Independent solution

How to solve this Binomial Distribution question

Setup

Setup

A stop occurs when the within-subject positive count is two or three.

K∼Binomial⁡(3,p)K\sim\operatorname{Binomial}(3,p)
q=Pr⁡(K≥2)q=\Pr(K\ge 2)

Model

Model

Add the two disjoint stopping cases, then view successive subjects as independent Bernoulli stopping trials.

q=(32)p2(1−p)+(33)p3q=\binom{3}{2}p^2(1-p)+\binom{3}{3}p^3
q=3p2(1−p)+p3q=3p^2(1-p)+p^3

Compute

Compute

Exactly n tests means n-1 continuations and then a stop.

Pr⁡(T=n)=(1−q)n−1q\Pr(T=n)=(1-q)^{n-1}q
=[1−3p2(1−p)−p3]n−1[3p2(1−p)+p3]=\left[1-3p^2(1-p)-p^3\right]^{n-1}\left[3p^2(1-p)+p^3\right]

Answer

Answer

The expression matches choice B.

[1−3p2(1−p)−p3]n−1[3p2(1−p)+p3](B)\boxed{\left[1-3p^2(1-p)-p^3\right]^{n-1}\left[3p^2(1-p)+p^3\right]\quad\text{(B)}}