Independent solution

How to solve this Bayes' Theorem question

Setup

Setup

Let S denote the first class and N the second. Their prior probabilities are 0.40 and 0.60, and the second class has Poisson mean three.

Pr⁡(S)=0.40,Pr⁡(N)=0.60\Pr(S)=0.40,\qquad \Pr(N)=0.60
λN=3\lambda_N=3

Model

Model

Translate the zero-count comparison into an equation for the unknown mean of the first class.

e−λS=3e−3e^{-\lambda_S}=3e^{-3}
λS=3−ln⁡3=1.9013877113…\lambda_S=3-\ln 3=1.9013877113\ldots

Compute

Compute

Evaluate the two likelihoods at count two and apply Bayes' theorem.

Pr⁡(X=2∣S)=e−λSλS22!=0.2699909325…\Pr(X=2\mid S)=e^{-\lambda_S}\frac{\lambda_S^2}{2!}=0.2699909325\ldots
Pr⁡(X=2∣N)=e−3322!=0.2240418077…\Pr(X=2\mid N)=e^{-3}\frac{3^2}{2!}=0.2240418077\ldots
Pr⁡(S∣X=2)=0.40(0.2699909325)0.40(0.2699909325)+0.60(0.2240418077)\Pr(S\mid X=2)=\frac{0.40(0.2699909325)}{0.40(0.2699909325)+0.60(0.2240418077)}
Pr⁡(S∣X=2)=0.4454901561…\Pr(S\mid X=2)=0.4454901561\ldots

Answer

Answer

The posterior probability rounds to 0.45.

0.45(D)\boxed{0.45\quad\text{(D)}}