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Bill's compact disc player randomly plays a song. If bill

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araspai
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Bill's compact disc player randomly plays a song. If bill [#permalink] New post   20 Aug 2003, 05:53
Bill's compact disc player randomly plays a song. If bill plays a disc with 14 songs, What are the chances that the third song he hears will be his favorite?
a. 1/14 b. 1/12 c. 1/11 d. 3/14 e. 1/3



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Re: Bill's compact disc player randomly plays a song. If bill [#permalink] New post   20 Aug 2003, 08:40
javropu wrote:
1/14?


That must be correct. In fact, the prob that ANY song he hears will be is favorite song is 1/14.
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Re: Bill's compact disc player randomly plays a song. If bill [#permalink] New post   20 Aug 2003, 22:40
I think the answer is B - 1/12.

The probability of any one song being his favourite is 1/14.

But in this case the CD player, randomly picks the songs. The probability of first song played being his favourite is 1/14. The second song played being his favourite is 1/13, as only 13 more are left after the first song has been played. Using the same logic, the third song played being his favourite is 1/12.

If the CD player is playing the songs serially (not random selection), what is the probability of 3rd song being his favourite?

1/14?!
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Re: Bill's compact disc player randomly plays a song. If bill [#permalink] New post   20 Aug 2003, 23:05
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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Re: Bill's compact disc player randomly plays a song. If bill [#permalink] New post   22 Aug 2003, 07:13
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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Re: Bill's compact disc player randomly plays a song. If bill [#permalink] New post   22 Aug 2003, 09:04
rshanm2 wrote:
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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Re: Bill's compact disc player randomly plays a song. If bill [#permalink]
22 Aug 2003, 09:04
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