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Suppose two players A and B are trying to hit a basket of basketball. The probability that A hit at a given pitch is p, and he insists on the throws until he hits r times. The probability that B matches a given release is mp, where m is an integer (m ≥ 2), with mp <1; and he insists on postings until hit mr times. Answer:

a) For which player is the expected number of releases less?

b) For which player is the variance of the number of entries less?

Thanks

From the comments:

We have mp<1, then 1/p>m. Then r/p>mr. Since A has Negative binomial distribution (r, p) then E[A]=r/p. Therefore, as r/p>mr, then E [A]>mr. Dividing both sides by mp we have E[A]/mp>mr/mp. Since B has Negative Binomial distribution (mr, mp) then E[B]=mr/mp. Therefore, E[A]/mp>E[B]. Therefore, E[A]/E[B]>mp. But as mp<1, we can not conclude anything about this division between hopes, for this division may be smaller than zero (which implies that E[A] is less than E[B]) and that it can be greater than 1 (implying that E[B] is greater than E[A]). I doubt this conclusion.

Shaun
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Wecsley Prates
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    What have you done so far? Where are you stuck? – gary Dec 05 '17 at 20:49
  • Here's a MathJax tutorial :) – Shaun Dec 05 '17 at 21:05
  • We have $mp<1$, then $1/p>m$. Then $r/p>mr$. Since A has Negative binomial distribution (r, p) then $E[A]=r/p$. Therefore, as $r/p>mr$, then $E [A]>mr$. Dividing both sides by $mp$ we have $E[A]/mp>mr/mp$. Since B has Negative Binomial distribution (mr, mp) then $E[B]=mr/mp$. Therefore, $E[A]/mp>E[B]$. Therefore, $E[A]/E[B]>mp$. But as $mp<1$, we can not conclude anything about this division between hopes, for this division may be smaller than zero (which implies that E[A] is less than E[B]) and that it can be greater than 1 (implying that E[B] is greater than E[A]). I doubt this conclusion. – Wecsley Prates Dec 05 '17 at 21:08

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