Independent solution

How to solve this Conditional Probability question

Setup

Setup

Let N denote the claim count and let H denote the event that the aggregate amount does not exceed the stated threshold. Only the two positive-count branches can enter the condition.

Pr⁡(N=1)=0.08,Pr⁡(N=2)=0.02\Pr(N=1)=0.08,\qquad \Pr(N=2)=0.02

Model

Model

Evaluate the conditional amount distribution at the threshold separately for each branch.

Pr⁡(H∣N=1)=1−5002000=34\Pr(H\mid N=1)=1-\frac{500}{2000}=\frac34
Pr⁡(H∣N=2)=1−10002000=12\Pr(H\mid N=2)=1-\frac{1000}{2000}=\frac12

Compute

Compute

Weight each likelihood by its claim-count probability and normalize the one-claim contribution.

Pr⁡(N=1,H)=0.08(34)=0.06\Pr(N=1,H)=0.08\left(\frac34\right)=0.06
Pr⁡(N=2,H)=0.02(12)=0.01\Pr(N=2,H)=0.02\left(\frac12\right)=0.01
Pr⁡(N=1∣N≥1,H)=0.060.06+0.01=67\Pr(N=1\mid N\ge1,H)=\frac{0.06}{0.06+0.01}=\frac67

Answer

Answer

The conditional probability is six-sevenths.

67(E)\boxed{\frac67\quad\text{(E)}}