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

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

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New post Updated on: 11 Aug 2013, 01:28
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Question Stats:

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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.
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Re: Re:  [#permalink]

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New post 11 Aug 2013, 01:31
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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.

:roll: :roll:

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 \(\frac{5!}{2!2!}\)).

Answer: B.

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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New post 09 Nov 2005, 15:54
(3c2*2c1*3c2)/8c5
=(3*2*3)/(8*7*6/3*2)
=18/56
=9/28

B
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Re: Prob/Perm  [#permalink]

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New post 10 Nov 2005, 00:45
(3C2*3C2*2C1)/8C5
= (3*3*2)/56=18/56=9/28
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New post 10 Nov 2005, 05:17
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.
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New post 11 Nov 2005, 21:44
would the answer be the same with replacement?
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New post 15 Nov 2005, 05:07
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
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New post 15 Nov 2005, 10:57
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.

:roll: :roll:
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Re: Prob/Perm  [#permalink]

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New post 15 Nov 2005, 16:18
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?
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New post 15 Nov 2005, 20:12
1
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.

:roll: :roll:


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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Re:  [#permalink]

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New post 09 Aug 2013, 23:32
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.

:roll: :roll:

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

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

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New post 29 Oct 2017, 07:23
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.

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

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New post 29 Oct 2018, 14:53
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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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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