Expected value.

Probability Level 3

Consider an archer whose probability of hitting a target increases with every attempt. Let $P(n)$ be the probability of hitting the target on $n$ th attempt such that it is given by

$\begin{cases} P(n) = \dfrac nN & \text{if } n \le N \\ P(n) = 1 & \text{if } n > N \end{cases}$

where $N$ is a constant.

What are the expected number of attempts required to hit the target $N$ times?

\frac{3N}{2}

\frac{3N-1}{2}

\frac{3N+1}{2}

\frac{4N+1}{3}

\frac{4N-1}{3}

\frac{4N}{3}

1 solution

Mark Hennings
Oct 6, 2020

If $Z$ is the number of times the target is hit in the first $N$ arrows, then the required number of arrows is $X = N + (N - Z) = 2N-Z$ , and so the expected number of arrows to hit the target $N$ times is $\mathsf{E}[X] = 2N - \mathsf{E}[Z]$ .

Let $Z_j$ be the random variable that is equal to $1$ if the $j$ th arrow hits its target, and $0$ otherwise, for $1 \le j \le N$ . Then $Z \; = \; \sum_{j=1}^N Z_j$ and hence $\mathsf{E}[Z] \; =\; \sum_{j=1}^N \mathsf{E}[Z_j] \; =\;\sum_{j=1}^N P[Z_j = 1] \; =\; \sum_{j=1}^N \frac{j}{N} \; =\; \tfrac12(N+1)$ and hence $\mathsf{E}[X] \; =\; 2N - \tfrac12(N+1) \; =\; \boxed{\tfrac12(3N-1)}$

Expected value.

1 solution

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