Question

A family has six children. The probability that there are fewer boys than girls, if the probability of any particular child being a boy is $\frac{1}{2}$ is

• A. $\frac{5}{32}$
• B. $\frac{7}{32}$
• C. $\frac{11}{32}$
• D. $\frac{9}{32}$

Answer 1

Answer: (C)

$\frac{11}{32}$

$P\left(B\right)=P\left(G\right)=\frac{1}{2}$
Now the probability of having fewer boys in a family of 6 children is
${}^6{C}_4\left(P\right(G\left){)}^4\right(P\left(B\right){)}^2+{}^6{C}_5\left(P\right(G\left){)}^5\right(P\left(B\right){)}^1+{}^6{C}_6\left(P\right(G\left){)}^6\right(P\left(B\right){)}^0$
$=\frac{1}{2^6}[15+6+1]$

$=\frac{22}{64}$

$=\frac{11}{32}$