At a party, he randomly chooses 5 of the disks and loads them in a 5-disk player. What is the probability that:

a. The first disk is Jazz?

b. None of the Hard-Rock disks is loaded?

c. All five disks are either Jazz or Blues?

d. The first disk is a Blues disk, and the fifth disk is not a Blues disk?

e. The middle three disks are Soft-Rock?

f. ONLY the middle three disks are Soft-Rock?

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William F. | Math & Computer Science TutorMath & Computer Science Tutor

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.

a. He can choose Jazz disk 15 ways. Overall there are 75 possible outcomes for the first disk. Therefore, probability is 15/75=1/5 or 20%.

b. He can pick 5 disks out of 75 by C_{5}^{75} ways, here C_{5}^{75} is the binomial coefficient. This is the total number of outcomes. If no hard-rock disks are chosen , then it can happen by C_{5}^{53} ways, since there are 53 non-hard-rock disks. The probability is C_{5}^{53}/C_{5}^{75}=(53!/75!)*(70!/48!)=(53*52*51*50*49)/(71*72*73*74*75)=(13*17*49*53)/(18*37*71*73)=573937/3451878≈0.166

c. Desired outcome can happen by C_{5}^{35} times, since there are 35 Blues or Jazz disks. The total number of outcomes has been found in the previous section. The probability is C_{5}^{35}/C_{5}^{75}=(35!/75!)*(70!/30!)=(35*34*33*32*31)/(71*72*73*74*75)=(4*7*11*17*31)/(5*9*37*71*73)≈0.0188.

The rest will come later.

Total disks: 15+20+18+22=75

a) Probability[Jazz] = 15/75 = 0.2

b) Probability[Hard-Rock] on first draw: 22/75

Probability[not Hard-Rock] on first draw: 1-22/75=(75-22)/75=53/75

Pr[HR] on 2nd draw if first disk was not HR: 22/74

Pr[HR] on 2nd draw if first disk was not HR: 22/74

Pr[not HR] on 2nd draw if first disk was not HR: 1-22/74=52/74

Pr[HR] on 3nd draw if previous disks were not HR: 22/73

Pr[not HR] on 3nd draw if previous disks were not HR: 1-22/73=51/73

So, we have five separate drawings with slightly decreasing odds of NOT drawing Hard-Rock:

53/75 * 52/74 * 51/73 * 50/72 * 49/71 = 1/3*13/37*17/73*53/6*49/71=573937/3451878=.166

c) 1st: (15+20)/75 = 35/75 = 7/15

2nd: 34/74 = 17/37

3rd: 33/73

4th: 32/72 = 4/9

5th: 31/71

So, 7/15*17/37*33/73*4/9*31/71 = .019

Whew, gotta go.

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