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Senior Manager  Joined: 11 May 2004
Posts: 420
Location: New York
Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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15 00:00

Difficulty:   25% (medium)

Question Stats: 79% (02:14) correct 21% (02:44) wrong based on 366 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

Originally posted by desiguy on 08 Nov 2005, 19:12.
Last edited by Bunuel on 11 Aug 2013, 01:28, edited 1 time in total.
Moved to PS forum and added the OA.
Math Expert V
Joined: 02 Sep 2009
Posts: 55804

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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.
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Manager  Joined: 05 Oct 2005
Posts: 78

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(3c2*2c1*3c2)/8c5
=(3*2*3)/(8*7*6/3*2)
=18/56
=9/28

B
Senior Manager  Joined: 15 Apr 2005
Posts: 386
Location: India, Chennai

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(3C2*3C2*2C1)/8C5
= (3*3*2)/56=18/56=9/28
Intern  Joined: 30 Oct 2005
Posts: 19

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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.
Senior Manager  Joined: 07 Jul 2005
Posts: 374

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would the answer be the same with replacement?
Intern  Joined: 30 Oct 2005
Posts: 19

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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
Senior Manager  Joined: 27 Jun 2005
Posts: 449
Location: MS

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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.  Manager  Joined: 11 Jul 2005
Posts: 78
Location: New York

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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?
SVP  Joined: 03 Jan 2005
Posts: 2034

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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.
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Manager  Joined: 30 May 2013
Posts: 152
Location: India
Concentration: Entrepreneurship, General Management
GPA: 3.82

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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.
Target Test Prep Representative G
Affiliations: Target Test Prep
Joined: 04 Mar 2011
Posts: 2823
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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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.

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GMATH Teacher P
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Joined: 12 Oct 2010
Posts: 936
Re: Richard has 3 green, 2 red, and 3 blue balls in a bag. He randomly...  [#permalink]

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jpfg259 wrote:
Richard 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, Richard 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

$$3G,2R,3B\,\,\,\, \to \,\,{\text{take}}\,\,5\,\,{\text{of}}\,\,{\text{them}}\,\,{\text{at}}\,{\text{random}}$$

$${\text{? = P}}\left( {1\,\,{\text{red}},\,\,2\,\,{\text{green}},\,\,2\,\,{\text{blue}}} \right) = \frac{{C\left( {2,1} \right) \cdot C\left( {3,2} \right) \cdot C\left( {3,2} \right)}}{{C\left( {8,5} \right)}} = \frac{{2 \cdot 3 \cdot 3}}{{8 \cdot 7}} = \frac{9}{{28}}$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Our high-level "quant" preparation starts here: https://gmath.net Re: Richard has 3 green, 2 red, and 3 blue balls in a bag. He randomly...   [#permalink] 29 Oct 2018, 14:53
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