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 $\dfrac{1}{2}$ is
a)$\dfrac{5}{32}$
b)$\dfrac{7}{32}$
c)$\dfrac{11}{32}$
d)$\dfrac{9}{32}$

Answer 1

Hint: As we know that the family has six children. We need to find the probability that the boys are fewer than girls. So, that means, either number of boys could be 0, 1, or 2. Also, we know that the probability of any particular child being a boy is $\dfrac{1}{2}$. So, find the probability if there are 0, 1, or 2 boys in the family. Then, add all the values which is the probability that there are fewer boys than girls in the family.

Complete step-by-step solution
We know that there are six children in the family.
Also, the probability of any particular child being a boy is $\dfrac{1}{2}$.
We need to find the probability if there are 0, 1, or 2 boys in the family.
So, by using the formula: $P(E)={}^{n}{{C}_{r}}\times {{\left( \text{probability of favorable outcome} \right)}^{n}}$ , where, n = 6, r = 0, 1 and 2 and probability of favorable outcome, i.e. probability of any particular child being boy is $\dfrac{1}{2}$.
We get:
Probability of having 0 boy child:
\[\begin{align}
  & P(0)={}^{6}{{C}_{0}}\times {{\left( \dfrac{1}{2} \right)}^{6}} \\
 & =\dfrac{6!}{0!\times 6!}\times \dfrac{1}{64} \\
 & =\dfrac{1}{64}......(1)
\end{align}\]
Probability of having 1 boy child:
\[\begin{align}
  & P(1)={}^{6}{{C}_{1}}\times {{\left( \dfrac{1}{2} \right)}^{6}} \\
 & =\dfrac{6!}{1!\times 6!}\times \dfrac{1}{64} \\
 & =\dfrac{6}{64}......(2)
\end{align}\]
Probability of having 2 boy children:
\[\begin{align}
  & P(2)={}^{6}{{C}_{2}}\times {{\left( \dfrac{1}{2} \right)}^{6}} \\
 & =\dfrac{6!}{2!\times 4!}\times \dfrac{1}{64} \\
 & =\dfrac{15}{64}......(3)
\end{align}\]
So, the probability of having fewer boys than girls in a family with 6 children is:
$\begin{align}
  & P(E)=\dfrac{1}{64}+\dfrac{6}{64}+\dfrac{15}{64} \\
 & =\dfrac{22}{64} \\
 & =\dfrac{11}{32}
\end{align}$
Hence, option (c) is the correct answer.

Note: It is mentioned that the probability of boys is fewer than girls. So, we took the number of boys as 0, 1, or 2. But it is not mentioned that the probability of boys could be equal to the girl child. So, we cannot take the number of boy children as 3.