# Probability of number of attempts given success

Consider various scenarios related to basketball tryouts. Specifically, for any given person trying out, let the event the person makes any free throw attempt be $X=1$ with $\mathsf P(X=1) =0.8$. Assume that successive attempts are independent and identically distributed (iid).

Let Y denote the number of attempts until the shooter makes a total of $10$ shots. Clearly, Sample Space = $\{ 10, 11, 12, 13,...\}$ . Compute $\mathsf P( Y \leq 12)$.

My Method: We know that the minimum attempts we need is $10$, and the maximum can, is infinity. So, $$\mathsf P(Y \leq 12) = \mathsf P(Y=10) + \mathsf P(Y=11) +\mathsf P(Y=12)$$
Hence

$$\mathsf P(Y \leq 12) = 0.8^{10} + {10\choose1} \cdot 0.8^{10}\cdot 0.2 + {11\choose2} \cdot 0.8^{10}\cdot 0.2^2$$

Is my approach correct? Is it more like conditional probability when we need to find Pr9 Y<=12) given that $X= 10$. $X$ was the success as defined in the beginning of the question. Thanks.

## 3 Answers

This probability model for this exercise is the Negative Binomial Distribution.

The probability that $x\geq 10$ is the number of shots it takes to make $10$ baskets is $$f(x\geq10,10,0.8)=\binom{x-1}{9}0.8^{10}\cdot0.2^{(x-10)}$$ To see this, consider the probability of hitting all ten shots is $0.8^{10}=0.107374$

In the probability density function defined above, this is $\binom{9}{9}0.8^{10}\cdot 0.2^{0}=0.8^{10}$

Yes, your answer is correct. Note that this is a negative binomial with $n$ trials given $k$ successes which takes the form

$${{n-1}\choose{n-k}}p^k(1-p)^{n-k}$$

We are only interested in the cases where $n\in\{10,11,12\}$ and $k$ stays fixed at $10$. Thus we get

$$P(Y\leq 12)={{9}\choose{0}}0.8^{10}+{{10}\choose{1}}\cdot0.8^{10}\cdot0.2+{{11}\choose{2}}\cdot0.8^{10}\cdot0.2^2\approx0.558$$

you are correct for X=10 case.
While doing for X=11, you have to select one attempt out of fist 10 which will be a failure, since last has to be a success.

Probability of failure=1-0.8=0.2
So, you need to take C(10,1)(0.8^10)(0.2^1).

Similarly for X=12, select 2 failures from first 11.
So take C(11,2)(0.8^10)(0.2^2).

Finally, (0.8^10) + C(10,1)(0.8^10)(0.2) + C(11,2)(0.8^10)(0.2^2) is the answer.

You are correct saying its like selecting 9 successes in first n-1 attempts if there are a total of n attempts; since nth attempt is definitely a success. go for n=12 and you get the answer.

• But the question says "until" 10 shots are made. Why would I take 11C1 in case Y=11? Because that would mean a case when I have 10 shots in first ten attempts and failure in the last. Why would I do that? The way to think this would be to take 9 success out of 10 attempts and then succeed in the 11th one, and hence 10C1. Please correct me if I am wrong. – Kong Sep 13 '18 at 5:32
• Yeah...i missed that info. Last attempt needs to be a success. So, your attempt is completely fine. – idea Sep 13 '18 at 5:37
• corrected my mistake... – idea Sep 13 '18 at 5:43