10 players are shooting at a target. Each player shoots twice and has a similar probability p of hitting the target. Assume that the different shots are independent. Let X denote the total number of shots that hit the target. Find the moment generating function of X.

Question

loser66 · Answer

|dw:1413038266971:dw|
I am studying this stuff, also. We can work together. I am not 100% sure that I am right. :)

loser66 · Answer

So that, the moment generating function is
$$M(t)=E[e^{tx}]=\sum_x e^{tx}P(X=x)=\dfrac{1}{10}\sum_xe^{tx}$$

loser66 · Answer

I don't use "twice" because it is exactly the same when counting the probability mass function. I mean 2/20 =1/10

kirbykirby · Answer

I think the situation is a bit more complicated than this. If $X$ is the total number of shots, then it's possible that the range of values for the number of shots is 0, 1, 2, ..., 20 (All the players miss the target, 1 player hits 1 target, ... all hit the target). 

Since the players all hit the target with probability $p$, then they all miss with probability $1-p$. 

Now, you need to find the probability distribution of $X$
For $X=0$: All the players miss, so player 1 misses, And player 2 misses, And, ... And player 20 misses), so 
$P(X=0)=\underbrace{(1-p)(1-p)\cdots (1-p)}_{\text{20 of these}} = (1-p)^{20}$

For $X=1$: Only 1 target is hit, but it could have been done on any of the 20 shots. So, 
$P(X=1)={20\choose 1}(1-p)^{19}p$, meaning that one target is hit with probability $p$, 19 are hit with probability $1-p$, but the target could have been hit on the, 1st target, Or the 2nd target, Or the 3rd target, Or the 4th target..., Or the 20th target.

If you find $P(X=2)$, you notice that 2 targets can be hit, and 18 are missed. But, you can arrange these probabilities in ${20 \choose 2}$ ways , meaning
$P(X=2)={20 \choose 2}(1-p)^{18}p^2$

Notice that in general, this amounts to a binomial distribution, so
$$ P(X=x)={20\choose x}(1-p)^{20-x}p^{x}$$And so, the moment generating function would be
$$ M_X(t)=E(e^{tX})=\sum_{x}e^{tx}\cdot P(X=x)\
=\sum_{x=0}^{20}e^{tx}{20\choose x}(1-p)^{20-x}p^{x} =\sum_{x=0}^{20}{20 \choose x}(1-p)^{20-x}(pe^t)^x $$
Notice that this sum is the binomial theorem.

kirbykirby · Answer

Hm I wrote there were 20 players in my response haha, but there are 10 sorry about that. But actually,  since their shots are all independent, it's the same result as if you assumed you have 20 players taking 1 shot each.

anonymous · Answer

thank you very much.. yeah its okay..  :)