Q.Determine $P(E|F)$. Mother, father and son line up at random for a family picture. $E$ : son on one end, $F$ : father in middle.

Condition on $F$ (father in the middle) and count the arrangements in which $E$ (son on an end) also holds.

With the father fixed in the centre, the two ends are filled by the mother and the son in $2!=2$ ways:

$(S,F,M)\quad\text{and}\quad(M,F,S).$ …

Set up

Three distinct people (Mother, Father, Son) line up in $3!=6$ equally likely orders. Let

• $E$: the son is on one of the two ends,
• $F$: the father is in the middle.

Event F

Fix the father in the centre position. The two end positions are filled by the mother and the son in $2!=2$ ways:

Event E and F together …

Method: Conditional probability on arrangements (permutation sample space)

Use this when distinct objects or people are placed in order at random and you must find $P(E\mid F)$ for events about positions ("on an end", "in the middle", "adjacent").

Steps

Step 1: Size the sample space as a permutation count.

Arranging $n$ distinct items in a row gives $n!$ equally likely orders (for three people, $3!=6$). Equally-likely outcomes let you count:

Step 2: Count $n(F)$ by fixing the positions $F$ constrains.

Lock each person that $F$ places, then the remaining people fill the remaining seats in (remaining)! ways. Fixing one person in a chosen seat leaves $2!=2$ arrangements of the other two.

Step 3: Impose $E$ on the arrangements already in $F$. …

Common Mistakes

Mistake 1: Dividing by $6$ (all line-ups) instead of by $2$.

Why it's wrong: once $F$ (father in the middle) is given, only the $2!=2$ arrangements $(S,F,M)$ and $(M,F,S)$ remain, so the denominator is $2$, not the full $6$. Correct approach: $P(E|F)=\frac{|E\cap F|}{|F|}=\frac{2}{2}=1$.

Mistake 2: Doubting that both remaining arrangements put the son on an end. …