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A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
29 Sep 2009, 00:14

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C

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Difficulty:

55% (hard)

Question Stats:

67% (02:27) correct
33% (01:27) wrong based on 130 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?

Re: 3 yellow balls and 5 black balls [#permalink]
29 Sep 2009, 00:34

5

This post received KUDOS

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.

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

Re: 3 yellow balls and 5 black balls [#permalink]
29 Sep 2009, 04:33

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).

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

Re: 3 yellow balls and 5 black balls [#permalink]
29 Sep 2009, 11:41

2

This post received KUDOS

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.

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

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

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

Re: 3 yellow balls and 5 black balls [#permalink]
07 Jun 2012, 02:37

6

This post received KUDOS

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.

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

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
07 Jun 2012, 11:36

1

This post received KUDOS

Expert's post

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).

Re: 3 yellow balls and 5 black balls [#permalink]
22 Nov 2012, 01:12

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.

I spent 2 minutes on drawing the tree. I don't understand much in probability but the tree shows that there is a total of 15 ways of drawing and 8 of them are black at the end. For certain, it cannot be 1/2 but it's neither 5/8!

Re: 3 yellow balls and 5 black balls [#permalink]
22 Nov 2012, 03:35

Expert's post

felixjkz 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.

I spent 2 minutes on drawing the tree. I don't understand much in probability but the tree shows that there is a total of 15 ways of drawing and 8 of them are black at the end. For certain, it cannot be 1/2 but it's neither 5/8!

The correct answer to this question is 5/8, AKProdigy87's solution is right.

Re: 3 yellow balls and 5 black balls [#permalink]
22 Nov 2012, 09:49

Bunuel wrote:

felixjkz 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.

I spent 2 minutes on drawing the tree. I don't understand much in probability but the tree shows that there is a total of 15 ways of drawing and 8 of them are black at the end. For certain, it cannot be 1/2 but it's neither 5/8!

The correct answer to this question is 5/8, AKProdigy87's solution is right.

Check similar questions to practice:

Hope it helps.

Well, it might be right, but I just don't know how to interpret this:

-------------- B 3-------B -------------- Y 2----B -------------- B 3-------Y --------------Y B

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
17 Feb 2014, 07:41

Bunuel 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?

Yes, but we don't know the results of these drawings. Have you checked the links in previous posts? They might help.

Hi Bunuel, could you please explain what happens in the case that we draw yellow in the first three draws, then the probability of the 4th ball being black would be 1 right? I understand the approach to the question just trying to figure out if such scenario could be applicable to this question in any way

Re: A box contains 3 yellow balls and 5 black balls. One by one [#permalink]
17 Feb 2014, 07:46

Expert's post

jlgdr wrote:

Bunuel 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?

Yes, but we don't know the results of these drawings. Have you checked the links in previous posts? They might help.

Hi Bunuel, could you please explain what happens in the case that we draw yellow in the first three draws, then the probability of the 4th ball being black would be 1 right? I understand the approach to the question just trying to figure out if such scenario could be applicable to this question in any way

Many thanks Cheers J

Obviously the probability of selecting black ball when there are only black balls in the box will be 1. Do not understand your question... _________________

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