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Question
Mother, father and son line up at random for a family picture. If A and B are two events given by A = Son on one end, B = Father in the middle, find P (A/B) and P (B/A).
Solution
Consider the given events.
A = Son standing on one end
B = Father standing in the middle
\[\text{ Clearly } , \]
\[S = \left\{ MFS, MSF, FSM, FMS, SMF, SFM \right\}\]
\[A = \left\{ MFS, FMS, SMF, SFM \right\}, \]
\[B = \left\{ MFS, SFM \right\}\]
\[\text{ Now } , \]
\[A \cap B = \left\{ MFS, SFM \right\} \]
\[\left( i \right) \text{ Required probability } = P\left( A/B \right) = \frac{n\left( A \cap B \right)}{n\left( B \right)} = \frac{2}{2} = 1\]
\[\left( ii \right) \text{ Required probability } = P\left( B/A \right) = \frac{n\left( A \cap B \right)}{n\left( A \right)} = \frac{2}{4} = \frac{1}{2}\]
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