Patients in a study are tested for sleep apnea, one at a time, until a patient is found to have this disease. Each patient independently has the same probability of having sleep apnea. Let [math]r[/math] represent the probability that at least four patients are tested.

Determine the probability that at least twelve patients are tested given that at least four patients are tested.

• [math]r^{\frac{11}{3}}[/math]
• [math]r^3[/math]
• [math]r^{\frac{8}{3}}[/math]
• [math]r^2[/math]
• [math]r^{\frac{1}{3}}[/math]

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

Solution: C

Let [math]X[/math] number of patients tested, which is geometrically distributed with constant “success” probability, say [math]p[/math].

[[math]] \operatorname{P}[X \geq n] = \operatorname{P}[\textrm{first n-1 patients do not have apnea}] = (1-p)^{n-1}. [[/math]]

Therefore,

[[math]] \begin{align*} r = \operatorname{P}[X \geq 4] &= (1-p)^3 \\ \operatorname{P}[X \geq 12 | X \geq 4] &= \frac{\operatorname{P}[X \geq 12]}{\operatorname{P}[X \geq 4]}\\ &= \frac{(1-p)^{11}}{(1-p)^3}\\ &= [(1-p)^3]^{\frac{8}{3}} \\ &= r^{\frac{8}{3}}. \end{align*} [[/math]]

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