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:

Whenever the sample space size is small (like 6 elements here), explicitly writing out the sets makes it visually obvious to find intersections and eliminates any logical counting mistakes.

Answer 1

Concept: This problem can be resolved using the concept of conditional probability. The conditional probability of an event \(E\) given that event \(F\) has already occurred is denoted as \(P(E/F)\) and is defined by the formula: \[ P(E/F) = \frac{P(E \cap F)}{P(F)} = \frac{n(E \cap F)}{n(F)} \] where \(n(F)\) represents the number of sample points in event \(F\), and \(n(E \cap F)\) represents the number of sample points common to both event \(E\) and event \(F\).

Step 1: Write down the complete sample space (\(S\)).
Let Mother be denoted by \(M\), Father by \(F_0\), and Son by \(S_0\).
The total number of ways these 3 individuals can line up in a row is given by \(3! = 3 \times 2 \times 1 = 6\) ways. Let us explicitly list all 6 arrangements in the sample space \(S\): \[ S = \{ (M, F_0, S_0), \, (M, S_0, F_0), \, (F_0, M, S_0), \, (F_0, S_0, M), \, (S_0, M, F_0), \, (S_0, F_0, M) \} \] So, total number of elements in sample space, \(n(S) = 6\).

Step 2: Identify the outcomes for event \(F\) (Father in the middle).
Let us inspect the sample space \(S\) and pick out all the ordered arrangements where the Father (\(F_0\)) occupies the second (middle) position:
• \((M, F_0, S_0)\) - Father is in the middle.
• \((S_0, F_0, M)\) - Father is in the middle. Therefore, the event set \(F\) is: \[ F = \{ (M, F_0, S_0), \, (S_0, F_0, M) \} \] The number of favorable outcomes for event \(F\) is \(n(F) = 2\).

Step 3: Identify the outcomes for event \(E\) (Son on one end) and determine \(E \cap F\).
Event \(E\) requires the Son (\(S_0\)) to be either at the absolute left end or the absolute right end.
Let us examine the elements that belong to event set \(F\) to check which of them also satisfy the condition for event \(E\):
• For the arrangement \((M, F_0, S_0)\): The Son (\(S_0\)) is at the right end. Thus, this element satisfies event \(E\).
• For the arrangement \((S_0, F_0, M)\): The Son (\(S_0\)) is at the left end. Thus, this element also satisfies event \(E\). Since both elements of \(F\) have the Son standing at one of the ends, the intersection set \(E \cap F\) is identical to the set \(F\): \[ E \cap F = \{ (M, F_0, S_0), \, (S_0, F_0, M) \} \] Hence, the number of elements in the intersection is \(n(E \cap F) = 2\).

Step 4: Compute the conditional probability \(P(E/F)\).
Using the definition of conditional probability: \[ P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1 \] This signifies that if the Father is standing in the middle, it is absolutely certain (a sure event) that the Son will be positioned at one of the two ends.