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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?

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    $\begingroup$ No, you should not use the Poisson distribution or any other distribution with a fancy name that you find listed in your book. Instead, the player should ponder the fact that (assuming that the various attempts at making a basket are independent) the probability of missing on each and every one of $n$ attempts is $\left(\frac 14\right)^n$ which decreases towards $0$ as $n \to \infty$. How large must $n$ be so that this probability is smaller than $0.01$? $0.01$? you ask in a bewildered tone. Where did that number come from..... $\endgroup$ Nov 16, 2014 at 14:57
  • $\begingroup$ @DilipSarwate I thought something like this before. Its like, probability of getting success in n trials is (3/4)^N . I should find some N for which its 0.99. I can never find such N. $\endgroup$ Nov 16, 2014 at 20:09
  • $\begingroup$ Hint: $\left(\frac 34\right)^n$ is the probability that the basketball player makes all $n$ shots. You want the probability that the player makes none of the $n$ shots because if this probability is very small, then the probability that the player makes at least one of the $n$ shots must be quite large, no? $\endgroup$ Nov 16, 2014 at 21:56
  • $\begingroup$ I think Binomial distribution should be used for this question. If we do it your way. Then for atleast one success in n trials probability is, (3/4)(1/4)^(n-1). Since that one shot can be in any of the n trials . We will multiply it by n.So, we get the same thing as we would get from binomial. $\endgroup$ Nov 17, 2014 at 10:35
  • $\begingroup$ No, it is the probability of making at least one basket in $n$ attempts that needs to exceed $0.99$. $\binom{n}{1}\left(\frac 34\right)\left(\frac 14\right)^{n-1}$ which you insist is the right answer is the probability of making exactly one basket in $n$ attempts. However, no matter. Since you believe that you need to find $n$ such that $\binom{n}{1}\left(\frac 34\right)\left(\frac 14\right)^{n-1}$ exceeds $0.99$, go right ahead. $\endgroup$ Nov 17, 2014 at 14:25

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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 $p$.

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    $\begingroup$ If the response satisfies you, accept it so others know it is solved. $\endgroup$
    – Tim
    Nov 16, 2014 at 9:19
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    $\begingroup$ But what is needed is the value of $n$ such that the player is sure that $P\{X \geq 1\} \geq 0.99$. Perhaps you could elaborate a little on how your answer helps in finding $n$? $\endgroup$ Nov 16, 2014 at 14:51
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    $\begingroup$ I am not sure that this is what was asked; see my answer $\endgroup$
    – tomka
    Aug 13, 2019 at 10:57
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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.

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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.

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