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A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

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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 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).

Re: probability question please solve [#permalink]

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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.
_________________

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?
_________________

Re: probability question please solve [#permalink]

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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).

Re: probability question please solve [#permalink]

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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.

Re: probability question please solve [#permalink]

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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?"

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.
_________________

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

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12 Feb 2014, 21:21

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: probability question please solve [#permalink]

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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? \(P=C\frac{8}{5}*\frac{5}{8} * \frac{3}{8}^7\)

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...

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

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04 May 2015, 20:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

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12 Jul 2016, 19:16

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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