Independent solution

How to solve this Geometric Distribution question

Setup

Setup

Let p be the common disease probability and let q=1-p be the probability of continuing past one patient.

Pr(Tn)=qn1\Pr(T\ge n)=q^{n-1}

Model

Model

Express the supplied probability r and the requested conditional tail in powers of q.

r=Pr(T4)=q3r=\Pr(T\ge4)=q^3
Pr(T12T4)=q11q3=q8\Pr(T\ge12\mid T\ge4)=\frac{q^{11}}{q^3}=q^8

Compute

Compute

Eliminate q using r=q cubed.

q=r1/3q=r^{1/3}
q8=(r1/3)8=r8/3q^8=\left(r^{1/3}\right)^8=r^{8/3}

Answer

Answer

The conditional probability is r to the eight-thirds power.

r8/3(C)\boxed{r^{8/3}\quad\text{(C)}}