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Ea n dF are two independent events. The ...

`Ea n dF` are two independent events. The probability that both `Ea n dF` happen is `1/12` and the probability that neither `Ea n dF` happens is `1/2`. Then, `A) P(E)=1//3, P(F)=1//4` `B) P(E)=1//4, P(F)=1//3` `C) P(E)=1//6, P(F)=1//2` `D) P(E)=1//2, P(F)=1//6`

A

`P(E )=1//3,P(F)=1//4`

B

`P(E )=1//2,P(E )=1//6`

C

`P(E )=1//6,P(F )=1//2`

D

`P(E )=1//4,P(F)=1//3`

Text Solution

Verified by Experts

The correct Answer is:
A, B
Both E and F happen `rArr P(E nn F) = (1)/(12)`
and neither E nor F happens `rArr P(bar(E) nn bar(F)) = (1)/(2)`
But for independent events, we have
`P(E nn F) = P(E) P(F) = (1)/(12) " ….(i)`
and `P(bar(E) nn bar(F)) = P(bar(E)) P(bar(F))`
`= {1-P(E)} {1-P(F)}`
` = 1-P(E) -P(F) + P(E) P(F)`
`rArr (1)/(2) = 1-{P(E) + P(F)}+ (1)/(12)`
`rArr P(E) + P(F) = 1-(1)/(2) + (1)/(12) = (7)/(12) " "... (ii)`
On solving Eqs. (i) and (ii), we get
either `P(E) = (1)/(3) "and " P(F) = (1)/(4)`
`"or " P(E) = (1)/(4) "and" P(F) = (1)/(3)`
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