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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]

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New post 29 Sep 2009, 00:14
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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

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

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

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

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New post 29 Sep 2009, 00:34
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The answer should be C: 5/8.

The initial probability of drawing a certain ball is the same as the probability of drawing that certain ball at any future point. Essentially, the "Expected" proportion of ball distribution for any round (3rd ball drawn, 4th ball drawn, etc.) is the same as the initial proportion.
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 29 Sep 2009, 02:54
AKProdigy87, that's correct but I wonder how many people would be quick to realize that the probability is that same as in the first draw :).
The other route of doing a tree diagram takes 4 mins at least !
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 29 Sep 2009, 04:33
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jeffm wrote:
AKProdigy87, that's correct but I wonder how many people would be quick to realize that the probability is that same as in the first draw :).


It's very intuitive. A formal proof would involve noticing that for each sequence of 4 balls that get pulled out (x1,x2,x3,x4) there is a corresponding sequence (x4,x2,x3,x1) that can be pull out with the same probability (we pull out 4 balls and then order them in a sequence) -> Prob(x4=black) = Prob(x1=black).
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 29 Sep 2009, 10:40
AKP : The initial probability of drawing a certain ball is the same as the probability of drawing that certain ball at any future point. Essentially, the "Expected" proportion of ball distribution for any round (3rd ball drawn, 4th ball drawn, etc.) is the same as the initial proportion.

question: will this hold for both with replacement and without replacement?
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 29 Sep 2009, 11:41
2
manojgmat wrote:
AKP : The initial probability of drawing a certain ball is the same as the probability of drawing that certain ball at any future point. Essentially, the "Expected" proportion of ball distribution for any round (3rd ball drawn, 4th ball drawn, etc.) is the same as the initial proportion.

question: will this hold for both with replacement and without replacement?


With replacement, it's pretty obvious that the probability will hold constant. Without replacement, it's a bit harder to see. Imagine if you had 3 balls (2 black, 1 yellow). What's the probability you draw a black ball on the 3rd draw? It will be 2/3. The reason is that 2/3 of the time, there will be black ball left. So the expected value of the colour of the third ball is 2/3 Black and 1/3 Yellow. The same principle applies to a 4th draw of 8 balls.
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 09 Feb 2010, 22:54
2
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: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 10 Feb 2010, 11:35
1
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: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 10 Feb 2010, 11:57
2
1
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: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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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]

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New post Updated on: 07 Jun 2012, 03:11
12
5
AKProdigy87 wrote:
The answer should be C: 5/8.

The initial probability of drawing a certain ball is the same as the probability of drawing that certain ball at any future point. Essentially, the "Expected" proportion of ball distribution for any round (3rd ball drawn, 4th ball drawn, etc.) is the same as the initial proportion.


That was simply amazing

My complicated version of your simple approach

Let the 5 black balls be BBBBB and 3 Red Balls be RRR

They can be arranged in 8 slots _ _ _ _ _ _ _ _
in (8!)/ (5!x3!)

If the fourth slot is Black ball then the arrangement will be to fill
_ _ _ B _ _ _ _
we have 7 slots and 4 Black (BBBB) and 3 Red (RRR)

They can be arranged in (7!)/ (4!x3!)

Hence required probability = [(7!)/ (4!x3!)] / [(8!)/ (5!x3!)]

Ans = 5/8

Originally posted by manulath on 07 Jun 2012, 02:37.
Last edited by manulath on 07 Jun 2012, 03:11, edited 1 time in total.
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 07 Jun 2012, 11:36
1
5
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. 1/2
D. 5/8
E. 2/3

There is a shortcut solution for this problem:

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).

Similar questions to practice:
a-box-contains-3-yellow-balls-and-5-black-balls-one-by-one-90272.html
a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html
each-of-four-different-locks-has-a-matching-key-the-keys-101553.html
if-40-people-get-the-chance-to-pick-a-card-from-a-canister-97015.html
new-set-of-mixed-questions-150204-100.html#p1208473
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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

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New post 27 Jun 2012, 01:06
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: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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New post 26 Jun 2013, 18:19
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sgclub wrote:
Bunuel, I don't get it even after reading the explanation. Aren't we reducing the total number of balls with each drawing?


Consider 2 black balls and 1 white ball

At the first draw, probability that the ball is black is 2/3

Once a ball is removed, remaining balls may be 1b and 1W, 1b and 1W or 2b

At the second draw probability that the ball is black is, 1/3*1/2 + 1/3*1/2 + 1/3*1 = 2/3

We can see that the probability does not change even without replacement.
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Re: A box contains 3 yellow balls and 5 black balls. One by one  [#permalink]

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