Independent solution

How to solve this Conditional Probability question

Setup

Setup

Let N be the event count. Once the stated count condition is imposed, only the two-count and three-count cases remain.

Pr⁡(N=2∣N≥2)=0.20.2+0.1=23\Pr(N=2\mid N\geq 2)=\frac{0.2}{0.2+0.1}=\frac{2}{3}
Pr⁡(N=3∣N≥2)=13\Pr(N=3\mid N\geq 2)=\frac{1}{3}

Model

Model

With two events, the threshold is reached only when both severities take their larger value. With three events, it is reached unless all three take their smaller value.

q2=(0.5)2=14q_2=(0.5)^2=\frac{1}{4}
q3=1−(0.5)3=78q_3=1-(0.5)^3=\frac{7}{8}

Compute

Compute

Average the two conditional threshold probabilities using the conditioned count weights.

Pr⁡(S≥2000∣N≥2)=23(14)+13(78)\Pr(S\geq 2000\mid N\geq 2)=\frac{2}{3}\left(\frac{1}{4}\right)+\frac{1}{3}\left(\frac{7}{8}\right)
Pr⁡(S≥2000∣N≥2)=1124=0.4583333333\Pr(S\geq 2000\mid N\geq 2)=\frac{11}{24}=0.4583333333

Answer

Answer

The conditional probability rounds to 0.46.

0.46(D)\boxed{0.46\quad\text{(D)}}