Independent solution

How to solve this Independence question

Setup

Setup

Split the target into the two mutually exclusive ways that exactly one event can occur.

{exactly one}=(EFc)(EcF)\{\text{exactly one}\}=(E\cap F^c)\cup(E^c\cap F)

Model

Model

Use independence for each event-complement pair.

Pr(EFc)=Pr(E)Pr(Fc)\Pr(E\cap F^c)=\Pr(E)\Pr(F^c)
Pr(EcF)=Pr(Ec)Pr(F)\Pr(E^c\cap F)=\Pr(E^c)\Pr(F)

Compute

Compute

Evaluate the two disjoint probabilities and add them.

Pr(EFc)=0.84(0.35)=0.294\Pr(E\cap F^c)=0.84(0.35)=0.294
Pr(EcF)=0.16(0.65)=0.104\Pr(E^c\cap F)=0.16(0.65)=0.104
Pr(exactly one)=0.294+0.104=0.398\Pr(\text{exactly one})=0.294+0.104=0.398

Answer

Answer

The exactly-one probability is 0.398.

0.398(B)\boxed{0.398\quad\text{(B)}}