Binomial distribution: Suppose that a basketball player sinks a basket from a certain position on the court with probability $0.35$

Question

2.3.15 Suppose that a basketball player sinks a basket from a certain position on the court with probability $0.35$.

(a) What is the probability that the player sinks three baskets in 10 independent throws?

(b) What is the probability that the player throws 10 times before obtaining the first basket?

(c) What is the probability that the player throws 10 times before obtaining two baskets?

Solution:

(a) $X$ ~ $Binomial(n = 10, \theta = 0.35)$

$$P(X = 3) = {10\choose3}(0.35)^3(0.65)^7$$

(b) $X$ ~ $NegativeBinomial(r = 1, \theta = 0.35)$

$$P(X = 10) = {10\choose10}(0.35)^1(0.65)^{10} \text{[Not right]}$$

(c) $X$ ~ $NegativeBinomial(r = 2, \theta = 0.35)$

$$P(X = 10) = {11\choose10}(0.35)^2(0.65)^{10} \text{[Not right]}$$

Could someone help me understand b and c? Thanks!

Answer 1

b is wrong because you calculated the chance he misses $10$ and then hits number $11$. The question did not require hitting number $11$.

For c you should just compute the chance he gets $0$ or $1$ baskets of the first $11$. In that case he misses $10$ before he makes $2$.

You did not explain your logic for the solutions, but it is clear you are not computing the correct thing.