Question 92276
p(E1)=0.25, P(E2)=0.75, P(F|E1)=0.05, P(F|E2)=0.12
E1 and E2 are mutually exclusive.

P(E2|F) = ____________

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P(E2|F) = [P(E2 and F)]/[P(F} 
=[P(F|E2)*P(E2)]/[P(F|E1)+P(F|E2)]

=[0.12*0.75]/[0.05+0.12]

= 0.09/0.17

= 0.5294....

Cheers,
Stan H.