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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.

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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 $\binom{9}{9}0.8^{10}\cdot 0.2^{0}=0.8^{10}$

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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

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

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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.

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    $\begingroup$ 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. $\endgroup$ – Kong Sep 13 '18 at 5:32
  • $\begingroup$ Yeah...i missed that info. Last attempt needs to be a success. So, your attempt is completely fine. $\endgroup$ – idea Sep 13 '18 at 5:37
  • $\begingroup$ corrected my mistake... $\endgroup$ – idea Sep 13 '18 at 5:43

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