It is currently 14 Dec 2017, 18:55

### 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

# Rich has 3 green, 2 red and 3 blue balls in a bag. He

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 11 May 2004
Posts: 453

Kudos [?]: 57 [1], given: 0

Location: New York
Rich has 3 green, 2 red and 3 blue balls in a bag. He [#permalink]

### Show Tags

08 Nov 2005, 18:12
1
KUDOS
11
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

81% (01:50) correct 19% (02:33) wrong based on 314 sessions

### HideShow timer Statistics

Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

A. 8/28
B. 9/28
C. 10/28
D. 10/18
E. 11/18
[Reveal] Spoiler: OA

Last edited by Bunuel on 11 Aug 2013, 00:28, edited 1 time in total.
Moved to PS forum and added the OA.

Kudos [?]: 57 [1], given: 0

Manager
Joined: 05 Oct 2005
Posts: 81

Kudos [?]: 1 [0], given: 0

### Show Tags

09 Nov 2005, 14:54
(3c2*2c1*3c2)/8c5
=(3*2*3)/(8*7*6/3*2)
=18/56
=9/28

B

Kudos [?]: 1 [0], given: 0

Senior Manager
Joined: 14 Apr 2005
Posts: 414

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

Location: India, Chennai

### Show Tags

09 Nov 2005, 23:45
(3C2*3C2*2C1)/8C5
= (3*3*2)/56=18/56=9/28

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

Intern
Joined: 30 Oct 2005
Posts: 19

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

### Show Tags

10 Nov 2005, 04:17
1. find combinations for each color separately:

green: xxo -> 3c2=3
blue: xxo -> 3c2=3
red: xo -> 2c1=2

2. find total combinations for picked balls:

xxxxxooo -> 5c8=56

3. find prob (selected over total cobmibations ):
P=3*3*2 / 56 = 9/28

B.

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

Senior Manager
Joined: 07 Jul 2005
Posts: 402

Kudos [?]: 62 [0], given: 0

### Show Tags

11 Nov 2005, 20:44
would the answer be the same with replacement?

Kudos [?]: 62 [0], given: 0

Intern
Joined: 30 Oct 2005
Posts: 19

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

### Show Tags

15 Nov 2005, 04:07
rigger wrote:
would the answer be the same with replacement?

I think it will be different.

Separate comb-s w/ replacemnt:
green: xxo -> 3c2=3*2*1=6
blue: xxo -> 3c2=3*2*1=6
red: xo -> 2c1=2*1=2

Total comb-s w/ replacement:
5c8 -> 8*7*6*5*4

Probability:
(6*6*2) / (8*7*6*5*4) = 3/280

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

Director
Joined: 27 Jun 2005
Posts: 501

Kudos [?]: 182 [0], given: 0

Location: MS

### Show Tags

15 Nov 2005, 09:57
ok guys !!! I am totally confused here about .With Replacement and Without Replacemetn Funda

..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2
2 Green = 3*2
2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....

Can any body pls explain the funda of With replacement and Without Replacement.

Kudos [?]: 182 [0], given: 0

Manager
Joined: 11 Jul 2005
Posts: 85

Kudos [?]: 1 [0], given: 0

Location: New York

### Show Tags

15 Nov 2005, 15:18
krisrini wrote:
(3C2*3C2*2C1)/8C5
= (3*3*2)/56=18/56=9/28

The event happens without replacement. Could you please explain how that factors into this answer? Maybe the answer for with replacement?

Kudos [?]: 1 [0], given: 0

SVP
Joined: 03 Jan 2005
Posts: 2227

Kudos [?]: 391 [1], given: 0

### Show Tags

15 Nov 2005, 19:12
1
KUDOS
cool_jonny009 wrote:
ok guys !!! I am totally confused here about .With Replacement and Without Replacemetn Funda

..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2
2 Green = 3*2
2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....

Can any body pls explain the funda of With replacement and Without Replacement.

Let's look at pick two green balls from three green balls. How many ways can you do it? Follow your train of thought, the first pick there's three possibility. The second pick there's only two left, so two possibility. Then you multiply them. However, notice that if the first pick you picked A, and then the second pick you picked B, you would count it as one outcome. If you picked B first, and then A second, you would count it as another outcome. While in fact, (A B) and (B A) is the same outcome, you picked the same two balls. So what you need to do in that case is that you need to divede your number by 2! to get rid of the ordering.

By the same logic for total outcome you have to divide 8*7*6*5*4 by 5!, and you'll get the same result as C(8,5).

Generally, picking n things one by one without replacement can be treated the same as picking n things all together.
_________________

Keep on asking, and it will be given you;
keep on seeking, and you will find;
keep on knocking, and it will be opened to you.

Kudos [?]: 391 [1], given: 0

Manager
Joined: 30 May 2013
Posts: 186

Kudos [?]: 90 [0], given: 72

Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

### Show Tags

09 Aug 2013, 22:32
cool_jonny009 wrote:
ok guys !!! I am totally confused here about .With Replacement and Without Replacemetn Funda

..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2
2 Green = 3*2
2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....

Can any body pls explain the funda of With replacement and Without Replacement.

Can somebody can put more light on this solution by probablit method with out using Combinations.

Thanks,
Rrsnathan.

Kudos [?]: 90 [0], given: 72

Math Expert
Joined: 02 Sep 2009
Posts: 42607

Kudos [?]: 135666 [1], given: 12705

### Show Tags

11 Aug 2013, 00:31
1
KUDOS
Expert's post
2
This post was
BOOKMARKED
rrsnathan wrote:
cool_jonny009 wrote:
ok guys !!! I am totally confused here about .With Replacement and Without Replacemetn Funda

..I thought With out replacement means we are taking out the balls and not putting it back.

so Total no of ways = 8*7*6*5*4

the no of ways the 1 Red can be drawn = 2
2 Green = 3*2
2 blue = 3*2

so prob = 6*6*2/8*7*6*5*4= 3/280.....

Can any body pls explain the funda of With replacement and Without Replacement.

Can somebody can put more light on this solution by probablit method with out using Combinations.

Thanks,
Rrsnathan.

Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

A. 8/28
B. 9/28
C. 10/28
D. 10/18
E. 11/18

$$P(RGGBB)=\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!2!}=\frac{9}{28}$$. We are multiplying by $$\frac{5!}{2!2!}$$ because RGGBB case can occur in several ways: RGGBB, GRGBBR, GGRBB, GGBRB, ... (basically it's # of permutations of 5 letters RGGBB, which is $$\frac{5!}{2!2!}$$).

The difference between with and without replacement cases is discussed here: rich-has-3-green-2-red-and-3-blue-balls-in-a-bag-he-55253.html#p637525

Hope it helps.
_________________

Kudos [?]: 135666 [1], given: 12705

Target Test Prep Representative
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 1810

Kudos [?]: 981 [0], given: 5

Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He [#permalink]

### Show Tags

29 Oct 2017, 06:23
desiguy wrote:
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag without replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

A. 8/28
B. 9/28
C. 10/28
D. 10/18
E. 11/18

The number of ways to select 1 red ball is 2C1 = 2.

The number of ways to select 2 green balls is 3C2 = 3.

The number of ways to select 2 blue balls is 3C2 = 3.

So, the number of ways to select 1 red, 2 green, and 2 blue balls is 2 x 3 x 3 = 18.

The number of ways to select 5 balls from 8 is 8C5:

8!/5![(8-5)!] = 8!/(5!x3!) = (8 x 7 x 6 x 5 x 4)/5! = (8 x 7 x 6 x 5 x 4)/(5 x 4 x 3 x 2 x 1) = 7 x 2 x 4 = 56

Thus, the probability is 18/56 = 9/28.

_________________

Jeffery Miller

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

Kudos [?]: 981 [0], given: 5

Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He   [#permalink] 29 Oct 2017, 06:23
Display posts from previous: Sort by