Question

Assume that each child born is equaly likely to be boy or a girl. IF a family has two children, what is the conditional probability that both are girls given that
the youngest is a girl?

atleast one is a girl?

Answer 1

Let B and g represent th boy and the girl child, respectively, If a family has two children, the sample space will be
S = (bb, bg, gb, gg)
Which contains four eqally likely sample points i.e. n (S) = 4
Let E : both hcildren are girls then E $=\left\{gg\right\}\Rightarrow n\left(E\right)=1$
Let F: the yungest is a girl, then $F=\{bg,gg\}\Rightarrow n\left(F\right)=2\Rightarrow E\cap F=\left\{gg\right\}\Rightarrow n(E\cap F)=1P\left(E\right)=\frac{1}{4},P\left(F\right)\frac{2}{4}=\frac{1}{2}andP(E\cap F)=\frac{1}{4}\therefore Requiredprobability=P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{P(E\cap F)}{P\left(F\right)}=\frac{P\left(E\right)}{P\left(F\right)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$

Let B and g represent th boy and the girl child, respectively, If a family has two children, the sample space will be
S = (bb, bg, gb, gg)
Which contains four eqally likely sample points i.e. n (S) = 4
Let E : both hcildren are girls then E $=\left\{gg\right\}\Rightarrow n\left(E\right)=1$
Let F: atleast one is a girl, then $F=\{bg,gb,gg\}\Rightarrow E\cap F=\left(gg\right)=E\Rightarrow n\left(F\right)=3,(E\cap F)=1\therefore Requiredprobability=P\left(\frac{E}{F}\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{P\left(E\right)}{P\left(F\right)}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}$