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Tom "makes"(scores) a basketball shot with the probability $0.8$. He stops when he has made 10 scores.

(A) what is the probability that he makes it in $13$ attempts

so here i used the negative binomial distribution: $\binom{k-1}{r-1}p^rq^{k-r}$ where $k$ = number of tries and $r$ = the number of "successes" (in our case the number of scores), $p =$ probability, $q= 1-p$

(B) how many attempts is it most likely that TOM needs ?

Now i could use set $p(k)= \binom{k-1}{r-1}p^rq^{k-r}$ and take the derivate w.r.t to $k$ and then find $p'(k) = 0 $ to obtain a maximum. The problem is i get a really nasty expression ..hopless. Is there another way?

Danny
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1 Answers1

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The probability $p(k)$ that he does it in exactly $k$ is, as you pointed out, $\binom{k-1}{9}(0.8)^{10}(0.2)^{k-10}$.

The probability $p(k+1)$ that he does it in exactly $k+1$ is for the same reason $\binom{k}{9}(0.8)^{10}(0.2)^{k-9}$.

We calculate the ratio $\frac{p(k+1)}{p(k)}$. Express the binomial coefficients in terms of factorials. There will be a lot of cancellation, and after a while we get $$\frac{p(k+1)}{p(k)}=\frac{k}{k-9}(0.2).$$ As long as this is $\gt 1$, we have that $p(k+1)\gt p(k)$, and so $k+1$ is more likely than $k$.

We get that the ratio is exactly $1$ at $k=\frac{9}{0.8}=11.25$. So for $k\le 11$, we have $p(k+1)\gt p(k)$, and after that there is decrease. It follows that the most likely number of throws is $12$.

Remark: We could instead have computed $p(10)$, $p(11)$, and so on, until the answer popped out. Instead we used a general procedure. It was done for your particular $r$ and $p$ only for the sake of concreteness. For some rather special combinations of $r$ and $k$, there will be two consecutive numbers at which the probability reaches a maximum.

André Nicolas
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