Question

$X$ has three children in his family. What is the probability of two or more boys in the family?

• A. $\frac{1}{8}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{3}{8}$

Answer 1

The correct option is B $\frac{1}{2}$

$X$ has $3$ children in his family.

All possible outcomes are if $G$ represents girl child and $B$ represent boy child, $\left\{\right(BBB),(BBG),(BGB),(BGG),(GBB),(GBG),(GGB),(GGG\left)\right\}$, hence $n\left(S\right)=8$

Out of which favorable outcomes for two or more boys are, $n\left(E\right)=4$

Hence, the probability that there are two or more boys in the family is $\frac{n\left(E\right)}{n\left(S\right)}=\frac{4}{8}=\frac{1}{2}$