Actuarial Science - Part 2

$$ We use the result from the solution Actuarial Science - Part 1 problem. Therefore, the probability that the policyholder is a good driver given that $\text{N}_1=0$ and $\text{N}_2=1$ is $$ $\begin{aligned} \text{Pr}[\text{G}|\text{N}=0\cap\text{N}=1]&=\frac{\text{Pr}[\text{N}=0|\text{G}]\cdot\text{Pr}[\text{N}=1|\text{G}]\cdot\text{Pr}[\text{G}]}{\text{Pr}[\text{N}=0\cap\text{N}=1]}\\ &=\frac{0.7\cdot0.2\cdot0.75}{0.1425}\\ &=\frac{14}{19}. \end{aligned}$ $$ Thus, $p+q=14+19=\boxed{\color{#3D99F6}{33}}$ . $$

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$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

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Likewise, I built on my solution to the other problem and simply divided 420 by 570.