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A couple has two children,
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
(i) Find the probability that both children are males, if it is known that at least one of the children is male.
(ii) Find the probability that both children are females, if it is known that the elder child is a female.
Solution:
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Verified Answer
(i) Let the events A and B are denoted as
$\mathrm{A}=$ Both children are male i.e., $\{\mathrm{M}, \mathrm{M}\}$
$\mathrm{B}=\mathrm{At}$ least one child is male
i.e., $\{\mathrm{MF}, \mathrm{FM}, \mathrm{MM}\}$
$\mathrm{A} \cap \mathrm{B}=\{\mathrm{M}, \mathrm{M}\}$
$P(A)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$, $P(B)=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$
$\begin{aligned} & P(A \cap B)=\frac{1}{4} \\ \therefore \quad & P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{3}}{\frac{3}{4}}=\frac{1}{3} \end{aligned}$
(ii) $\mathrm{A}=$ Both the children are female $=\{\mathrm{FF}\}$
$P(A)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$
$\mathrm{B}=$ Elder child is a female $=\{\mathrm{FF}, \mathrm{FM}\}$
$\mathrm{P}(\mathrm{B})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}, \quad \mathrm{~A} \cap \mathrm{B}=\{\mathrm{FF}\}$
$\mathrm{P}=(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$
$\mathrm{P}$ (Both the children are female if elder child is female)
$$
=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}
$$
$\mathrm{A}=$ Both children are male i.e., $\{\mathrm{M}, \mathrm{M}\}$
$\mathrm{B}=\mathrm{At}$ least one child is male
i.e., $\{\mathrm{MF}, \mathrm{FM}, \mathrm{MM}\}$
$\mathrm{A} \cap \mathrm{B}=\{\mathrm{M}, \mathrm{M}\}$
$P(A)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$, $P(B)=\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}$
$\begin{aligned} & P(A \cap B)=\frac{1}{4} \\ \therefore \quad & P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{3}}{\frac{3}{4}}=\frac{1}{3} \end{aligned}$
(ii) $\mathrm{A}=$ Both the children are female $=\{\mathrm{FF}\}$
$P(A)=\frac{1}{2} \times \frac{1}{2}=\frac{1}{4}$
$\mathrm{B}=$ Elder child is a female $=\{\mathrm{FF}, \mathrm{FM}\}$
$\mathrm{P}(\mathrm{B})=\frac{1}{4}+\frac{1}{4}=\frac{1}{2}, \quad \mathrm{~A} \cap \mathrm{B}=\{\mathrm{FF}\}$
$\mathrm{P}=(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$
$\mathrm{P}$ (Both the children are female if elder child is female)
$$
=P(A / B)=\frac{P(A \cap B)}{P(B)}=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{2}
$$
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