May 24 10:00 PM PDT  11:00 PM PDT Join a FREE 1day workshop and learn how to ace the GMAT while keeping your fulltime job. Limited for the first 99 registrants. May 25 07:00 AM PDT  09:00 AM PDT Attend this webinar and master GMAT SC in 10 days by learning how meaning and logic can help you tackle 700+ level SC questions with ease. May 27 01:00 AM PDT  11:59 PM PDT All GMAT Club Tests are free and open on May 27th for Memorial Day! May 27 10:00 PM PDT  11:00 PM PDT Special savings are here for Magoosh GMAT Prep! Even better  save 20% on the plan of your choice, now through midnight on Tuesday, 5/27
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 24 Jul 2004
Posts: 14

A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
Updated on: 07 Apr 2018, 12:39
Question Stats:
51% (01:25) correct 49% (01:18) wrong based on 1058 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
Official Answer and Stats are available only to registered users. Register/ Login.
Originally posted by Keen on 26 Jul 2004, 10:33.
Last edited by Bunuel on 07 Apr 2018, 12:39, edited 3 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.




Math Expert
Joined: 02 Sep 2009
Posts: 55231

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
23 Sep 2013, 07:53
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. Answer: D.
_________________




Senior Manager
Joined: 19 May 2004
Posts: 287

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
26 Jul 2004, 10:41
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



Intern
Joined: 24 Jul 2004
Posts: 14

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
26 Jul 2004, 11:14
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.



Joined: 31 Dec 1969
Location: Russian Federation
Concentration: Entrepreneurship, International Business
WE: Supply Chain Management (Energy and Utilities)

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
31 Jul 2004, 07:53
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



Senior Manager
Joined: 07 Sep 2010
Posts: 256

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
23 Sep 2013, 20:22
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/richhas3gr ... 55253.htmlCan you provide a high level conceptual knowledge as in when to consider cases and when not to? Pls help. Posted from my mobile device



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9230
Location: Pune, India

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
27 Sep 2013, 20:21
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/richhas3gr ... 55253.htmlCan 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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Manager
Joined: 20 Jan 2014
Posts: 140
Location: India
Concentration: Technology, Marketing

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
22 Sep 2014, 06:33
VeritasPrepKarishma wrote: 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/richhas3gr ... 55253.htmlCan 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. Thank You Karishma. I got a key concept here But i am more comfortable by Combination method 1C3*1C2/ (1C9 * 1C9) = 6/81 = 2/27 Now we can get this in two ways (as described by u) 2* 2/27 = 4/27
_________________
Consider +1 Kudos Please



SVP
Joined: 06 Nov 2014
Posts: 1877

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
16 Jul 2016, 12:45
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 Please explain your answer. I came up with but that is wrong. 3R, 4B, 2W balls. P(1R, 1W) = (3/9)*(2/9) = 1/3*2/9 = 2/27 Now the Red and the While balls can be drawn in any order Hence probability = 2*2/27 = 4/27 Correct Option: D



Intern
Joined: 14 Aug 2017
Posts: 43
Concentration: Operations, Social Entrepreneurship

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



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9230
Location: Pune, India

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
26 Oct 2017, 22:58
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
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Target Test Prep Representative
Status: Head GMAT Instructor
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2823

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#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. Answer: D
_________________
5star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews
If you find one of my posts helpful, please take a moment to click on the "Kudos" button.



EMPOWERgmat Instructor
Status: GMAT Assassin/CoFounder
Affiliations: EMPOWERgmat
Joined: 19 Dec 2014
Posts: 14188
Location: United States (CA)

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
25 Nov 2017, 15:15
Hi All, We're told that a bag contains 3 red, 4 black and 2 white balls. We're asked for the probability of drawing a red and a white ball in two successive draws, while replacing each ball after it is drawn. The prompt does NOT state that the red ball has to be drawn first, so there are two options that we have to consider: red first, white second and white first, red second. The probability of pulling a red first and a white second (with replacement) = (3/9)(2/9) = 6/81 The probability of pulling a while first and a red second (with replacement) = (2/9)(3/9) = 6/81 The total probability of pulling a red and white ball is 6/81 + 6/81 = 12/81 = 4/27 Final Answer: GMAT assassins aren't born, they're made, Rich
_________________
760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com*****Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*****
Rich Cohen
CoFounder & GMAT Assassin
Special Offer: Save $75 + GMAT Club Tests Free
Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/



NonHuman User
Joined: 09 Sep 2013
Posts: 10980

Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
Show Tags
26 Nov 2018, 08:27
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). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: A bag contains 3 red, 4 black and 2 white balls. What is the probabili
[#permalink]
26 Nov 2018, 08:27






