# Thread: probability problem concerning expected value

1. ## probability problem concerning expected value

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 0.6, 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.

I have no idea how to deal with this question.....

2. ## Re: probability problem concerning expected value

This is a terrible problem.

It states that 10 hunters are independently shooting at an average of 6 ducks and yet somehow they are independent of one another.

That's just not possible. Well it's possible if you allow for multiple hunters to shoot a single duck but the problem specifically refers to "his duck".

How do you interpret this problem? Can hunters shoot at the same duck?

3. ## Re: probability problem concerning expected value

Originally Posted by romsek
This is a terrible problem.

It states that 10 hunters are independently shooting at an average of 6 ducks and yet somehow they are independent of one another.

That's just not possible. Well it's possible if you allow for multiple hunters to shoot a single duck but the problem specifically refers to "his duck".

How do you interpret this problem? Can hunters shoot at the same duck?
It says "his target".

I would interpret it as follows:
Probability of zero ducks hit: $0.4^{10}$
Probability of one duck hit:
$\displaystyle 6\sum_{n=1}^{10} \dbinom{10}{n}(1-0.4^n)\left( \dfrac{1}{6} \right)^n(0.4)^{10-n}\left(\dfrac{5}{6}\right)^{10-n}$
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4. ## Re: probability problem concerning expected value

I can't say I understand your formula yet but let's see if we agree on the problem.

There are $n$ ducks with $n$ Poisson distributed $\lambda=6$

There are 10 hunters. Independently they select a target, one of the ducks, and have a $0.6$ probability of hitting the target they select.

Multiple hunters may select the same duck, they are independent.

Find the expected number of ducks that are hit.

Does this jibe with your take on the problem?

5. ## Re: probability problem concerning expected value

Originally Posted by romsek
I can't say I understand your formula yet but let's see if we agree on the problem.

There are $n$ ducks with $n$ Poisson distributed $\lambda=6$

There are 10 hunters. Independently they select a target, one of the ducks, and have a $0.6$ probability of hitting the target they select.

Multiple hunters may select the same duck, they are independent.

Find the expected number of ducks that are hit.

Does this jibe with your take on the problem?
Yes. So my formula is:
Choose the duck that is hit: 6 (or $\lambda$)
Sum over the number of hunters firing at that duck (at least one): n hunters firing at chosen duck
Choose the hunters: 10 choose n
Probability those hunters chose the chosen duck: (1/6)^n
Probability at least one of them hit is complement of probability that none of them hit: 1-0.4^n
For the remaining 10-n hunters, probability none of them picked the chosen duck: (5/6)^(10-n)
Probability they all missed: 0.4^(10-n)

6. ## Re: probability problem concerning expected value

For 2 ducks:
Choose two ducks that are hit: $\dbinom{6}{2}$
Sum over the number of hunters firing at first hit duck (at least one, at most 9): $n_1$ hunters firing at first duck.
Choose the hunters: $\dbinom{10}{n_1}$
Probability those hunters chose the first chosen duck: $\left( \dfrac{1}{6} \right)^{n_1}$
Probability that at least one of them hit: $1-(0.4)^{n_1}$
Sum over the number of hunters firing at the second hit duck (at least 1 at most $10-n_1$): n_2 hunters firing at second duck.
Choose the hunters: $\dbinom{10-n_1}{n_2}$
Probability those hunters chose the second chosen duck: $\left( \dfrac{1}{6} \right)^{n_2}$
Probability at least one of them hit: $1-(0.4)^{n_2}$
For the remaining $10-n_1-n_2$ hunters, probability none of them picked the chosen ducks: $\left(\dfrac{4}{6}\right)^{10-n_1-n_2}$
Probability they all missed: $0.4^{10-n_1-n_2}$

Formula:
$\displaystyle \dbinom{6}{2}\sum_{n_1=1}^9 \sum_{n_2=1}^{10-n_1}\dfrac{10!}{n_1!n_2!(10-n_1-n_2)!}\left(\dfrac{1}{6}\right)^{n_1+n_2}\left(1-0.4^{n_1}\right)\left(1-0.4^{n_2}\right)\left(\dfrac{4}{6}\right)^{10-n_1-n_2}0.4^{10-n_1-n_2}$

Formula for exactly 3 ducks:
$\displaystyle \dbinom{6}{3}\sum_{n_1=1}^8 \sum_{n_2=1}^{9-n_1} \sum_{n_3=1}^{10-n_1-n_2} \dfrac{10!}{n_1!n_2!n_3!(10-n_1-n_2-n_3)!} \left( \dfrac{1}{6} \right)^{n_1+n_2+n_3} \left(1-0.4^{n_1} \right) \left( 1-0.4^{n_2} \right) \left( 1-0.4^{n_3} \right) \left( \dfrac{3}{6} \right)^{10-n_1-n_2-n_3}0.4^{10-n_1-n_2-n_3}$

Formula for exactly 4 ducks:
$\displaystyle \dbinom{6}{4}\sum_{n_1=1}^7 \sum_{n_2=1}^{8-n_1} \sum_{n_3=1}^{9-n_1-n_2} \sum_{n_4=1}^{10-n_1-n_2-n_3} \dfrac{10!}{n_1!n_2!n_3!n_4!(10-n_1-n_2-n_3-n_4)!} \left( \dfrac{1}{6} \right)^{n_1+n_2+n_3+n_4} \left(1-0.4^{n_1} \right) \left( 1-0.4^{n_2} \right) \left( 1-0.4^{n_3} \right)\left( 1-0.4^{n_4} \right) \left( \dfrac{2}{6} \right)^{10-n_1-n_2-n_3-n_4}0.4^{10-n_1-n_2-n_3-n_4}$

Formula for exactly 5 ducks:
$\displaystyle \dbinom{6}{5}\left( \dfrac{1}{6} \right)^{10}\sum_{n_1=1}^6 \sum_{n_2=1}^{7-n_1} \sum_{n_3=1}^{8-n_1-n_2} \sum_{n_4=1}^{9-n_1-n_2-n_3} \sum_{n_5=1}^{10-n_1-n_2-n_3-n_4} \dfrac{10!}{n_1!n_2!n_3!n_4!n_5!(10-n_1-n_2-n_3-n_4-n_5)!} \left(1-0.4^{n_1} \right) \left( 1-0.4^{n_2} \right) \left( 1-0.4^{n_3} \right)\left( 1-0.4^{n_4} \right)\left( 1-0.4^{n_5} \right)0.4^{10-n_1-n_2-n_3-n_4-n_5}$

Formula for exactly 6 ducks:
$\displaystyle \dbinom{6}{6}\left( \dfrac{1}{6} \right)^{10} \sum_{\sum_{i=1}^6 n_i = 10, n_i\ge 1} \dfrac{10!}{n_1!n_2!n_3!n_4!n_5!n_6!}(1-0.4^{n_1})(1-0.4^{n_2})(1-0.4^{n_3})(1-0.4^{n_4})(1-0.4^{n_5})(1-0.4^{n_6})$

7. ## Re: probability problem concerning expected value

You've used 6 for the number of ducks in the flock throughout these formulae.

6 is just the average number, the actual number is Poisson distributed.

Wouldn't you have to derive these for a general n and then take the weighted average to get the expectation?

Are you allowed to push through the expectation like this? I'm not sure you are.

8. ## Re: probability problem concerning expected value

Originally Posted by romsek
You've used 6 for the number of ducks in the flock throughout these formulae.

6 is just the average number, the actual number is Poisson distributed.

Wouldn't you have to derive these for a general n and then take the weighted average to get the expectation?

Are you allowed to push through the expectation like this? I'm not sure you are.
I am more of a combinatorist than a statistician. In general, if there are $k$ ducks, the probability that exactly $r$ of them are shot is given by:
$\displaystyle \dbinom{k}{r}\sum_{\begin{matrix}\sum_{i=1}^{r+1}n _i = 10 \\ \forall 1\le i \le r, n_i \ge 1 \\ n_{r+1} \ge 0\end{matrix} }10!\left( \prod_{i=1}^r \dfrac{(1-0.4^{n_i})}{n_i!} \right)\dfrac{0.4^{n_{r+1}}}{n_{r+1}!}\left( \dfrac{1}{k} \right)^{\sum_{i=1}^r n_i} \left( \dfrac{k-r}{k} \right)^{n_{r+1}}$

Note: To make it bigger and easier to see, for the first sum, it is the sum over all possible integral solutions to the Diophantine equation: $\displaystyle \sum_{i=1}^{r+1} n_i = 10$ where $\forall 1 \le i \le r, n_i \ge 1$, but $n_{r+1} \ge 0$.

Also, we are using the notation that $0^0 = 1$.

9. ## Re: probability problem concerning expected value

Here's an easier solution: A duck is either hit or it is not. This should be a binomial distribution problem. There are $k$ ducks. For a single duck, the probability that it is hit is the complement of the probability that it is not hit. Since every hunter acts independently, we can calculate the probability that a single hunter misses a single duck. If it aims at a different duck, it will miss. If the hunter aims at the duck we care about, there is a $\dfrac{2}{5}$ chance the hunter will still miss.

So, the probability that a single hunter will miss is given by:

$\dfrac{k-1}{k}+\dfrac{2}{5}\cdot \dfrac{1}{k} = \dfrac{5k-3}{5k}$

Now, the probability that our duck is hit:
$1-\left(\dfrac{5k-3}{5k}\right)^{10}$

The probability that exactly $r$ ducks are hit is going to be given by:
$\dbinom{k}{r}\left[1-\left(\dfrac{5k-3}{5k}\right)^{10}\right]^r\left(\dfrac{5k-3}{5k}\right)^{10(k-r)}$

Let $P(k)$ be the probability that the flock has $k$ ducks. Then, the expected value of ducks shot will be:

$\displaystyle \displaystyle \sum_{k\ge 1} \left[\sum_{r=0}^k r\dbinom{k}{r}\left[1-\left(\dfrac{5k-3}{5k} \right)^{10} \right]^r \left(\dfrac{5k-3}{5k} \right)^{10(k-r)}\right]P(k)$

But, the sum in the middle is just the expected value of ducks that are hit given there are $k$ ducks and the probability of hitting one is $1-\left(\dfrac{5k-3}{5k}\right)^{10}$. The expected value of the binomial distribution is $k\left[1-\left(\dfrac{5k-3}{5k}\right)^{10}\right]$

So, we can replace that in the summation to get the expected value of number of ducks shot:

$\displaystyle \displaystyle \sum_{k\ge 1} k\left[1-\left(\dfrac{5k-3}{5k}\right)^{10}\right]P(k)$

Finally, we know that $P(k) = \dfrac{6^ke^{-6}}{k!}$

Plugging that in, we have:

$\displaystyle \displaystyle \sum_{k\ge 1} k\left[1-\left(\dfrac{5k-3}{5k}\right)^{10}\right]\dfrac{6^ke^{-6}}{k!} \approx 3.698898753351333707335919841492432062088016365370 228702514$

http://www.wolframalpha.com/input/?i...infinity%7D%5D