Independent solution

How to solve this Bayes' Theorem question

Setup

Setup

Let F identify the first-year group and let H denote at least one reported claim. The complementary group has prior probability 0.85.

Pr⁡(F)=0.15,Pr⁡(Fc)=0.85\Pr(F)=0.15,\qquad \Pr(F^c)=0.85

Model

Model

For a Poisson count, complement the zero-count probability to obtain the likelihood of the observed event.

Pr⁡(H∣F)=1−e−0.50=0.3934693403…\Pr(H\mid F)=1-e^{-0.50}=0.3934693403\ldots
Pr⁡(H∣Fc)=1−e−0.20=0.1812692469…\Pr(H\mid F^c)=1-e^{-0.20}=0.1812692469\ldots

Compute

Compute

Weight the two likelihoods by their prior group probabilities and normalize.

Pr⁡(F∣H)=0.15(1−e−0.50)0.15(1−e−0.50)+0.85(1−e−0.20)\Pr(F\mid H)=\frac{0.15(1-e^{-0.50})}{0.15(1-e^{-0.50})+0.85(1-e^{-0.20})}
=0.2769620166…=0.2769620166\ldots

Answer

Answer

The conditional probability rounds to 0.277.

0.277(B)\boxed{0.277\quad\text{(B)}}