Mother,father and son line up at random for a family photo.If A and B are two events given by A=Son on one end,  B=Father in the middle,find P\left ( \frac{B}{A} \right ).

 

 

 

 
 
 
 
 

Answers (1)

given
A: son on one end,   B:Father is the middle &= Sample space.
If mother(M),father(F) and son(s)'line up for the family picture,then the sample space.will be
S_{1}= \left \{ MPS,MSF, FMS,FSM,SMF,SFM\right \}
A= \left \{ MFS,FMS,SMF,SFM \right \}
P\left ( A\cap B \right )= \frac{2}{6}= \frac{1}{3}\: \:\: P\left ( B \right )= \frac{2}{6}= \frac{1}{3}
P\left ( A \right )= \frac{4}{6}= \frac{2}{3}
P\left ( \frac{B}{A} \right )= \frac{P\left ( A\cap B \right )}{P\left ( A \right )}= \frac{\frac{1}{\not{3}}}{\frac{2}{\not{3}}}= \frac{1}{2}

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