a. 15/75
b. 1[(22/75)(22/74)(22/73)(22/72)(22/71)]
c. (35/75)(35/74)(35/73)(35/72)(35/71)
d. (20/75)(55/71)
e. (18/74)(18/73)(18/72)
f. (57/75)(18/74)(18/73)(18/72)(57/71)
Note: Suppose there are 5 disk and 3 are jazz while 2 are blues. Suppose I
select 3 discs. Then, by the logic above, the chance that they are all jazz is
(3/5)(3/4)(3/3). However, there's a chance that I got the 3 jazz discs as my first, second and third disc; so the probability would be (3/5)(2/4)(1/3). Furthermore, there's a chance that the first 2 discs selected are jazz ,so the probability would be
(3/5)(2/4)(2/3). These three probabilities are vastly different, and the desire to incorporate each possibility into one's hypothesis by adding them together could have dire consequences, because doing so may be incorrect. Simply adding together these probabilities
could result in an overestimation of the true probability.
However, if we have, say 75 discs, and 3 are jazz while 2 are blues; the new probabilities are (3/75)(3/74)(3/73), (3/75)(2/74)(2/73), and (3/75)(2/74)(1/73). Yet, 3/74 is not significantly larger than 2/74; Nor is (3/73) significantly larger than
2/73 or 1/73. Thus, if the size of the population in which we draw our sample is large, then the variation from the true probability will be too insignificant to dramatically alter one's initial hypothesis.
My answers could be underestimations because they barely account for the impact of permutation and order on probability.
Oct 7

William F.