Independent solution

How to solve this Conditional Probability question

Setup

Setup

Represent the number of adverse outcomes among the three independent trials by N.

N∼Binomial⁡(3,p)N\sim\operatorname{Binomial}(3,p)

Model

Model

Because exactly one adverse outcome implies at least one, the conditional numerator reduces to the probability that N equals one.

Pr⁡(N=1∣N≥1)=Pr⁡(N=1)Pr⁡(N≥1)\Pr(N=1\mid N\ge1)=\frac{\Pr(N=1)}{\Pr(N\ge1)}

Compute

Compute

Expand both probabilities as polynomials, then cancel their common factor p.

Pr⁡(N=1)=3p−6p2+3p3\Pr(N=1)=3p-6p^2+3p^3
Pr⁡(N≥1)=3p−3p2+p3\Pr(N\ge1)=3p-3p^2+p^3
Pr⁡(N=1∣N≥1)=3−6p+3p23−3p+p2=3(1−p)2p2−3p+3\Pr(N=1\mid N\ge1)=\frac{3-6p+3p^2}{3-3p+p^2}=\frac{3(1-p)^2}{p^2-3p+3}

Answer

Answer

The simplified expression is equivalent to the expression in choice E.

3(1−p)2p2−3p+3(E)\boxed{\frac{3(1-p)^2}{p^2-3p+3}\quad\text{(E)}}