Question

Ten hunters are waiting for ducks to fly by. When a flock of ducks flies overhead, the hunters fire at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability .$6$, compute the expected number of ducks that are hit. Assume that the number of ducks in a flock is a Poisson random variable with mean $6$.

Short answer

$E\left(X\right)=\sum _{n=0}^{\infty }n\left[1-{\left(1-\frac{0.6}{n}\right)}^{10}\right]·\frac{6^n}{n!}{e}^{-6}$

Step-by-step solution

Given information

Given in the question that, Ten hunters are waiting for ducks to fly by. When a flock of ducks flies overhead, the hunters fire at the same time, but each chooses his target at random, independently of the others. If each hunter independently hits his target with probability $6$.

Explanation

Characterize arbitrary variable $X$as the quantity of ducks that have been hit and characterize $N$as the quantity of ducks in a flock. We know that $N~Pois\left(6\right)$. Assuming we are given data that $N=n$, we can compose

$X=\sum _{k=1}^n{I}_k$

where $I_k$is pointer arbitrary variable which demonstrates regardless of whether $k$th duck in a flock has been hit. See that

Law of the total expectation 

Since the principal duck will be remembered fondly assuming each hunter miss that duck and the likelihood that a specific hunter shots that duck is equivalent to $\frac{1}{n}·0.6$since he needs to pick that duck and hit it. Utilizing the law of the total expectation, we have that

$=\sum _{n=0}^{\infty }nE\left(I_1\right)P(N=n)$

$=\sum _{n=0}^{\infty }n\left[1-{\left(1-\frac{0.6}{n}\right)}^{10}\right]·\frac{6^n}{n!}{e}^{-6}$

Final answer

Expected number of ducks that are hit is,

$E\left(X\right)=\sum _{n=0}^{\infty }n\left[1-{\left(1-\frac{0.6}{n}\right)}^{10}\right]·\frac{6^n}{n!}{e}^{-6}$