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Suppose that the probability of making a three pointer is $p=35\%$.

(c) Calculate the expected value and variance for the number of misses that you have prior to making your third shot.

(d) Calculate the expected value for the total number of shot attempts that you take in order to make three shots.

I understand that this is a negative binomial distribution so the distribution is as follows:

$$\Pr(M=k)=\binom{2+k}{2} p^3 q^k \text{ for } k=0,1,2,3,\ldots\tag{1}$$

where M is number of misses prior to rth success (in this case $r=3$).

To calculate (c), I did the following:



I'm not sure whether I understand the question to part (d). In a negative binomial distribution, you can have an infinite number of shot attempts to make 3 shots. For example, you can have the following sequences:

$$ \begin{align} SSS \\ \{S,S,M\}S \\ \{S, S, M, M \}S \\ \{S, S, M, M, M \}S \\ \vdots\qquad{} \end{align} $$

And to calculate the expected value of each case, you just apply equation (2). I don't really know whether I understand part (d) and how to solve. Any help is appreciated. Thank you.

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I would read he total number of shot attempts being 3 + number of misses. – deinst Jul 26 '13 at 19:04
up vote 1 down vote accepted

EDIT: in the first case if your $n^{\text{th}}$ shot is the last, you need 'the number of failures', so your random variable $X$ takes values $\{0,1, \ldots \}$, in the second case you count the 'total number of tosses', so your random variable takes values $\{3,4, \ldots \}$. In the first case the expectation is $$ \sum_{k=0}^{\infty}k \binom{k+2}{2}p^2 q^k $$ in the second case it is $$ \sum_{k=3}^{\infty}k \binom{k-1}{k-3}p^3q^{k-3} $$

Does this help now?

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No sorry I don't understand your response at all... – user1527227 Jul 26 '13 at 22:36
pls see the edit – Alex Jul 26 '13 at 23:01
Ohhhh. Thank you sooo much @Alex!!! Now that makes more sense. I think though in your first equation though you should have a $p^3$ since there are 3 shots made right? – user1527227 Jul 27 '13 at 0:47
You are welcome. – Alex Jul 27 '13 at 2:56
Not quite, because you count the number of tosses $before$ the 3rd one, and in this set there are only 2 successes and $k$ failures – Alex Jul 27 '13 at 2:57

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