What is the probability that the first Beatles song heard is the fifth song played?

Cade

1. The problem statement, all variables and given/known data

A particular iPod playlist contains 100 songs, 10 of which are by the Beatles. What is the probability that the
first Beatles song heard is the fifth song played?

2. Relevant equations

Combinatorics: permutation, combinations.

3. The attempt at a solution

The number of desired outcomes for the event is:
90*89*88*87 (choose 4 of 90 songs that aren't Beatles)
* 10 (choose 1 song that is Beatles)

The sample space is:
100*99*98*97*96 (choose 5 of 100 songs from the playlist)

The correct answer is (90*89*88*87*10)/(100*99*98*97*96).

How would I solve this problem using the formulae for combinations? I tried it this way, but the answer was incorrect: C(90,4)*C(10,1)/C(100,5)

As I understand it, the order within the non-Beatles set is irrelevant, but using P(90,4)P(10,1)/P(100,5) works. What am I missing?

Ray Vickson

In this problem the order is important, so combinations are misleading or downright incorrect. Rather than using canned formulas, I prefer to think directly about such problems: the first song must be any of the 90 non-beatles songs, then the second one must be any of the 89 remaining non-beatles songs, etc.

RGV

Cade

How do you recognize that order is important in this problem? Could songs 1-4 not be any of the 90 non-Beatles song?

Or, have I misunderstood the question, and is the question actually asking ways entire the playlist can be ordered that the first Beatles song is #5?

Ray Vickson

Read the question again. It said the first Beatles song is song #5. Yes, indeed, songs 1-4 are any of the 90 non-beatles songs (presumably played only once, not repeated); that is exactly what I said in my first reply.

RGV

Cade

I meant, for songs 1-4, given that they are non-Beatles, does individual order within the set of non-Beatles matter?

HallsofIvy

There are 100 songs, 10, or 10%, of which are Beatles songs. The probabilty the first song is not a Beatles song is 90/100= 9/10. That leaves 99 songs for the second song, 89 of which are not Beatles songs. The probabilty that second song is not a Beatle song is 89/99. Similarly, the probability the third and fourth songs are not Beatles songs is 88/98 and 87/97. The probability that the fifth song is a Beatle's song is 10/96. The probability that the fifth song is the first Beatles song is (90/100)(89/99)(88/98)(87/97)(10/96) just as you say. Now, we can write that numerator, 90(89)(88)(87)(10) as [tex](90!/86!)(10)= (_{90}P_{86})(10)[/tex] and the denominator 100(99)(98)(97)(96) as [tex]100!/95!= _{100}P_{95}[/tex]. Therefore, we can write the probability as [tex]\frac{_{90}P_{86}(10)}{_{100}P_{95}}[/tex]. I don't know that there is any benefit to writing it that way.

Ray Vickson

No, the order of songs 1-4 does not matter.

RGV

Cade

Oh, I see. Thanks for your help, I understand the problem now.