Probability of multiple variables, geometric distribution? You are on a basketball team, and at the end of every practice, you shoot half-court shots until you make one. Once you make a shot, you go home. Each half-court shot, independent of all other shots, has a 0.1 probability of going in. Your team has 100 practices per season. Estimate the probability that you shoot more than 1111 half-court shots after practices this season.
So I set X equal to the number of shots until finished per practice, and since $X\sim\mathrm{Geo}(0.1)$, I know that the $P(X)$ is $(1-p)^{x-1}p$ but I don't know what to set $x$ as.
I also set $Y$ equal to the number of shots until finished per season so that $Y = 100X$. I'm not sure how to go about finding the probability though? I understand that $P(Y) = P(X)^{100}$, but how can I find $P(X)$?
 A: Outline Let $X_1,X_2,\dots, X_n$, where $n=100$, be the number of shots taken on practice $1, 2,\dots,n$.
Let $Y=X_1+X_2+\cdots+X_n$.  We want the probability that $Y\gt 1111$.
Each $X_i$ has geometric distribution. You know the mean of $X_i$. Call it $\mu$. Find or look up what the variance is. Call it $\sigma^2$.
By the Central Limit Theorem, $Y$ has roughly normal distribution, with mean $n\mu$ and variance $n\sigma^2$. Now you can calculate.
Remark: The post refers to $100X$. This is an entirely different sort of random variable than our random variable $Y$, and its distribution is not relevant to our problem.
You asked about the distribution of $Y$. This random variable measures the number of shots until the $100$-th success. It has negative binomial distribution. It is not hard to write down an explicit formula for the probability that $Y=y$.  
Then we could solve our problem by finding the sum of $\Pr(Y=y)$, from $y=100$ to $y=1111$, and subtracting the result from $1$. This involves an unpleasant amount of computation, though the right kind of software could handle it.  That's the reason that we find the normal approximation path more appealing.
