Skateboarders A and B practice one difficult stunt until becoming injured while attempting the stunt. On each attempt, the probability of becoming injured is [math]p[/math], independent of the outcomes of all previous attempts.

Let [math]F(x, y)[/math] represent the probability that skateboarders A and B make no more than [math]x[/math] and [math]y[/math] attempts, respectively, where [math]x[/math] and [math]y[/math] are positive integers. It is given that [math]F(2, 2) = 0.0441[/math].

Calculate [math]F(1, 5)[/math].

• 0.0093
• 0.0216
• 0.0495
• 0.0551
• 0.1112

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

Solution: C

The probability that a skateboarder makes no more than two attempts is the probability of being injured on the first or second attempt, which is

[[math]] p + (1-p)p = 2p - p^2. [[/math]]

Then,

[[math]] \begin{align*} 0.0441 = F(2,2) = (2p-p^2)^2 \\ 0.21 = 2p-p^2 \\ p^2-2p + 1 = 0.79 \\ (p-1)^2 = 0.79 \\ p-1 = ±0.88882 \\ p = 0.11118 \end{align*} [[/math]]

The probability that a skateboarder makes no more than one attempt is [math]p[/math] while the probability of making no more than five attempts is the complement of having no injuries on the first five attempts. Hence,

[[math]] F(1,5) = p[1-(1-p)^2] = 0.0495. [[/math]]

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