Determine an expression for the probability that exactly n patients are tested

Question

A group of 100 patients is tested, one patient at a time, for three risk factors for a certain disease until either all patients have been tested or a patient tests positive for more than one of these three risk factors. For each risk factor, a patient tests positive with probability p, where 0 < p < 1. The outcomes of the tests across all patients and all risk factors are mutually independent. Determine an expression for the probability that exactly n patients are tested, where n is a positive integer less than 100.

try

This problem "seems" easy. Let $X$ be the number of patients tested. $X$ is geometric seems it says that we stop when one patient tests positive and the probability that patient is positive is just $p$. Hence,

$$ P(X=n) = (1-p)^n p $$

am I misunderstanding this question? or is it that easy?

Answer 1

Here $n<100$ and let $X_2$ denote the random variable $i=1,2,3$ which represents $3$ different risk factors.

$$X_2=\begin{cases} 1 &\mbox{with probability $p$ (test +ve) } \\ 0& \mbox{with probability $1-p$ (test -ve)}\end{cases}$$

Here we test the next person only if the previous test is +ve at most in $1$ risk factor. The probability in testing +ve in at most $1$ risk factor is $$(1-p)^3+p(1-p)^2=q \mbox{ or } q=(1-p)^2(1-p+p)=(1-p)^2$$

Here $(1-p)^3$ is Prob$(\mbox{No +ve in any risk factor})$

$p(1-p)^2$ is prob$(\mbox{+ve in anyone of the risk factor})$

Now $$\mbox{Prob(n patients are tested)}=q^{n-1}\cdot p \begin{cases}\mbox{ p=1-q}\end{cases}$$ $$=(1-p)^{2n-2}\cdot(1-(1-p)^2)$$

Answer 2

It's given that $p$ is the probability that a patient tests positive for any single risk factor.

But each patient is tested for $3$ risk factors, and the experiment ends when some patient tests positive for at least $2$ of the $3$ risk factors.

Hence, for $1 \le n < 100$, the probability that the experiment ends on the $n$-th patient is $$(1-p')^{n-1}p'$$ where $$p'=3p^2(1-p)+p^3=3p^2-2p^3$$ is the probability that a patient tests positive for at least $2$ of the $3$ risk factors.

Explanation:

• The probability that a patient tests positive for exactly $2$ of the $3$ risk factors is $3p^2(1-p)$, and the probability that a patient tests positive for all $3$ risk factors is $p^3$, hence $p'=3p^2(1-p)+p^3$.
• The probability that a patient tests positive for at most $1$ of the $3$ rsik factors is $1-p'$, hence the probability that both
  • The first $n-1$ patients test positive for at most $1$ of the $3$ risk factors.
  • The $n$-th patient tests positive for at least $2$ of the $3$ risk factors.
  
  is $(1-p')^{n-1}p'$.