Question

A father has three children with at least one boy. The probability that he has two boys and one girl is

• A. $\frac{1}{4}$
• B. $\frac{3}{7}$
• C. $\frac{1}{3}$
• D. $\frac{2}{5}$

Answer 1

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

Let A be the event that father has at least one boy and B be the event that he has 2boys and one girl.
$P\left(A\right)=P(1boy,2girl)+P(2boy,1girl)+P(3boy,nogirl)=\frac{3C_1+3C_2+3C_3}{2^3}=\frac{7}{8}$
$P(A\bigcap B)=P(2boy,1girl)=\frac{3}{8}$
$HenceP\left(\frac{B}{A}\right)=\frac{P(A\bigcap B)}{P\left(A\right)}=\frac{3}{7}.$