Question

Mother, father and son line up at random for a family picture
$E:$ son on one end,

$F:$ father in middle.
Determine $P\left(E|F\right)$

Answer 1

If mother $\left(M\right)$, father $\left(F\right)$ and son $\left(S\right)$ line up for a family picture, then the sample space will be
$S=\{MFS,MSF,FMS,FSM,SMF,SFM\}$

$\Rightarrow E=\{MFS,FMS,SMF,SFM\}$

$F=\{MFS,SFM\}$

$\therefore E\cap F=\{MFS,SFM\}$

$P(E\cap F)=\frac{2}{6}=\frac{1}{3}$

$P\left(F\right)=\frac{2}{6}=\frac{1}{3}$

$\therefore P\left(E|F\right)=\frac{P(E\cap F)}{P\left(F\right)}=\frac{\frac{1}{3}}{\frac{1}{3}}=1$