Question

Consider the following events for a family with children
$A=\left\{ofbothsexes\right\}$; $B=\left\{atmostoneboy\right\}$
In which of the following $\left(are/is\right)$ the events $A$ and $B$ are independent
(a)If a family has $3$ children (b) If a family has $2$ children
Assume that the birth of a boy or a girl is equally likely mutually exclusive and exhaustive

Answer 1

(A)

$P\left(A\right)=\frac{n\left(S\right)n\left(notbothsexes\right)}{n\left(S\right)}=\frac{2^3-2}{2^3}=\frac{3}{4}$

favourable outcomes for not both sexes=(GGG), (BBB)$

$P\left(B\right)=\frac{4}{8}=\frac{1}{2}\left[\right(GGG),(BGG),(GBG),(GGB\left)\right]$

$P(A\cap B)=\frac{3}{8}\left[\right(GBG),(BGG),(GGB\left)\right]$

$P\left(A\right).P\left(B\right)=\frac{3}{4}-\frac{1}{2}=\frac{3}{8}=P(A\cap B)\Rightarrow A$ & $B$ are independent

(B)

$P\left(A\right)=\frac{4-2}{4}=\frac{1}{2}$ [Non-favourable outcomes $\left(GG\right),\left(BB\right)$]

$P\left(B\right)=\frac{3}{4}\left[\right(GG),(GB),(BG\left)\right]$

$P(A\cap B)=\frac{2}{4}=\frac{1}{2}\left[\right(GB,\left(BG\right)]$

$P\left(A\right).P\left(B\right)=\frac{1}{2}\times \frac{3}{4}=\frac{3}{8}\ne P(A\cap B)\Rightarrow A$ & $B$ are not independent