Question

If atleast one child in a family with $3$ children is a boy then the probability that $2$ of the children are boys, is

• A. $\frac{3}{7}$
• B. $\frac{4}{7}$
• C. $\frac{1}{3}$
• D. $\frac{3}{8}$

Answer 1

The correct option is A $\frac{3}{7}$

total no of ways are 8 but given that at least 1 boy is present so 1 case is excluded in which all three are girls
so total no ways in which at least one boy is present is 7
total no ways in which 1 girl and 2 boys are present are 3 ways
so the probability is $\frac{3}{7}$