In a large population of patients, 20% have early stage cancer, 10% have advanced stage cancer, and the other 70% do not have cancer. Six patients from this population are randomly selected.

Calculate the expected number of selected patients with advanced stage cancer, given that at least one of the selected patients has early stage cancer.

• 0.403
• 0.500
• 0.547
• 0.600
• 0.625

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

Solution: C

Let [math]X[/math] and [math]Y[/math] represent the number of selected patients with early stage and advanced stage cancer, respectively. We need to calculate [math]\operatorname{E}(Y | X ≥ 1) [/math].

From conditioning on whether or not X ≥ 1 , we have

[[math]] \operatorname{E}(Y ) =\operatorname{P}[ X =0]\operatorname{E}(Y | X =0) + \operatorname{P}[ X ≥ 1]\operatorname{E}(Y | X ≥ 1) . [[/math]]

Observe that [math]\operatorname{P}[X=0] = (1-0.2)^6 = (0.8)^6 [/math], [math]\operatorname{P}[X \geq 1] = 1- \operatorname{P}[X = 0] = 1-(0.8)^6[/math], and [math]\operatorname{E}[Y] = 6(0.1) = 0.6 [/math]. Also, note that if none of the 6 selected patients have early stage cancer, then each of the 6 selected patients would independently have conditional probability [math]\frac{0.1}{1-0.2} = \frac{1}{8}[/math] of having late stage cancer, so [math]\operatorname{E}[Y | X = 0] = 6(1/8) = 0.75 [/math].

Therefore

[[math]] \operatorname{E}(Y | X \geq 1) = \frac{\operatorname{E}(Y) − \operatorname{P}[ X= 0]\operatorname{E}(Y | X= 0)}{\operatorname{P}[X \leq 1]} = \frac{0.6 − (0.8)6 (0.75)}{1-(0.8)^6} = 0.547. [[/math]]

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