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Question: Let *E* and *F* be two independent events. The probability that both *E* and *F* happens is \(\frac ...

Let E and F be two independent events. The probability that both E and F happens is 112\frac { 1 } { 12 } and the probability that neither E nor F happens is 12\frac { 1 } { 2 }, then

A

P(E)=13,P(F)=14P ( E ) = \frac { 1 } { 3 } , P ( F ) = \frac { 1 } { 4 }

B

P(E)=12,P(F)=16P ( E ) = \frac { 1 } { 2 } , P ( F ) = \frac { 1 } { 6 }

C

P(E)=16,P(F)=12P ( E ) = \frac { 1 } { 6 } , P ( F ) = \frac { 1 } { 2 }

D

None of these

Answer

P(E)=13,P(F)=14P ( E ) = \frac { 1 } { 3 } , P ( F ) = \frac { 1 } { 4 }

Explanation

Solution

We are given P(E∩F)=112P ( E \cap F ) = \frac { 1 } { 12 } and P(Eˉ∩Fˉ)=12P ( \bar { E } \cap \bar { F } ) = \frac { 1 } { 2 }

Ž …..(i)

and P(Eˉ)⋅P(Fˉ)=12P ( \bar { E } ) \cdot P ( \bar { F } ) = \frac { 1 } { 2 } …..(ii)

Ž {1−P(E)}{(1−P(F)}=12\{ 1 - P ( E ) \} \left\{ ( 1 - P ( F ) \} = \frac { 1 } { 2 } \right. Ž 1+P(E)P(F)−P(E)−P(F)=121 + P ( E ) P ( F ) - P ( E ) - P ( F ) = \frac { 1 } { 2 } Ž 1+112−[P(E)+P(F)]=121 + \frac { 1 } { 12 } - [ P ( E ) + P ( F ) ] = \frac { 1 } { 2 } Ž P(E)+P(F)=712P ( E ) + P ( F ) = \frac { 7 } { 12 } …..(iii)

On solving (i) and (iii), we get P(E)=13,14P ( E ) = \frac { 1 } { 3 } , \frac { 1 } { 4 } and P(F)=14,13P ( F ) = \frac { 1 } { 4 } , \frac { 1 } { 3 } .