which distribution should be used in this question? A basketball player succeeds in making a basket three tries out of four. How many
times must he try for a basket in order to have greater than 0.99 probability of making 
at least one basket?
In this question, should we use the Poisson distribution?
 A: Sounds like a Binomial:
$$\Pr(X = k) = {n\choose k}p^k(1-p)^{n-k}$$
It models $k$ successes out of $n$ tries with a probability of success $p$.
A: There seems to be a slight ambiguity in your problem proposition. It does not make sense to say 'how often do you have to try to make at least one basket'. Because as soon as there is one success the criterion is met and you would stop to sample new throw attempts, the question can be rephrased as 'how often do you have to try until the player makes the first basket with probability larger than $\pi$'? 
The distribution you need to answer this question is the cumulative geometric distribution which gives the probability that the first success occurs on or before trial $k$.
$$\pi = P(X \le k) = 1-(1-p)^k $$
so $k$ is the ceiling of $\log(1-\pi)/\log(1-p)$, in your case $p=3/4$, $\pi=0.99$ and we have $k=4$. So you have to to trial four times to have larger than 99% probability to have the first basket. 
If you trial $k=4$ times he of course may make more than one basket. The probability distribution for the random number $Y$ of baskets made within $k$ trials then is binomial, as indicated by @Tim. Note however that the geometric distribution acknowledges the sequence of throws, whereas the binomial does not. 
A: Maybe try out the negative binomial distribution:
https://en.wikipedia.org/wiki/Negative_binomial_distribution
It counts the number of successes until r failures are reached. Simply flip the probabilities of success and error in your case (and set r=1) and you have a match for your problem.
