Independent solution

How to solve this Bayes’ Theorem question

Setup

Setup

For each arrival class, multiply its prior share by its class-specific survival probability.

wC=0.10(0.60)=0.06w_C=0.10(0.60)=0.06
wS=0.30(0.90)=0.27w_S=0.30(0.90)=0.27
wT=0.60(0.99)=0.594w_T=0.60(0.99)=0.594

Model

Model

The three classes are mutually exclusive and exhaustive, so their survival weights sum to the overall survival probability.

Pr(survive)=0.06+0.27+0.594=0.924\Pr(\text{survive})=0.06+0.27+0.594=0.924

Compute

Compute

The serious-class survival weight is 0.27 and the total is 0.924; normalizing gives 0.27/0.924=0.292208.

Pr(Ssurvive)=0.27/0.924=0.292208\Pr(S\mid\text{survive})=0.27/0.924=0.292208

Answer

Answer

Among survivors, the serious-class probability rounds to 0.29, corresponding to choice B.

0.29(B)\boxed{0.29\quad\text{(B)}}