Question

Suppose that a family has exactly n children (n ≥ 2). Assume that the probability that any child will be a girl is 1/2 and that all births are independent. Given that the family has at least one girl, determine the probability that the family has at least one boy.

Short answer

The required probability is

\(P\left( {B \ge 1\left| {G \ge 1} \right.} \right) = \frac{{1 - {\textstyle{1 \over {{2^{n - 1}}}}}}}{{1 - {\textstyle{1 \over {{2^n}}}}}}\).

Step-by-step solution

Given information

The family has exactly n children

The probability of any child being a girl\(P\left( G \right) = \frac{1}{2}\)

All births are independent events

The family has at least one girl

compute the probability

The probability that the family has at least one boy given that it has atleast one girl is given by

\(1 - probability\,of\,having\,no\,boys\)

It is denoted by,

\(P\left( {B \ge 1\left| {G \ge 1} \right.} \right) = 1 - P\left( {B = 0\left| {G \ge 1} \right.} \right)\)

Now,

\(\begin{aligned}{}P\left( {G \ge 1} \right) = 1 - P\left( {G = 0} \right)\\ = 1 - \frac{1}{{{2^n}}}\end{aligned}\)

Therefore,

\(\begin{aligned}{}1 - P\left( {B = 0\left| {G \ge 1} \right.} \right)& = 1 - \frac{{P\left( {B = 0\left| {G \ge 1} \right.} \right)P\left( {G \ge 1} \right)}}{{P\left( {G \ge 1} \right)}}\\& = 1 - \frac{{P\left( {B = 0 \cap G \ge 1} \right)}}{{P\left( {G \ge 1} \right)}}\\ &= 1 - \frac{{{\textstyle{1 \over {{2^n}}}}}}{{1 - {\textstyle{1 \over {{2^n}}}}}}\\ &= \frac{{1 - {\textstyle{1 \over {{2^{n - 1}}}}}}}{{1 - {\textstyle{1 \over {{2^n}}}}}}\end{aligned}\)

The required probability is

\(P\left( {B \ge 1\left| {G \ge 1} \right.} \right) = \frac{{1 - {\textstyle{1 \over {{2^{n - 1}}}}}}}{{1 - {\textstyle{1 \over {{2^n}}}}}}\).