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Question

Mother, father and son line up at random for a family picture E: son on one end, F: father in middle

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Solution

It is given that when three persons Mother (M), father (F) and son (S) line up, then possible sample space is 6.

s={ MFS,MSF,FMS,FSM,SFM,SMF }.

Let E be the event that son is at one end,

E={ MFS,FMS,SFM,SMF } P( E )= 4 6 = 2 3

Let F be the event that father is in middle,

F={ SFM,MFS } P( F )= 2 6 = 1 3

The common outcome between E and F is,

EF={ MFS,SFM } P( EF )= 2 6 = 1 3

The probability P( E|F ) is calculated as,

P( E|F )= P( EF ) P( F ) = 1 3 1 3 =1

Therefore, the value of P( E|F ) is 1.


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