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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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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
[Reveal] Spoiler: OA

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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: 3 yellow balls and 5 black balls [#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: 3 yellow balls and 5 black balls [#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: 3 yellow balls and 5 black balls [#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: 3 yellow balls and 5 black balls [#permalink]

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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: 3 yellow balls and 5 black balls [#permalink]

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New post 10 Aug 2010, 17:35
AKProdigy87 wrote:
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.


So you are saying it does not matter if with or without replacement? I mean the probablity of drawing a ball from a mixture of colored balls in the nth drawing is simply the fraction of that color?
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Re: 3 yellow balls and 5 black balls [#permalink]

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New post 07 Jun 2012, 00:56
tejal777 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?
1/4
1/2
1/2
5/8
2/3
3/4



Can some explain this concept please :

How does the fourth ball does not depend of the first 3 balls ?

if the first 3 are yellow then the 4 one has to be black , right?

on the other hand if we have BBB or BYB then the fourth one can be Y so it looks like there is some dependence on the first 3 balls .

can some one please explain this with clear concept. Highly appreciated .

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Re: 3 yellow balls and 5 black balls [#permalink]

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New post 07 Jun 2012, 02:14
Hi,

My approach,

_ _ _ _ (representing 4 draw)
total no. of ways in which 4 balls can be drawn = 8P4
_ _ _ B (when 4th ball is black) = 7P4

Thus probability of 4th ball being black = 7P4 / 8P4
=1/8
Now, I believe since 5 black balls are there, answer would be 5/8
but all black balls being same, so the answer should be 1/8.

Any comments?

Regards,

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Re: 3 yellow balls and 5 black balls [#permalink]

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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

Last edited by manulath on 07 Jun 2012, 03:11, edited 1 time in total.

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Re: 3 yellow balls and 5 black balls [#permalink]

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New post 07 Jun 2012, 03:08
manulath wrote:
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 = [(8!)/ (5!x3!)]/[(7!)/ (4!x3!)]

Ans = 5/8


@ manunath +1


[(7!)/ (4!x3!)] / [(8!)/ (5!x3!)] = 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 07 Jun 2012, 11:36
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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. 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).

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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, 12:54
Bunuel, I don't get it even after reading the explanation. Aren't we reducing the total number of balls with each drawing?

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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, 12:58
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?


Yes, but we don't know the results of these drawings. Have you checked the links in previous posts? They might help.
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Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]

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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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New post 04 Nov 2015, 20:09
I simply calculated the total possibilities 8!/3!5! = 56

Then fixed the 4th position as Black, then calculated the possibilities - 7!/3!4! = 35

therefore 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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SravnaTestPrep wrote:
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.



This explanation is pretty much clear. I was grappling with confusion, as I could not figure out why the probability remains the same even without replacement. Thanks a lot

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Re: A box contains 3 yellow balls and 5 black balls. One by one   [#permalink] 10 Oct 2017, 23:10
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