Question

A father has 3 children with atleast one boy. The probability that he has 2 boys and one girl is:

• A. $\frac{1}{4}$
• B. $\frac{3}{7}$
• C. $\frac{1}{3}$
• D. $None$

Answer 1

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

Let A be the event that father has a least one boy and B be the event that he has 2 boys and one girl. Then
we have to find
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}$
P(A) = P (1 boy, 2girls) + P (2boys, 1girl) + P (3boys, no girl)
${=}^3{C}_1{\left(\frac{1}{2}\right)}^1{\left(\frac{1}{2}\right)}^2{+}^3{C}_2{\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^1{+}^3{C}_2{\left(\frac{1}{2}\right)}^3$
$=\frac{7}{8}$
and $P(A\cap B)=P(2boys,1girl)=\frac{3}{8}$.
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}$
$=\frac{3}{7}$
.