Question

If birth to male child and birth to female child are equal probable, then what is the probability that atleast one of the three children born to a couple is male?

• A. $\raisebox{1ex}{$4$}\!\left/ \!\raisebox{-1ex}{$5$}\right.$
• B. $\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$
• C. $8/7$
• D. $1/7$

Answer 1

The correct option is B

$\raisebox{1ex}{$7$}\!\left/ \!\raisebox{-1ex}{$8$}\right.$

Explanation for the correct option:

Step 1. Find the required probability:

Given,

total number of children $n=3$

Let $"p"$ be the probability of male child $=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

Hence, $\begin{array}{rcl}q& =& 1-\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\\ & =& \left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)\end{array}$

($q$ be the probability of female child)

Step 2. By using binomial distribution,

$\Rightarrow f\left(x\right)={}^{\mathrm{n}}c_{\mathrm{x}}{p}^{\mathrm{x}}{q}^{\mathrm{n}-\mathrm{x}}$

$={}^{3}c_x{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{\mathrm{x}}{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{3-\mathrm{x}}$

Now, Put $x=1$

$\begin{array}{rcl}& \Rightarrow & f\left(1\right)={}^{3}c_1\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{3-1}\\ & =& {}^{3}c_1\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2\end{array}$

Put $x=2$

$\begin{array}{rcl}& \Rightarrow & f\left(2\right)={}^{3}c_2{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{3-2}\\ & =& {}^{3}c_2{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^1\end{array}$

Put $x=3$

$\begin{array}{rcl}& \Rightarrow & f\left(3\right)={}^{3}c_3{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^3{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^{3-3}\\ & =& {}^{3}c_3{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^3\end{array}$

Step 3. Probability that a couple get at least one of the three child is male

$$\begin{gathered} =f\left(1\right)+f\left(2\right)+f\left(3\right) \\ ={}^{3}c_1\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2+{}^{3}c_2{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^2{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^1+{}^{3}c_3{\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}^3 \\ =\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)+\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right)+\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$8$}\right.\right) \\ =\mathbf{\left(}\raisebox{1ex}{$\mathbf{7}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{8}$}\right.\mathbf{\right)} \end{gathered}$$

Hence, option (B) is correct.