GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 17 Dec 2018, 07:38

# R1 Decisions:

Michigan Ross Chat (US calls are expected today)  |  UCLA Anderson Chat  (Calls expected to start at 7am PST; Applicants from Asia will hear first)

### 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 December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• ### 10 Keys to nail DS and CR questions

December 17, 2018

December 17, 2018

06:00 PM PST

07:00 PM PST

Join our live webinar and learn how to approach Data Sufficiency and Critical Reasoning problems, how to identify the best way to solve each question and what most people do wrong.
• ### R1 Admission Decisions: Estimated Decision Timelines and Chat Links for Major BSchools

December 17, 2018

December 17, 2018

10:00 PM PST

11:00 PM PST

From Dec 5th onward, American programs will start releasing R1 decisions. Chat Rooms: We have also assigned chat rooms for every school so that applicants can stay in touch and exchange information/update during decision period.

# M32-04

Author Message
TAGS:

### Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 51262

### Show Tags

13 Jun 2016, 05:40
1
5
00:00

Difficulty:

65% (hard)

Question Stats:

49% (00:36) correct 51% (00:42) wrong based on 88 sessions

### HideShow timer Statistics

A bag contains only blue and red balls. What is the probability that the second ball we pick will be blue, if the first ball we picked was red?

(1) Initially, the number of red balls was one more than the number of blue balls.

(2) Initially, the number of red balls was 7.

_________________
Math Expert
Joined: 02 Sep 2009
Posts: 51262

### Show Tags

13 Jun 2016, 05:40
1
1
Official Solution:

A bag contains only blue and red balls. What is the probability that the second ball we pick will be blue, if the first ball we picked was red?

Say, initially there were $$b$$ blue balls and $$r$$ red balls. The probability that the second ball we pick will be blue, if the first ball we picked was red will therefore be $$\frac{b}{b+r-1}$$

(1) Initially, the number of red balls was one more than the number of blue balls. This implies that $$r=b+1$$, hence the probability is $$\frac{b}{b+r-1}=\frac{b}{b+(b+1)-1}=\frac{b}{2b}=\frac{1}{2}$$. Sufficient.

(2) Initially, the number of red balls was 7. Clearly insufficient.

_________________
Intern
Joined: 30 Mar 2016
Posts: 1

### Show Tags

17 Jun 2016, 20:12
I think this is a poor-quality question and I don't agree with the explanation. I chose answer E because the question doesn't mention whether or not the first ball picked was put back in the bag.
Math Expert
Joined: 02 Sep 2009
Posts: 51262

### Show Tags

19 Jun 2016, 09:24
edgargee wrote:
I think this is a poor-quality question and I don't agree with the explanation. I chose answer E because the question doesn't mention whether or not the first ball picked was put back in the bag.

If it were with replacement it would have been explicitly mentioned.
_________________
Intern
Joined: 10 Aug 2013
Posts: 47
Location: United Kingdom
Concentration: Strategy, Marketing
GMAT 1: 710 Q49 V38
GMAT 2: 730 Q49 V40
GPA: 3
WE: Consulting (Consulting)

### Show Tags

22 Aug 2016, 20:34
This probability defined for picking blue given that first red doesnt seem right to me as its a conditional probability and should not be straight up b/(r+b-1). I might be wrong but would be good to confirm this.
Math Expert
Joined: 02 Sep 2009
Posts: 51262

### Show Tags

24 Aug 2016, 03:57
sharmaanurag wrote:
This probability defined for picking blue given that first red doesnt seem right to me as its a conditional probability and should not be straight up b/(r+b-1). I might be wrong but would be good to confirm this.

No, it's right. When you pick 1 red you are left with b blue balls and r red balls. The probability that the second ball we pick will be blue, if the first ball we picked was red will therefore be $$\frac{b}{b+(r-1)}$$
_________________
Intern
Joined: 06 May 2016
Posts: 14

### Show Tags

06 Sep 2016, 10:23
Bunuel,

I think I am missing something here. I tried with an example:

Lets blue balls be 3 and red balls be 4 :

4/7 * 3/6 = 2/7

but if

B=5 and R=6 then it will be 3/11

How you are getting 1/2 as the answer?
Intern
Joined: 22 Aug 2016
Posts: 1

### Show Tags

07 Sep 2016, 07:12
3
gmatravi wrote:
Bunuel,

I think I am missing something here. I tried with an example:

Lets blue balls be 3 and red balls be 4 :

4/7 * 3/6 = 2/7

but if

B=5 and R=6 then it will be 3/11

How you are getting 1/2 as the answer?

What you've calculated here is conditional probability.

3/11 would've been correct if we were asked to calculate the probability of getting the FIRST ball red and THEN picking the second ball blue.

But here we're given that the FIRST has already been picked and it was red. That event has already happened, so we don't need to worry about that.

In mathematical terms,

P[first ball red] = 1 , as there is no other possible outcome other than what already came

Let blue balls be 3 and red balls be 4 :

1/1 * 3/6 = 1/2

Hence, it's correct.
Current Student
Joined: 18 Jun 2015
Posts: 41

### Show Tags

11 Sep 2016, 04:00
3
We can solve this question while reading the question ...
Given one red ball was picked.

condition 1 > red ball was one more than the blue ball.
as we have picked one red ball now so red balls equals blue ball.

So the probability of picking the blue ball is half.
Hence condition 1 is sufficient.

condition 2 > not sufficient.
Intern
Joined: 17 Feb 2016
Posts: 2

### Show Tags

22 Mar 2017, 03:42
Hi Bunuel. I don't agree with the explanation,

Statement (1): What's up if there are zero blue balls and only one red ball in the bag? The probability of getting a blue ball would be zero (because there are no blue balls in the bag!) - I agree that for any other case, the P is 1/2.

So we need statement (2) to solve.

Math Expert
Joined: 02 Sep 2009
Posts: 51262

### Show Tags

22 Mar 2017, 04:08
jcriedel wrote:
Hi Bunuel. I don't agree with the explanation,

Statement (1): What's up if there are zero blue balls and only one red ball in the bag? The probability of getting a blue ball would be zero (because there are no blue balls in the bag!) - I agree that for any other case, the P is 1/2.

So we need statement (2) to solve.

Since we are told that "A bag contains only blue and red balls", then I think it's safe to assume that the number of blue balls is more than 0. If it were otherwise then we could say that the bag contains 0 number of any other color balls.
_________________
Intern
Joined: 10 Jan 2018
Posts: 13
Location: United States (NY)
GMAT 1: 710 Q49 V38
GMAT 2: 770 Q51 V44
GPA: 3.36

### Show Tags

14 Mar 2018, 05:07
I don't agree with the explanation. The probability of first red, and then blue should be [r][/b+r]*[b][/b+r-1]. How could you guarantee you always get red ball first?
Math Expert
Joined: 02 Sep 2009
Posts: 51262

### Show Tags

14 Mar 2018, 05:14
1
Canteenbottle wrote:
I don't agree with the explanation. The probability of first red, and then blue should be [r][/b+r]*[b][/b+r-1]. How could you guarantee you always get red ball first?

This is called conditional probability. The question asks to find the probability that the second ball we pick will be blue, PROVIDED the first ball we picked is red.
_________________
Manager
Joined: 03 Feb 2016
Posts: 69
Location: India
Concentration: Technology, Marketing
GMAT 1: 650 Q48 V32
GPA: 4
WE: Sales (Computer Software)

### Show Tags

21 Apr 2018, 13:45
As per conditional probability,

P(A|B) = P(A+B) / p(B)

if b=x and r=x+1 are the numbers.

Then = P(A+B) / P(B) = P(1st ball red and second ball blue) / p(1st ball red) = [(x+1)/(2x+1)]*[1/2] / [(x+1)/(2x+1)] = 1/2.

So, this is correct. Unfortunately missed this question due to a silly mistake.
_________________

GMAT1 650 Q48 V32.

Intern
Joined: 15 May 2018
Posts: 2

### Show Tags

10 Sep 2018, 15:54
I think this is a high-quality question and I agree with explanation.
Intern
Status: All our dreams can come true, if we have the courage to pursue them
Joined: 03 Jul 2015
Posts: 23
Location: India
Concentration: Technology, Finance
WE: Information Technology (Computer Software)

### Show Tags

21 Nov 2018, 09:07
I think this is a high-quality question and I agree with explanation. This is a nice trap question. We need to understand the difference between getting first one red and first is red.
Re M32-04 &nbs [#permalink] 21 Nov 2018, 09:07
Display posts from previous: Sort by

# M32-04

Moderators: chetan2u, Bunuel

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.