What is the probability of a basketball player making 10 free-throw shots in a row and only doing so once in 100 attempts, assuming the player has a 75% success rate?
The "in a row" part is what caused me to have trouble applying the binomial distribution.
At least we all know that the probability of making the first 10 shots in a row would be $$ P(\text{achieve first 10 shots in a row})=\left(\frac 34 \right)^{10} $$
But if we were given 100 attempts, then the basketball player can begin his 10-shot streak in, e.g. his first 10 attempts ($n=1$), or miss at his $(n-1)$-th attempt and begin his 10-in-a-row streak at his $n$-th attempt, provided that $1 \le n \le 91$ (because if he begins at $n=91$ that would be his last chance of making 10 shots in a row in 100 attempts).
Since beginning a 10-in-a-row streak at his $n$-th attempt, I was thinking that these were disjoint events. Since there are $90$ possible events, I thought I could simply multiply by $90$ and "conclude" that $$ P(\text{achieve 10 shots in a row one time in 100 attempts})=\left(\frac 34 \right)^{10} \cdot 90 $$ ...except that $\left(\frac 34 \right)^{10} \cdot 90 > 1$.
I believe I have yet to apply the binomial distribution here somehow?