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araspai
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If the CD player had already played two songs that were not his favorite, and the randomizer does not repeat songs until they have all been played, then your answer is correct - 1/12.
But as it stands, if the randomizer is random, then the odds are always 1/14.

With your logic, what are the odds that the 14th song played is his favorite? 1? Does that make sense?

What if he is on the airplane for 8 hrs. and the CD has been playing the whole time...what are the odds that the 60th song will be his favorite? By your logic, this would be negative!! And the correct answer is clearly 1/14.
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Is this the right way?

Total songs = 14
Not favorites = 13
Favorites = 1

P(Song 3 is favorite) = P(Song 1 not fav) * P(Song 2 not fav) * P(Song 3 fav)

= 13/14 * 12/13 * 1/12

= 1/14
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rshanm2
Is this the right way?

Total songs = 14
Not favorites = 13
Favorites = 1

P(Song 3 is favorite) = P(Song 1 not fav) * P(Song 2 not fav) * P(Song 3 fav)

= 13/14 * 12/13 * 1/12

= 1/14


I would say that this is NOT the right way, but I could be wrong. Nowhere in the question does it say that the first two songs played are not his favorites. Thus the prob of this event is independant of any other event. But, your answer works, and in this case you would get the points.



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