Independent solution

How to solve this Geometric Distribution question

Setup

Setup

Let T be the trial number of the first positive result. Its probability mass follows a geometric distribution on the positive integers.

Pr⁡(T=k)=(1−p)k−1p,k=1,2,…\Pr(T=k)=(1-p)^{k-1}p,\qquad k=1,2,\ldots

Model

Model

The event of stopping by m is contained in the event of stopping by n. Therefore the intersection is just the smaller event.

{T≤m}⊆{T≤n}\{T\le m\}\subseteq\{T\le n\}
Pr⁡(T≤m∣T≤n)=Pr⁡(T≤m)Pr⁡(T≤n)\Pr(T\le m\mid T\le n)=\frac{\Pr(T\le m)}{\Pr(T\le n)}

Compute

Compute

Use the complement of consecutive failures to write both cumulative probabilities.

Pr⁡(T≤r)=1−Pr⁡(T>r)=1−(1−p)r\Pr(T\le r)=1-\Pr(T>r)=1-(1-p)^r
Pr⁡(T≤m∣T≤n)=1−(1−p)m1−(1−p)n\Pr(T\le m\mid T\le n)=\frac{1-(1-p)^m}{1-(1-p)^n}

Answer

Answer

The ratio of the two geometric cumulative probabilities is the listed expression in choice E.

1−(1−p)m1−(1−p)n(E)\boxed{\frac{1-(1-p)^m}{1-(1-p)^n}\quad\text{(E)}}