Step 1 — Write the sample space.

Denote Mother by M, Father by F and Son by S. All possible line-ups of three people:

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

$$n(S) = 3! = 6$$

Step 2 — List the outcomes in event F (Father in the middle).

Scanning the sample space for arrangements with Father in position 2:

$$F = \{\text{MFS},\ \text{SFM}\}$$

$$n(F) = 2 \quad \Longrightarrow \quad P(F) = \frac{2}{6} = \frac{1}{3}$$

Step 3 — List the outcomes in $E \cap F$ (Son on one end AND Father in the middle).

• MFS : Son is at the right end ✓
• SFM : Son is at the left end ✓

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

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

Step 4 — Apply the conditional probability formula.

$$P(E|F) = \frac{P(E \cap F)}{P(F)} = \frac{1/3}{1/3}$$

$$\boxed{P(E|F) = 1}$$

Step 5 — Interpret.

This makes sense: once the Father occupies the middle position, only Mother and Son can take the two end positions, so the Son is certain to be at one end.