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

It is currently 20 Feb 2019, 07:40

Close

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
Your Progress

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

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Close

Request Expert Reply

Confirm Cancel
Events & Promotions in February
PrevNext
SuMoTuWeThFrSa
272829303112
3456789
10111213141516
17181920212223
242526272812
Open Detailed Calendar
  • Free GMAT Prep Hour

     February 20, 2019

     February 20, 2019

     08:00 PM EST

     09:00 PM EST

    Strategies and techniques for approaching featured GMAT topics. Wednesday, February 20th at 8 PM EST
  • Online GMAT boot camp for FREE

     February 21, 2019

     February 21, 2019

     10:00 PM PST

     11:00 PM PST

    Kick off your 2019 GMAT prep with a free 7-day boot camp that includes free online lessons, webinars, and a full GMAT course access. Limited for the first 99 registrants! Feb. 21st until the 27th.

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  
Author Message
TAGS:

Hide Tags

 
CEO
CEO
User avatar
Joined: 21 Jan 2007
Posts: 2584
Location: New York City
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post Updated on: 01 Feb 2019, 04:21
12
32
00:00
A
B
C
D
E

Difficulty:

(N/A)

Question Stats:

72% (01:38) correct 28% (01:28) wrong based on 141 sessions

HideShow timer Statistics

1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?
Spoiler: :: OA
14/15



2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?
Spoiler: :: OA
1/15



3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Spoiler: :: OA
49/60



4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
Spoiler: :: OA
21/40

Originally posted by bmwhype2 on 06 Dec 2007, 14:24.
Last edited by Bunuel on 01 Feb 2019, 04:21, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
Most Helpful Expert Reply
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 53020
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 21 Jan 2010, 00:35
10
6
Bullet wrote:
Can any body please explain Question No.3 using probability

Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15

Thanks and appreciated


The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

Probability of at least two marble are blue is the sum of the two probabilities:

A. Two marbles are blue and one is red - BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\);

B. All three marbles are blue - BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\)

So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\)
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Most Helpful Community Reply
Manager
Manager
User avatar
Joined: 22 Jul 2009
Posts: 168
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 29 Aug 2009, 14:50
35
12
I would like to make a recap of this problem, as in the prior posts there are 3 different answers for questions 3 and 4.

Have 3 red and 7 blue.

1) Pick 2. P of picking at least 1 blue?
Combinatorics: P(1b&1r)+P(2b) -> (7C1 x 3C1 + 7C2)/10C2
Combinatorics shortcut: 1-P(2r) -> 1 - 3C2/10C2
Probability: P(BR)+P(RB)+P(BB) = 7/10*3/9+ 3/10*7/9+ 7/10*6/9
Answer: 14/15

2) Pick 2. P of picking 0 blue?
Combinatorics: P(2r) -> 3C2/10C2
Probability: P(RR) = 3/10*2/9
Answer: 1/15

3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Probability: P(BBR)+P(BRB)+P(RBB)+P(BBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8+ 7/10*6/9*5/8 = 7/40+7/40+7/40+7/24
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)

4) Pick 3. P of picking 2 blue?
Combinatorics: P(2b&1r) -> (7C2 x 3C1)/10C3
Probability: P(BBR)+P(BRB)+P(RBB) = 7/10*6/9*3/8+ 7/10*3/9*6/8+ 3/10*7/9*6/8
Answer: 21/40 (8/15 and 7/40 are wrong; do the math)

Hope this may save new visitors time.
_________________

Please kudos if my post helps.

General Discussion
Senior Manager
Senior Manager
avatar
Joined: 06 Mar 2006
Posts: 454
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 06 Dec 2007, 14:40
2
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

1- [3!/(2!1!)]/[10!/(2!8!)]=14/15

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

[3!/(2!1!)]/[10!/(2!8!)]=1/15


3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that two marbles are blue?

[7!/(2!5!)]/[10!/(2!8!)]=7/15


Personally, I really don't know what is the difference between question 3 & 4.
SVP
SVP
User avatar
Joined: 29 Aug 2007
Posts: 2349
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 06 Dec 2007, 23:54
14
12
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

At least 1 is blue = (1b and 1r) + 2 blue
total = 10c2
prob = (7c1 x 3c1 + 7c2)/10c2 = 42/45 = 14/15



2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?


prob = 3c2/10c2 = 3/45 = 1/15


3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

= (7c2 x 3c1 + 7c3)/10c3
= (63 + 35)/10c3
= 98/120
= 49/60

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?


= (7c2 x 3c1 )/10c3
= (63)/10c3
= 63/120
= 21/40
SVP
SVP
avatar
Joined: 29 Mar 2007
Posts: 2407
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 10 Jan 2008, 11:25
1
bmwhype2 wrote:
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?


1: at least one is blue: how bout 1-none are blue? 3/10*2/9 --> 6/90 1-1/15 = 14/15
2: 3/10*2/9 = 1/15
3: At least 2 are blue: 1-1 is blue none are blue 3/10*2/9*1/8 =6/720
3/10*2/9*7/8(3) b/c we have 3!/2! ways to arrange RRB

378/720 --> 384/720 192/360 --> 96/180--> 48/90--> 8/15 --> 7/15 is our answer

4: 7/10*6/9*3/8 * (3) again b/c 3!/2! 378/720 = 8/15


hrmmm appears I am wrong on 3 and 4
Manager
Manager
avatar
Joined: 15 Nov 2007
Posts: 131
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 10 Jan 2008, 18:32
1
1
1. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that at least one marble is blue?

scenario 1: 1 red, 1 blue
scenario 2: 2 blue

total possibility
10C2
(scenario 1 + scenario 2)/total possibilities = (3C1 x 7C1 +7C2)/10C2

2. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two marbles are selected at random, what is the probability that none is blue?

scenario 2/total possibilities

3c2/10c2

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

3c1*7c2+7c3/10c3

4. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?

7c2*3c1/10c3
Senior Manager
Senior Manager
User avatar
Joined: 21 Apr 2008
Posts: 260
Location: Motortown
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 04 Oct 2008, 09:23
1
Thanks bmwhype2 for posting all these questions.
And thanks walker for compiling them very well. I am going through your list right now...

My answer :

•1-(all red) = 1-(3/10*2/9) = 1-(1/15) = 14/15
•All red = 3/10*2/9 = 1/15
•3 blue + 3(2 blue, 1red) = RRR+RRB+BRR+RBR = 7/24+7/40+7/40+7/40 = 98/120 = 49/60
•2blue,1 red = 7/10*6/9*3/8 = 7/40 (You can’t look for 3 Blue, because the questions asks for 2 Blue, so the 3rd one has to be red)
SVP
SVP
User avatar
Joined: 07 Nov 2007
Posts: 1650
Location: New York
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 06 Feb 2009, 13:08
1
djveed wrote:
I'm not getting it. For A, I would assume the chance of a marble on the first pull is 7/10, and the chance on the second is 6/9. Multiply those two to get the chance that one is blue and you get a 7/15 chance that one of the two marbles pulled are blue. Which is different than what you all have.

And for B, chance of red is 3/10 on first and 2/9 on second, and when multiplied is 1/15 chance. What am I doing wrong?


there are lot ways to do this problem..
You are trying to solve the problem using below method.. but missing some steps..
you considered only p(2B) but missed p(1B,1R)+ p(1R,1B)

1. At least one Blue Marble =
= p(1B,1R)+ p(1R,1B)+[color=#0000FF]p(2B)[/color] = (7/10)*(3/9)+(3/10)*(7/9)+ (7/10)*(6/9)
= 14/45

Another Way. = 7C1*3C1/10C2 + 7C2/10C2
= 21/45 + 21/45 = 14/45

best way = 1- P(2R) = 1- 3C2/10C2 = 14/15.

Is it clear now..
_________________

Your attitude determines your altitude
Smiling wins more friends than frowning

Intern
Intern
User avatar
Joined: 17 Nov 2009
Posts: 32
Schools: University of Toronto, Mcgill, Queens
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 21 Jan 2010, 01:49
Bunuel wrote:
Bullet wrote:
Can any body please explain Question No.3 using probability

Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15

Thanks and appreciated


The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

Probability of at least two marble are blue is the sum of the two probabilities:

A. Two marbles are blue and one is red - BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\);

B. All three marbles are blue - BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\)

So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\)


Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.

So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?

Thanks
_________________

--Action is the foundational key to all success.

Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 53020
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 21 Jan 2010, 02:56
5
7
Bullet wrote:
Thanks Bunuel. I posted the solution from the thread as i was confused with the solution.

So what you're saying is that we need to multiply the permutations with the probability. Having said that we will do only when we need to choose at least two marbles or any other thing from the whole?

Thanks


To make it simple: suppose we have the jar of 10 marbles - 5 red, 2 blue and 3 green. If five marbles are selected at random, what is the probability that two will be red, one blue and two green?

We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is \(\frac{5!}{2!2!}\).

Hence the answer for the above question would be \(\frac{5!}{2!2!}*\frac{5}{10}*\frac{4}{9}*\frac{2}{8}*\frac{3}{7}*\frac{2}{6}\).

If the question were: three marbles are selected at random, what is the probability that all three will be red?

RRR can occur only in one way: RRR, so the probability would be \(\frac{5}{10}*\frac{4}{9}*\frac{3}{8}\).

You can check the Probability and Combination chapters in the Math Book (link below) for more.
Also check my posts at:
probability-colored-balls-55253.html#p637525
4-red-chips-and-2-blue-chips-85987.html#p644603
probability-qs-attention-88945.html#p671958
p-c-88431.html?highlight=probability+of+occurring+event
probability-88069.html?highlight=probability+of+occurring+event
combination-problem-princenten-review-2009-bin-4-q2-87673.html?highlight=probability+of+occurring+event
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Intern
Intern
avatar
Joined: 13 Jan 2012
Posts: 37
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 22 Feb 2012, 13:11
powerka wrote:
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)


#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues)

P (exactly 2 blues)= \(7C2/10C3\)=\(21/120\)
P (3 blues) = \(7C3/10C3\)=\(35/120\)

Therefore,
P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15

I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply \(7C2\) by 3C1 to get P (exactly 2 blues)? Or, is my answer \(7/15\)correct?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 53020
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 22 Feb 2012, 13:22
1
fxsunny wrote:
powerka wrote:
3) Pick 3. P of picking at least 2 blue?
Combinatorics: P(2b&1r)+P(3b) -> (7C2 x 3C1 + 7C3)/10C3
Answer: 49/60 (7/15 and 42/90 are wrong; do the math)


#3. P(at least 2 blues) = P(exactly 2 blues) + P (3 blues)

P (exactly 2 blues)= \(7C2/10C3\)=\(21/120\)
P (3 blues) = \(7C3/10C3\)=\(35/120\)

Therefore,
P(at least 2 blues) = 21/120 + 35/120 = 56/120 = 7/15

I see that the commented section above has an extra x 3C1 that's probably the cause for my discrepancy. Why do I need to multiply \(7C2\) by 3C1 to get P (exactly 2 blues)? Or, is my answer \(7/15\)correct?


That's because you are picking 3 marbles and if you pick 2 blue then the third one must be red: BBR. Ways to pick one red marble out of 3 is \(C^1_3\).

Complete solution using combinations: \(P(R\geq{2})=P(R=2)+P(R=3)=\frac{C^2_7*C^1_3}{C^3_{10}}+\frac{C^3_7}{C^3_{10}}\).

Hope it's clear.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 07 Nov 2009
Posts: 241
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 07 Mar 2012, 15:59
The below questions are discussed at the below thread, but i have a silly doubt so posting that here:
a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html

I understand the solution provided, but is there anything wrong with the simple method as below as i am getting different ans :( ?

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Case 1: Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
Case 2: All 3 marbles are blue. Prob: 7/10 * 6/9 * 5/8
Add 1 and 2

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that two marbles are blue?
Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 53020
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 07 Mar 2012, 16:10
rohitgoel15 wrote:
A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?
Case 1: Only 2 marbles are blue. Prob: 7/10 * 6/9 * 3/8
Case 2: All 3 marbles are blue. Prob: 7/10 * 6/9 * 5/8
Add 1 and 2


The point is that BBR case can occur in 3 different ways: BBR, BRB, and RBB. So you should multiply 7/10 * 6/9 * 3/8 by 3.

This is discussed here: a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html#p677059 and here: a-bag-of-10-marbles-contains-3-red-marbles-and-7-blue-56728.html#p677083
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 15 Aug 2013
Posts: 244
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 20 Apr 2014, 14:05
Bunuel wrote:
We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is \(\frac{5!}{2!2!}\).

RRR can occur only in one way: RRR, so the probability would be \frac{5}{10}*\frac{4}{9}*\frac{3}{8}.




Can someone please talk about this a little? I understand that the marbles can be arranged in various ways RBRGG etc. but how do we come up with the \(\frac{5!}{2!2!}\) equation?

EDIT: To add to the question above, why can RRR only occur in one way, isn't that 3 separate ways as well? Meaning, R_1, R_2, R3 -- there are 3 different R's and therefore 3 different ways?
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 53020
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 20 Apr 2014, 23:44
1
russ9 wrote:
Bunuel wrote:
We are looking for all the cases with 2R, 1B and 2G. We can draw these marbles like: RRBGG or GGBRR or RBGGR ... So how many combinations of the drawing of these marbles are there? The answer is as many as there is permutation of the letters GGBRR, which is \(\frac{5!}{2!2!}\).

RRR can occur only in one way: RRR, so the probability would be \frac{5}{10}*\frac{4}{9}*\frac{3}{8}.




Can someone please talk about this a little? I understand that the marbles can be arranged in various ways RBRGG etc. but how do we come up with the \(\frac{5!}{2!2!}\) equation?

EDIT: To add to the question above, why can RRR only occur in one way, isn't that 3 separate ways as well? Meaning, R_1, R_2, R3 -- there are 3 different R's and therefore 3 different ways?


THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Manager
Manager
avatar
Joined: 12 May 2013
Posts: 63
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 21 Apr 2014, 00:48
Bunuel wrote:
Bullet wrote:
Can any body please explain Question No.3 using probability

Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15

Thanks and appreciated


The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

Probability of at least two marble are blue is the sum of the two probabilities:

A. Two marbles are blue and one is red - BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\);

B. All three marbles are blue - BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\)

So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\)


Bunuel, i just have 1 silly question, here why cant we do: 3!/2!*0.7^2*0.3 , i mean it is not specified in the question whether we replace the ball after selecting it
i get confused in such situations, a lot !
Math Expert
User avatar
V
Joined: 02 Sep 2009
Posts: 53020
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 21 Apr 2014, 01:05
adymehta29 wrote:
Bunuel wrote:
Bullet wrote:
Can any body please explain Question No.3 using probability

Why we need to add twice
7/10*6/9*3/8 + 7/10*6/9*5/8 = 7/15

Thanks and appreciated


The solution you are posting for the third question is not right. Below is the solution of this question using the "probability".

3. A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If three marbles are selected at random, what is the probability that at least two marbles are blue?

Probability of at least two marble are blue is the sum of the two probabilities:

A. Two marbles are blue and one is red - BBR. This can occur in \(\frac{3!}{2!}=3\) # of ways, which is basically the # of permutations of three letters B, B, and R: BBR, BRB, RBB. \(3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}\);

B. All three marbles are blue - BBB. This can occur only one way, namely BBB. \(\frac{7}{10}*\frac{6}{9}*\frac{5}{8}\)

So \(P=3*\frac{7}{10}*\frac{6}{9}*\frac{3}{8}+\frac{7}{10}*\frac{6}{9}*\frac{5}{8}=\frac{49}{60}\)


Bunuel, i just have 1 silly question, here why cant we do: 3!/2!*0.7^2*0.3 , i mean it is not specified in the question whether we replace the ball after selecting it
i get confused in such situations, a lot !


If it were with replacement it would be specified. Or let me put it this way: proper GMAT question would make it clear whether it's with or without replacement case.
_________________

New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


What are GMAT Club Tests?
Extra-hard Quant Tests with Brilliant Analytics

Senior Manager
Senior Manager
User avatar
Status: The Final Countdown
Joined: 07 Mar 2013
Posts: 283
Concentration: Technology, General Management
GMAT 1: 710 Q47 V41
GPA: 3.84
WE: Information Technology (Computer Software)
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two  [#permalink]

Show Tags

New post 29 Nov 2014, 23:43
In a similar question.

In a blue jar there are red, white and green balls. The probability of drawing a red ball is 1/5. The probability of drawing a red ball, returning it, and then drawing a white ball is 1/10. What is the probability of drawing a white ball?

a) 1/5.
b) ½.
c) 1/3.
d) 3/10.
e) ¼.


the OA is 1/2..which is just 1/10=1/5*x,so x=1/2..
my question is that returning the ball does not figure into the question in anyway?
GMAT Club Bot
Re: A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two   [#permalink] 29 Nov 2014, 23:43

Go to page    1   2    Next  [ 25 posts ] 

Display posts from previous: Sort by

A bag of 10 marbles contains 3 red marbles and 7 blue marbles. If two

  new topic post reply Question banks Downloads My Bookmarks Reviews Important topics  


Copyright

GMAT Club MBA Forum Home| About| Terms and Conditions and Privacy Policy| GMAT Club Rules| Contact| Sitemap

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®.