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

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

Solution:

We assume that $\mathrm{M}$ denote mother, $\mathrm{F}$ denote father and $\mathrm{S}$ denote son.

Then, the sample space for the given experiment will be:

S= ${$ MFS, SFM, FSM, MSF, SMF, FMS}

The events given are E: Son on one end

& F: Father in middle
Sample space of the following are:

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

$\Rightarrow \mathrm{E} \cap \mathrm{F}={\mathrm{MFS}, \mathrm{SFM}}$

The probability of the event,

$P(E)=\frac{4}{6}=\frac{2}{3}, P(F)=\frac{2}{6}=\frac{1}{3}, P(E \cap F)=\frac{2}{6}=\frac{1}{3}$

Now, we know that by definition of conditional probability,

$P(E \mid F)=\frac{P(E \cap F)}{P(F)}$

Now by substituting the values we get

$\Rightarrow \mathrm{P}(\mathrm{E} \mid \mathrm{F})=\frac{1 / 3}{1 / 3}=1$

$\Rightarrow P(E \mid F)=1$

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