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).
Show Hint
This explains why the conditional probability is 1 (a certain event under the given condition).
Solution and Explanation
• Sample space \( S \) for \( n \) objects is \( n! \).
• Conditional Probability: \( P(E/F) = \frac{n(E \cap F)}{n(F)} \).
Step 1: List the sample space and identify event F
Let M = Mother, F = Father, S = Son.
Total arrangements = \( 3! = 6 \).
\( S = \{ MFS, MSF, FMS, FSM, SMF, SFM \} \).
Event F (Father in the middle):
\( F = \{ MFS, SFM \} \).
So, \( n(F) = 2 \).
Step 2: Identify event E and the intersection
Event E (Son on one end):
\( E = \{ SMF, SFM, MFS, FMS \} \).
The intersection \( E \cap F \) includes outcomes where Father is in the middle AND Son is on an end:
\( E \cap F = \{ MFS, SFM \} \).
So, \( n(E \cap F) = 2 \).
Step 3: Calculate conditional probability
\[ P(E/F) = \frac{n(E \cap F)}{n(F)} = \frac{2}{2} = 1 \]
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