Question

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 ti is known that the elder child is a female.

• A. $0.55,0.38$
• B. $0.33,0.50$
• C. $0.67,0.78,$
• D. $0.56,0.67$

Answer 1

The correct option is B $0.33,0.50$

If a couple has two children, then the sample space is
$S=\{(b,b),(b,g),(g,b),(g,g)\}$
(i) Let $E$ and $F$ respectively denote the events that both children are males and at least one of the children is a male.
$\therefore E\cap F=\left\{(b,b)\right\}\Rightarrow P(E\cap F)=\frac{1}{4}$
$P\left(E\right)=\frac{1}{4}$
$P\left(F\right)=\frac{3}{4}$
$\Rightarrow P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}=0.33$
(ii) Let $A$ and $B$ respectively denote the events that both children are females and the elder child is a female.
$A=\left\{(g,g)\right\}\Rightarrow P\left(A\right)=\frac{1}{4}$
$B=\{(g,b),(g,g)\}\Rightarrow P\left(B\right)=\frac{2}{4}$
$A\cap B=\left\{(g,g)\right\}\Rightarrow P(A\cap B)=\frac{1}{4}$
$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{\frac{1}{4}}{\frac{2}{4}}=\frac{1}{2}=0.50$