Let E and F...

Question

Let E and F be events such that Pr[E]=0.5, Pr[F]= 0.53, and Pr[E intersect F]=0.24. Find.

 

Pr(E l F complement)

 

 

Pr(F l E complement)

Answer 1

1. P(E|F')=P(EF')/P(F')                        definition of conditional probability

 

2. P(EF')=P(E)-P(EF)=.5-.24=.26         since P(EF)+P(EF')=P(E)

 

3. P(F')=1-P(F)=1-.53=.47                  since P(F)+P(F')=1

 

therefore from line 1: P(E|F') = .26/.47=.553

Answer 2

Shelby,

 

Pr(E)=Pr(E | F') + P (E intersect F)

.5=Pr(E | F') + .24

Pr (E | F')=.26

 

Pr(F) = Pr(E' | F) + Pr (E intersect F)

.53=Pr(E' | F) + .24

Pr (E' | F)=.29

 

Hope that helps!