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Home » Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?
CBSE | Maths | NCERT | Probability
Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is a girl?

Solution

Assume B denotes boy and G denote girl.

Then, the sample space of the given experiment is S={G G, G B, B G, B B} Let \mathrm{E} be the event that ‘both are girls’.

\Rightarrow \mathrm{E}={\mathrm{GG}}
\Rightarrow \mathrm{P}(\mathrm{E})=\frac{1}{4}

(i) Let \mathrm{F} be the event that ‘the youngest is a girl’.

\Rightarrow \mathrm{F}={\mathrm{GG}, \mathrm{BG}}

\Rightarrow \mathrm{P}(\mathrm{F})=\frac{2}{4}=\frac{1}{2}

Now, E \cap F={G G}

\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{4}

Now, we know that by relation of conditional probability,

P(E \mid F)=\frac{P(E \cap F)}{P(F)}

\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1 / 4}{1 / 2}=\frac{2}{4}=\frac{1}{2}[\mathrm{Using}(\mathrm{i}) and (ii)]
\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1}{2}

(ii) Assume \mathrm{H} be the event that ‘at least one is a girl’.

\Rightarrow \mathrm{H}={\mathrm{GG}, \mathrm{GB}, \mathrm{BG}}

\Rightarrow \mathrm{P}(\mathrm{H})=\frac{3}{4}

Now, E \cap H={G G}
\Rightarrow \mathrm{P}(\mathrm{E} \cap \mathrm{H})=\frac{1}{4}

Now, we know that by relation of conditional probability,

P(E \mid F)=\frac{P(E \cap F)}{P(F)}

    \[\begin{aligned}&\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{H})=\frac{\mathrm{P}(\mathrm{E} n \mathrm{H})}{\mathrm{P}(\mathrm{H})}=\frac{1 / 4}{3 / 4}=\frac{1}{3} &\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{H})=\frac{1}{3}\end{aligned}\]

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