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A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
 09 Feb 2010, 22:00
Question Stats:
88% (01:11) correct
11% (03:05) wrong based on 0 sessions
A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? A. 1/4 B. 1/2 C. 5/8 D. 2/3 E. 3/4
Last edited by Bunuel on 22 Jun 2012, 02:28, edited 1 time in total.
Added the answer choices and the OA.
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Re: probability question please solve [#permalink]
09 Feb 2010, 23:54
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3 yellow 5 black
possible outcomes for the first 3 tries and 4th is black
yyyb - 3/8 * 2/7 * 1/6 * 5/5 = 30/(8*7*6*5)
yybb - 3/8 * 2/7 * 5/6 * 4/5 = 120/(8*7*6*5)
ybyb - 3/8 * 5/7 * 2/6 * 4/5 = 120/(8*7*6*5)
ybbb -3/8 * 5/7 * 4/6 * 3/5 = 180/(8*7*6*5)
bbbb - 5/8 * 4/7 * 3/6 * 2/5 = 120/(8*7*6*5)
bbyb - 5/8 * 4/7 * 3/6 * 3/5 = 180/(8*7*6*5)
bybb - 5/8 * 3/7 * 4/6 * 3/5 = 180/(8*7*6*5)
byyb - 5/8 * 3/7 * 2/6 * 4/5 = 120/(8*7*6*5)
total = 1050/(8*7*6*5)= 5/8
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Re: probability question please solve [#permalink]
10 Feb 2010, 05:56
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dmetla wrote: A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results). Hope it's clear.
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Re: probability question please solve [#permalink]
10 Feb 2010, 12:35
Bunuel wrote: The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
Hope it's clear. Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black.
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Re: probability question please solve [#permalink]
10 Feb 2010, 12:57
BarneyStinson wrote: Bunuel wrote: The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
Hope it's clear. Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black. Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8. Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change? Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade? I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it?
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Re: probability question please solve [#permalink]
10 Feb 2010, 21:36
Thanks for the explanation. Answer is 5/8
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Re: probability question please solve [#permalink]
11 Feb 2010, 00:35
Bunuel wrote: dmetla wrote: A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black? The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results). Hope it's clear. Nice explanation....Thanks
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Re: probability question please solve [#permalink]
12 Feb 2010, 07:49
I didnt realize it was so straight forward --
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Re: probability question please solve [#permalink]
16 Feb 2010, 10:38
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I thought of it this way:
All possibilities with fourth ball fixed as black XXXBXXXX
7 spots, three filled by yellow, four filled by black:
(7!)/(3!*4!) = 35
All total possibilities of arranging 3 yellow and 5 black balls:
(8!)/(3!*5!) = 56
35/56 = 5/8
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Re: probability question please solve [#permalink]
16 Feb 2010, 11:19
Bunuel wrote: BarneyStinson wrote: Bunuel wrote: The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
Hope it's clear. Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black. Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8. Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change? Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade? I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it? Nice explanation didn't thought after reading questin that it was so simple.
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Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
22 Jun 2012, 02:23
I got answer is 5/8 but choices are not displaying!
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Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
22 Jun 2012, 02:28
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Re: probability question please solve [#permalink]
26 Jun 2012, 05:39
Bunuel wrote: BarneyStinson wrote: Bunuel wrote: The initial probability of drawing black ball is 5/8. Without knowing the other results, the probability of drawing black ball will not change for ANY successive drawing: second, third, fourth... The same for yellow ball probability of drawing yellow ball is 3/8, the probability that 8th ball drawn is yellow is still 3/8. There is simply no reason to believe WHY is any drawing different from another (provided we don't know the other results).
Hope it's clear. Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black. Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8. Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change? Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade? I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it? @Bunuel please correct me if I am wrong "if it is said that first card is sprade, then what is the probability of the 3rd card would be a sprade?-- in this case we wouldn't consider initial probability?"
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Re: probability question please solve [#permalink]
27 Jun 2012, 02:06
marzan2011 wrote: Bunuel wrote: BarneyStinson wrote: Selected balls aren't replaced. The first ball to be black is 5/8 and to be yellow is 3/8. But, for the second ball selected to be black, it could still be 5/7 or 4/7. So, for the probability of third ball to be black, the first two could be black, or just one, or none could be black at all. Going by this counting technique..., I don't see how to calculate the fourth ball's probability to be black. Consider this: there are 8 cards, 5 spades, 3 hearts. What is the probability that the first card you pick will be spade? Clearly 5/8. Now next question, suppose I throw away three cards at random, not telling you which cards I thrown away. What is the probability NOW that you pick a spade out of 5 remaining cards? I've just reduced the sample from which you pick, but does the probability changed? WHY should it change? Another question, the same cards 5 spades and 3 hearts: I say that you can pick one card but only out of randomly selected 5 cards. What is the probability that the card you pick will be spade? I think there is no difference in ALL above cases and the probability will remain 5/8. If it changed, increased (or decreased), it would mean that the probability of picking heart on the other hand decreased (or increased). Why would it? @Bunuel please correct me if I am wrong "if it is said that first card is sprade, then what is the probability of the 3rd card would be a sprade?-- in this case we wouldn't consider initial probability?" If the first card is spade then we are left with 4 spades and 3 hearts and the probability that 3rd (or any other card) will be a spade is 4/7.
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Re: probability question please solve
[#permalink]
27 Jun 2012, 02:06
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