A study found that 3% of children under 18 live with their father only. The question asks to calculate the probability that at most 2 children selected randomly from a group of 12 children live with their father only. The solution uses the binomial distribution formula to calculate the probability as 0.04380299, which is the probability that exactly 2 children live with their father only.

1. A study determined that 3% of children under 18 years of age live with their only. find the
probability that at most 2 persons selected at random from 12 children under 18 years of age
lived with their father only. The probability that 2 children lived with their father only is
_________ .
(Do not rounduntil the final answer. then round to the nearest thousandth as needed)
Solution
Given X~Binomial(n=12, p=0.03)
P(X=x)=xC12*(0.03^x)*(0.97^(12-x))
So the probability that 2 children lived with their father only is
P(X=2)=2C12*(0.03^2)*(0.97^(12-2)) =0.04380299