Independent solution

How to solve this Independence question

Setup

Setup

Combine the mutually exclusive life-policy choices into one event L and denote the health-policy event by H.

Pr⁡(H)=0.45,Pr⁡(H∩Lc)=0.18\Pr(H)=0.45,\qquad \Pr(H\cap L^c)=0.18

Model

Model

Use independence to recover the probability of wanting some life policy.

0.18=Pr⁡(H)Pr⁡(Lc)=0.45Pr⁡(Lc)0.18=\Pr(H)\Pr(L^c)=0.45\Pr(L^c)
Pr⁡(Lc)=0.40,Pr⁡(L)=0.60\Pr(L^c)=0.40,\qquad \Pr(L)=0.60

Compute

Compute

Exactly one policy is either health only or one life policy without health.

Pr⁡(L∩Hc)=0.60(0.55)=0.33\Pr(L\cap H^c)=0.60(0.55)=0.33
Pr⁡(exactly one policy)=0.33+0.18=0.51\Pr(\text{exactly one policy})=0.33+0.18=0.51

Answer

Answer

The exactly-one probability is 0.51.

0.51(A)\boxed{0.51\quad\text{(A)}}