Independent solution

How to solve this Set Probability question

Setup

Setup

Liability claims occur with probability 0.04, property claims with probability 0.10, and liability without property with probability 0.01.

P(L)=0.04,P(P)=0.10P(L)=0.04,\quad P(P)=0.10
P(L∩Pc)=0.01P(L\cap P^c)=0.01

Model

Model

Subtract the liability-only probability from the liability marginal to recover the intersection probability 0.03.

P(L∩P)=0.04−0.01=0.03P(L\cap P)=0.04-0.01=0.03

Compute

Compute

Inclusion–exclusion gives union probability 0.11. Its complement, the probability of neither claim type, is 0.89.

P(L∪P)=0.04+0.10−0.03=0.11P(L\cup P)=0.04+0.10-0.03=0.11
P(Lc∩Pc)=1−0.11=0.89P(L^c\cap P^c)=1-0.11=0.89

Answer

Answer

The probability of neither claim type is 0.890, corresponding to choice E.

0.890(E)\boxed{0.890\quad\text{(E)}}