Question

In a family with $4$ children, the probability that there are at least two girls is

• A. $\frac{1}{2}$
• B. $\frac{9}{16}$
• C. $\frac{3}{4}$
• D. $\frac{11}{16}$

Answer 1

The correct option is D $\frac{11}{16}$

Each of the $4$ children can either be a boy or a girl. Therefore, total number of outcomes $=2^4=16$.

No. of outcomes such that atleast $2$ out of $4$ children are girls $={}^4{C}_2+{}^4{C}_3+{}^4{C}_4=6+4+1=11$

Therefore, probability $=\frac{\text{No. of favorable outcomes}}{\text{Total no. of outcomes}}=\frac{11}{16}$