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 [math]p[/math], where [math]0 \lt p \lt 1[/math]. The outcomes of the tests across all patients and all risk factors are mutually independent.

Determine an expression for the probability that exactly [math]n[/math] patients are tested, where [math]n[/math] is a positive integer less than 100.

• [math][1-3p^2(1-p)]^{n-1}[3p^2(1-p)][/math]
• [math][1-3p^2(1-p) - p^3]^{n-1}[3p^2(1-p) + p^3][/math]
• [math][1-3p^2(1-p) - p^3]^{n-1}[3p^2(1-p) + p^3]^{n-1}[/math]
• [math]n[1-3p^2(1-p) - p^3]^{n-1}[3p^2(1-p) + p^3][/math]
• [math]3[(1 − p )^{n −1} p][1-(1-p)^{n-1}p] + [(1-p)^{n-1}p]^3[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

From the binomial distribution formula, the probability P that a given patient tests positive for at least 2 of these 3 risk factors is

[[math]] P = \binom{3}{2}p^2(1-p)^{3-2} + \binom{3}{3}p^3(1-p)^{3-3} = 3p^2(1-p) + p^3. [[/math]]

Using the geometric distribution formula with probability of success

[[math]] P= 3 p^2 (1 − p ) + p^3, [[/math]]

The probability that exactly [math]n[/math] patients are tested is

[[math]] (1-p)^{n-1}P = [1-3p^2(1-p) - p^3]^{n-1}]3p^2(1-p) + p^3]. [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.