of getting exactly 8 boys in selected 10 births is 0.044