In a basketball shooting workout, a player keeps shooting until she makes 10 baskets. Suppose the probability that she makes any given shot is 0.7, and let X be the total number of shots she takes, calculate E[X], V(X)?
In a basketball shooting workout, a player keeps shooting until she makes 10 baskets. Suppose the probability that she makes any given shot is 0.7, and let X be the total number of shots she takes, calculate E[X], V(X)?
Sorry, I misunderstood the original problem.
Let's calculate variance a little differently. Since you have $E[X] = 14$, then
$$V[X] = E[X^2]-14^2$$
$$E[X^2] = (0.7)^{10}\sum_{n\ge 0}\dbinom{9+n}{n}(0.3)^n(10+n)^2 = \dfrac{10(11-0.7)}{0.7^2}$$
So, you have:
$$V[X] = \dfrac{696}{49}$$
Lets change the question.
Toss a die repeatedly until a six appears the three times.
Further, let $X$ be the number of the toss upon which the third six appears.
Is it clear that $X\ge 3~?$ and $\mathcal{P}(X=3)=6^{-3}$. Moverover, $\mathcal{P}(X=9)=\dbinom{8}{2}\cdot 5^6\cdot 6^{-9}$
In general, $n\ge 3\to \mathcal{P}(X=n)=\dbinom{n-1}{2}\cdot 5^{n-3}\cdot 6^{-n}$
What are $\mathcal{E[X]}~\&~\mathcal{V[X]}~?$