Independent solution

How to solve this Conditional Probability question

Setup

Setup

Form the conditioning group F complement by subtracting the 312 records in F from the total of 937.

N(Fc)=937312=625N(F^c)=937-312=625

Model

Model

The numerator must count records in H but not F, so subtract the 102 records in both H and F from the 210 records in H.

N(HFc)=210102=108N(H\cap F^c)=210-102=108

Compute

Compute

There are 108 qualifying H records among 625 records in F complement; their ratio is 108/625=0.1728.

Pr(HFc)=108625=0.1728\Pr(H\mid F^c)=\frac{108}{625}=0.1728

Answer

Answer

Rounding the conditional probability to three decimals gives 0.173, which is choice B.

0.173(B)\boxed{0.173\quad\text{(B)}}