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:
$$E[M]=\cfrac{rq}{p}=\cfrac{3\cdot0.65}{0.35}=5.57\tag{2}$$
$$Var[M]=\cfrac{rq}{p^2}=\cfrac{3\cdot0.65}{0.35^2}=15.92\tag{3}$$
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.
Best Answer
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?