Question

Mother, father and son line up at random for a family picture. $E$ is the event of son on one end, $F$ is the event of father in the middle. Then $P\left(\frac{E}{F}\right)$ is equal to:

Answer 1

Let $M$ denotes mother and $F$ denotes father snd $S$ denotes son.
Here $E$: Son on one end
And $F$: Father in the middle
$\Rightarrow E=\{MFS,SFM,SMF,FMS\}$ and $F=\{MFS,SFM\}$
$\Rightarrow E\cap F=\{MFS,SFM\}$
The total elements in the sample space is $6$
So, $P\left(E\right)=\frac{4}{6}=\frac{2}{3},P\left(F\right)=\frac{2}{6}=\frac{1}{3},P(E\cap F)=\frac{2}{6}=\frac{1}{3}$
Now, we know that by definition of conditional probability,
$P\left(E/F\right)=\frac{P(E\cap F)}{P\left(F\right)}$
Now by substituting the values we get
$\Rightarrow P\left(E/F\right)=\frac{1/3}{1/3}=1$