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A box contains 3 yellow balls and 5 black balls. One by one

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A box contains 3 yellow balls and 5 black balls. One by one [#permalink] New post 09 Feb 2010, 21:00
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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
[Reveal] Spoiler: OA

Last edited by Bunuel on 22 Jun 2012, 01:28, edited 1 time in total.
Added the answer choices and the OA.
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Re: probability question please solve [#permalink] New post 09 Feb 2010, 22:54
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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] New post 10 Feb 2010, 04:56
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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] New post 10 Feb 2010, 11:35
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] New post 10 Feb 2010, 11:57
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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] New post 10 Feb 2010, 20:36
Thanks for the explanation. Answer is 5/8
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Re: probability question please solve [#permalink] New post 10 Feb 2010, 23:35
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] New post 12 Feb 2010, 06:49
I didnt realize it was so straight forward --
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Re: probability question please solve [#permalink] New post 16 Feb 2010, 09:38
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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] New post 16 Feb 2010, 10:19
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] New post 22 Jun 2012, 01:23
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] New post 22 Jun 2012, 01:28
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Re: probability question please solve [#permalink] New post 26 Jun 2012, 04:39
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] New post 27 Jun 2012, 01:06
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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: A box contains 3 yellow balls and 5 black balls. One by one [#permalink] New post 12 Feb 2014, 21:21
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Re: probability question please solve [#permalink] New post 20 Apr 2014, 12:13
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?


Hi Bunuel,

How can we say that the probability stays the same -- aren't there unknowns involved?

What I mean is, if the first ball was randomly selected to be Black, wouldn't that change the probability of the second ball being black to 4/7? What I mean by that is -- how is probability remaining constant if *some* card is being chosen, even if we don't know what the card being chose is?

I tried to follow this approach and was completely off -- can you please explain why? P=C\frac{8}{5}*\frac{5}{8} * \frac{3}{8}^7
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Re: probability question please solve [#permalink] New post 21 Apr 2014, 00:02
Expert's post
russ9 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?


Hi Bunuel,

How can we say that the probability stays the same -- aren't there unknowns involved?

What I mean is, if the first ball was randomly selected to be Black, wouldn't that change the probability of the second ball being black to 4/7? What I mean by that is -- how is probability remaining constant if *some* card is being chosen, even if we don't know what the card being chose is?

I tried to follow this approach and was completely off -- can you please explain why? P=C\frac{8}{5}*\frac{5}{8} * \frac{3}{8}^7


If we knew that the firs ball was black, then yes, the probability that the second ball will be black would be 3/7/ Nut notice that for our original question we don't know the results of 1st, 2nd and 3rd draws...

Hope that similar questions will help to understand better:
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
a-medical-researcher-must-choose-one-of-14-patients-to-127396.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html#p747637
new-set-of-mixed-questions-150204-100.html#p1208473
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Re: probability question please solve   [#permalink] 21 Apr 2014, 00:02
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