SOLUTION: 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) = ___________

Algebra ->  Probability-and-statistics -> SOLUTION: 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) = ___________       Log On


   

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) = ____________

Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
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) = ____________
----------
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.