Question of Conditional Expectation? Please help me with solution?

Question

A population has two classes of drivers. The number of accidents per individual driver has a geometric distribution. For a driver selected at random from Class I, the geometric distribution parameter has a uniform distribution over the interval (0,1). Twenty-ve percent of the drivers are in Class I. All drivers in Class II have expected number of claims 0.25. For a driver selected at random from this population, determine the probability of exactly two accidents.

Answer 1

$N\sim \mathsf{Geom}(p)$, i.e. with pdf $(1-p)^kp$ for $k\in\{0,1,2,\ldots\}$, $p=\frac{1}{1+\beta}$, $\Bbb E(N)=\beta\sim\mathsf{U}(0,1)$ \begin{align} \Bbb P(N=2|\text{Class I}) &=\int_0^1 \Bbb P(N=2|\beta)\cdot \underbrace{1}_{\text{density of }\mathsf{U}(0,1)}\cdot \mathrm d\beta=\int_0^1 \underbrace{\frac{\beta^2}{(1+\beta)^3}}_{(1-p)^2p}\, \mathrm d\beta\\ &=\int_1^2 \frac{(t-1)^2}{t^3} \mathrm dt=\int_1^2 \left(\frac{1}{t}-\frac{2}{t^2}+ \frac{1}{t^3}\right)\mathrm dt\\ &=\left[\log |t|+\frac{2}{t}-\frac{1}{2t^2}\right]=0.068\\ \\ \Bbb P(N=2|\text{Class II}) &= \frac{\beta^2}{(1+\beta)^3}= \frac{0.25^2}{1.25^3}=0.032\\ \\ \Bbb P(N=2) &= \Bbb P(N=2|\text{Class I}) \underbrace{\Bbb P(\text{Class I})}_{0.25} +\Bbb P(N=2|\text{Class II}) \underbrace{\Bbb P(\text{Class II})}_{0.75}\\ &=0.068\times 0.25+0.032\times 0.75=0.041 \end{align}