It is currently 20 Nov 2017, 14:48

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# A bag contains 3 red, 4 black and 2 white balls. What is the

Author Message
TAGS:

### Hide Tags

Intern
Joined: 24 Jul 2004
Posts: 14

Kudos [?]: 26 [2], given: 0

A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

26 Jul 2004, 10:33
2
KUDOS
23
This post was
BOOKMARKED
00:00

Difficulty:

65% (hard)

Question Stats:

50% (01:05) correct 50% (00:55) wrong based on 674 sessions

### HideShow timer Statistics

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9
[Reveal] Spoiler: OA

Last edited by Bunuel on 25 Oct 2017, 01:56, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.

Kudos [?]: 26 [2], given: 0

Senior Manager
Joined: 19 May 2004
Posts: 291

Kudos [?]: 16 [2], given: 0

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

26 Jul 2004, 10:41
2
KUDOS
A is correct if you want to get a Red ball First and then a White one.
If the order doesn't matter, then the answer is D.

(2/27)*2 = 4/27

Kudos [?]: 16 [2], given: 0

Manager
Joined: 19 Jul 2004
Posts: 75

Kudos [?]: [0], given: 0

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

26 Jul 2004, 10:46
agree with Dookie....here the events are independent....
_________________

To Strive, To Seek and Not to Yield

Kudos [?]: [0], given: 0

Manager
Joined: 18 Jun 2004
Posts: 102

Kudos [?]: 121 [0], given: 0

Location: san jose , CA
Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

26 Jul 2004, 11:02
Karthik wrote:
agree with Dookie....here the events are independent....

can you post a detail explanation ? thanks
_________________

---- Hero never chooses Destiny
Destiny chooses Him ......

Kudos [?]: 121 [0], given: 0

Intern
Joined: 24 Jul 2004
Posts: 14

Kudos [?]: 26 [0], given: 0

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

26 Jul 2004, 11:14
1
This post was
BOOKMARKED
rahul wrote:
Karthik wrote:
agree with Dookie....here the events are independent....

can you post a detail explanation ? thanks

This is how I came up with 2/27. the probability of getting one red is 3/9 (nine is the total number of balls). The probability of getting a white ball is 2/9 (9 again because the ball is put back after each draw) so 3/9*2/9 + 6/81 = 2/27

according to Dookie (who is right) if they are asking for the balls to draw one of the the other, which they are (successive draws) you have to multiply 2/27 by 2 = 4/27.

Kudos [?]: 26 [0], given: 0

Joined: 31 Dec 1969

Kudos [?]: [1], given:

Location: Russian Federation
GMAT 3: 740 Q40 V50
GMAT 4: 700 Q48 V38
GMAT 5: 710 Q45 V41
GMAT 6: 680 Q47 V36
GMAT 9: 740 Q49 V42
GMAT 11: 500 Q47 V33
GMAT 14: 760 Q49 V44
WE: Supply Chain Management (Energy and Utilities)
Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

31 Jul 2004, 07:53
1
KUDOS
Here is how I solved it

First of all we have
Probability of drawing a Red ball is 3/9
Probability of drawing a White ball is 3/9

There are two ways in which the balls can be drawn

Case 1: Red ball in the first draw and white in the second draw
Hence the combined Probability is 3/9*2/9=6/81

Case 2: White ball in the first draw and red in the second draw
Hence the combined Probability is 2/9*3/9=6/81

both these cases satisfy our requirement
Hence either of them will do i.e OR
Hence the final probability comes to be
Case 1 OR Case 2 = 6/81 + 6/81 (OR means addition)
Hence the Ans is 12/81=4/27

Kudos [?]: [1], given:

Intern
Joined: 27 Jul 2004
Posts: 3

Kudos [?]: [0], given: 0

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

31 Jul 2004, 07:59
Sorry for the Typo above
the Probability of drawing the white ball is 2/9

I posted the above explanation but just forgot to login

Kudos [?]: [0], given: 0

Manager
Joined: 19 Jun 2003
Posts: 151

Kudos [?]: 31 [0], given: 0

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

05 Aug 2004, 19:51
Dookie wrote:
A is correct if you want to get a Red ball First and then a White one.
If the order doesn't matter, then the answer is D.

(2/27)*2 = 4/27

Dookie-

First red, then white:
drawing red: 3/9
drawing white: 2/9

3/9 * 2/9 = 2/27.

First white, the red:
2/9 * 3/9 = 2/27

how come order being matter would change the result?

Kudos [?]: 31 [0], given: 0

Manager
Joined: 28 Jul 2004
Posts: 179

Kudos [?]: 18 [0], given: 0

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

05 Aug 2004, 19:58
D is the right answer, without any mention about the other, we need to assume that the order is independent so it is

3/9 * 2/9 * 2 ( last 2 is for the order ) = 4/27

Kudos [?]: 18 [0], given: 0

Senior Manager
Joined: 07 Sep 2010
Posts: 330

Kudos [?]: 1053 [0], given: 136

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

23 Sep 2013, 07:36
Bumping this thread as I am looking for experts to provide an explanation and confirm the OA.

Thanks
_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Kudos [?]: 1053 [0], given: 136

Math Expert
Joined: 02 Sep 2009
Posts: 42269

Kudos [?]: 132827 [1], given: 12378

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

23 Sep 2013, 07:53
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
imhimanshu wrote:
Bumping this thread as I am looking for experts to provide an explanation and confirm the OA.

Thanks

A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

This is with replacement case.

$$P=2*\frac{3}{9}*\frac{2}{9}=\frac{4}{27}$$

We are multiplying by 2 as there are two possible wining scenarios RW and WR.

_________________

Kudos [?]: 132827 [1], given: 12378

Senior Manager
Joined: 07 Sep 2010
Posts: 330

Kudos [?]: 1053 [1], given: 136

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

23 Sep 2013, 20:22
1
KUDOS
Hello Bunuel,
I m a bit confused about when to consider and when not to consider. I am having a tough time understanding this concept. I was under the impression that in "with replacement" cases, we dont need to consider the cases, however in without replacement cases, scenarios needs to be considered.

In addition,I found this link, where the question is also testing the same concept, but we didn't consider the multiple cases here. Please clarify.
http://gmatclub.com/forum/rich-has-3-gr ... 55253.html

Can you provide a high level conceptual knowledge as in when to consider cases and when not to?
Pls help.

Posted from my mobile device
_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Kudos [?]: 1053 [1], given: 136

Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 7738

Kudos [?]: 17819 [2], given: 235

Location: Pune, India
Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

27 Sep 2013, 20:21
2
KUDOS
Expert's post
imhimanshu wrote:
Hello Bunuel,
I m a bit confused about when to consider and when not to consider. I am having a tough time understanding this concept. I was under the impression that in "with replacement" cases, we dont need to consider the cases, however in without replacement cases, scenarios needs to be considered.

In addition,I found this link, where the question is also testing the same concept, but we didn't consider the multiple cases here. Please clarify.
http://gmatclub.com/forum/rich-has-3-gr ... 55253.html

Can you provide a high level conceptual knowledge as in when to consider cases and when not to?
Pls help.

Posted from my mobile device

Responding to a pm:

The status of "replacement" has nothing to do with the "sequence". It only changes the probability.

Say we have 2 red and 3 white balls in a bag. We pull out two one after another with replacement. What is the probability that one is red and the other is white.
Now note that there are 4 ways in which you can pull out two balls from the bag:
1. You pull a Red and then a Red again RR - (2/5)*(2/5) (Note that it is with replacement)
2. You pull a Red and then a White RW - (2/5)*(3/5)
3. You pull a White and then a Red WR - (3/5)*(2/5)
4. You pull a White and then a White WW - (3/5)*(3/5)

Total probability = (2/5)*(2/5) + (2/5)*(3/5) + (3/5)*(2/5) + (3/5)*(3/5) = 1

In how many cases do we have a red and a white ball? In case 2 and case 3.
Probability of picking a red and a white with replacement = (2/5)*(3/5) + (3/5)*(2/5) = (3/5)*(2/5) * 2
Since the probability of picking a red and then a white is same as probability of picking a white and then a red, you simply write down one case and multiply it by 2. You do the same in case of 'without replacement' too. The only thing that changes is the probability.

Without Replacement:
1. You pull a Red and then a Red again RR - (2/5)*(1/4)
2. You pull a Red and then a White RW - (2/5)*(3/4)
3. You pull a White and then a Red WR - (3/5)*(2/4)
4. You pull a White and then a White WW - (3/5)*(2/4)
Probability of picking a red and a white WITHOUT replacement = (2/5)*(3/4) + (3/5)*(2/4) = (3/5)*(2/4) * 2

As for the link you have mentioned, this is exactly what is done there too. Check it out - I will show you how it is done there.
_________________

Karishma
Veritas Prep | GMAT Instructor
My Blog

GMAT On Demand Course $299 Free Online Trial Hour Kudos [?]: 541 [0], given: 23 Intern Joined: 14 Aug 2017 Posts: 17 Kudos [?]: 0 [0], given: 7 Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink] ### Show Tags 25 Oct 2017, 01:52 karishma Bunuel - Can you please explain as to where my concept is headed in the wrong direction? At first i saw that there are 9 balls and 2 balls are to be drawn, i selected the 2 balls out of 9 (9C2) giving me the total outcomes and next when i am supposed to draw the balls in the manner of white and red it clicked to me that there are 4 case, 1 - RR 2 - RW 3 - WR 4 - WW and i applied the same logic to the above 4 cases, but the answer did not match? Can you please help me with this. Thanks in advance. Kudos [?]: 0 [0], given: 7 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7738 Kudos [?]: 17819 [1], given: 235 Location: Pune, India Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink] ### Show Tags 26 Oct 2017, 22:58 1 This post received KUDOS Expert's post siddyj94 wrote: karishma Bunuel - Can you please explain as to where my concept is headed in the wrong direction? At first i saw that there are 9 balls and 2 balls are to be drawn, i selected the 2 balls out of 9 (9C2) giving me the total outcomes and next when i am supposed to draw the balls in the manner of white and red it clicked to me that there are 4 case, 1 - RR 2 - RW 3 - WR 4 - WW and i applied the same logic to the above 4 cases, but the answer did not match? Can you please help me with this. Thanks in advance. Note that you cannot use 9C2 here because you are drawing balls with replacement. 9C2 means draw 2 balls out of 9 or draw 1 out of 9 and then 1 out of 8. But here, we need to draw 1 out of 9 and then again 1 out of 9 (since the first ball is put back) Also, why only 4? Wouldn't we have other cases too such as a Red and a Black? Look at the solutions above to see how to solve it. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

Veritas Prep Reviews

Kudos [?]: 17819 [1], given: 235

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1684

Kudos [?]: 906 [0], given: 5

Re: A bag contains 3 red, 4 black and 2 white balls. What is the [#permalink]

### Show Tags

29 Oct 2017, 07:41
Keen wrote:
A bag contains 3 red, 4 black and 2 white balls. What is the probability of drawing a red and a white ball in two successive draws, each ball being put back after it is drawn?

(A) 2/27
(B) 1/9
(C) 1/3
(D) 4/27
(E) 2/9

There are 9 balls in the bag, so the probability of drawing a red ball is P(Red) = 3/9 = ⅓, and the probability of drawing a white ball is P(White) = 2/9. We will draw two balls, replacing each ball after it is drawn.

The probability of drawing a red ball first and then a white ball is: P(Red) x P(White) = ⅓ x 2/9 = 2/27. But we can also draw a white ball first and then a red ball: P(White) x P(Red) = 2/9 x ⅓ = 2/27. Either of these outcomes satisfies our outcome of interest, and so we add the two probabilities: 2/27 + 2/27 = 4/27.

_________________

Jeffery Miller

GMAT Quant Self-Study Course
500+ lessons 3000+ practice problems 800+ HD solutions

Kudos [?]: 906 [0], given: 5

Re: A bag contains 3 red, 4 black and 2 white balls. What is the   [#permalink] 29 Oct 2017, 07:41
Display posts from previous: Sort by