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.