Question

Mother, Father and Son line up at random for a family picture. Let events \(E\): Son on one end and \(F\): Father in the middle. Find \(P(E/F)\).

Hint:

Conditional probability is calculated using the formula \(P(A/B) = \frac{P(A \cap B)}{P(B)}\). It represents the probability of event \(A\) occurring given that event \(B\) has already occurred.

Answer 1

Step 1: Understand the total arrangements.
There are three persons: Mother (M), Father (F), and Son (S).
The total number of possible arrangements is \[ 3! = 6 \] The arrangements are \[ MFS,\; MSF,\; FMS,\; FSM,\; SMF,\; SFM \] Step 2: Define event $F$.
Event \(F\): Father is in the middle.
Possible arrangements with Father in the middle are \[ MFS,\; SFM \] Thus, \[ n(F) = 2 \] Step 3: Define event $E$.
Event \(E\): Son is at one end (first or last position).
Now check arrangements where Father is in the middle and Son is on an end.
The valid arrangements are \[ MFS,\; SFM \] In both cases, Son is on one end.
Thus, \[ n(E \cap F) = 2 \] Step 4: Apply conditional probability formula.
\[ P(E/F) = \frac{P(E \cap F)}{P(F)} \] \[ = \frac{2/6}{2/6} \] \[ = 1 \] Step 5: Final conclusion.
Whenever the Father stands in the middle position, the remaining two positions are automatically occupied by Mother and Son at the ends. Therefore, the Son must be on one of the ends.
Final Answer:
\[ \boxed{P(E/F) = 1} \]