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Current Student Joined: 31 Aug 2007
Posts: 362
Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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5
27 00:00

Difficulty:   65% (hard)

Question Stats: 59% (02:11) correct 41% (02:10) wrong based on 446 sessions

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Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

a) $$(\frac{2}{8}) * (\frac{3}{8})^2 * (\frac{3}{8})^2 *\frac{5!}{(2! * 2!)}$$
b) $$(\frac{2}{8}) * (\frac{3}{8})^2 * (\frac{3}{8})^2 *5!$$
c) $$(\frac{2}{8}) * (\frac{3}{8})^2 * (\frac{3}{8})^2$$
d) $$(\frac{2}{8}) * (\frac{3}{8}) * (\frac{3}{8})^2$$
e) $$(\frac{3}{8})^2 * (\frac{3}{8})^2$$
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Math Expert V
Joined: 02 Sep 2009
Posts: 55277
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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22
39
Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: $$\frac{2}{8}*(\frac{3}{8}*\frac{3}{8})*(\frac{3}{8}*\frac{3}{8})*\frac{5!}{2!2!}=14.8%$$

CASE II.
If the problem was WITHOUT replacement:

Then the probability would be: $$\frac{2}{8}*(\frac{3}{7}*\frac{2}{6})*(\frac{3}{5}*\frac{2}{4})*\frac{5!}{2!*2!}=\frac{9}{28}=32.14%$$

OR different way of counting: $$\frac{2C1*3C2*3C2}{8C5}=\frac{9}{28}$$

The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.

We are then multiplying this, in BOTH CASES, by $$\frac{5!}{2!2!}$$, because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.

The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.

Hope now it's clear.
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Senior Manager  Joined: 13 Dec 2006
Posts: 469
Location: Indonesia
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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I got (3^4/8^4)/4, I am not too sure...

Calulation

The probability of getting 1 red ball is 2/8, while the probability of getting 2 green and 2 blue balls are 3^2/8^2 each

on multiplying all we will get the answer mentioned above.

Amardeep
Senior Manager  Joined: 01 Sep 2006
Posts: 279
Location: Phoenix, AZ, USA
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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young_gun wrote:
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

GGBBR prob = 3/8*3/8*3/8*3/8*2/8=3^4*2/8^5
GGBBR can have 3C2*3C2*2C1 combination = 3*3*2=18

Total prob = 18*3^4*2/8^5

Originally posted by Damager on 08 Nov 2007, 09:28.
Last edited by Damager on 08 Nov 2007, 10:22, edited 1 time in total.
Manager  Joined: 01 Nov 2007
Posts: 68
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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Here is my solution

Total no of way 5 ball can be selected from 8 = 8C5 = 56

now 1 out of 2 red balls can be selected in 2 way

2 out of 3 green ball can be selected in 3 ways

and 2 out of 3 blue ball can be selected in 3 ways

so total number of ways = 2*3*3 = 18

so the probability = 18 / 56 = 9/28

Am I right ?
VP  Joined: 22 Nov 2007
Posts: 1026
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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sportyrizwan wrote:
young_gun wrote:
sportyrizwan wrote:
Here is my solution

Total no of way 5 ball can be selected from 8 = 8C5 = 56

now 1 out of 2 red balls can be selected in 2 way

2 out of 3 green ball can be selected in 3 ways

and 2 out of 3 blue ball can be selected in 3 ways

so total number of ways = 2*3*3 = 18

so the probability = 18 / 56 = 9/28

Am I right ?

No, that's without replacement.

ooh sorry...

in that case damager is right

i think damager is absolutely right. we have replacements, so the total number of combinations is 8^5 (n*n*n*n*n) and not 8c5...what's OA?
CEO  B
Joined: 17 Nov 2007
Posts: 3407
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40 Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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1
N=(2/8)*(3/8)^2*(3/8)^2*5P5/(2P2*2P2)=2*3^4*5!/(8^5*2*2)=20*3^5/8^5

Damager wrote:
GGBBR can have 3C2*3C2*2C1 combination = 3*3*2=18

I think combination=5C2*3C2*1C1=5!/(3!*2!)*3!/(2!)*1=5!/2!^2
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VP  Joined: 22 Nov 2007
Posts: 1026
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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walker wrote:
N=(2/8)*(3/8)^2*(3/8)^2*5P5/(2P2*2P2)=2*3^4*5!/(8^5*2*2)=20*3^5/8^5

Damager wrote:
GGBBR can have 3C2*3C2*2C1 combination = 3*3*2=18

I think combination=5C2*3C2*1C1=5!/(3!*2!)*3!/(2!)*1=5!/2!^2

your reasoning sounds better...what's OA?
SVP  Joined: 29 Mar 2007
Posts: 2373
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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young_gun wrote:
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

2r 3g 3b

a total of 8 balls.

2/8*3/8*3/8*3/8*3/8 *5!/2!*2! (we have 5!/2!2! b/c we have 2 green picks and 2 blue picks and 1 red pick but we don't need to worry about that one--> 5*4*3*2/2*2=30) --> 30(2*3^4/8^5) 81*2*30 -->
162*30= 4860/32768 -->~14%

This was a beast, took me about 6minutes.
Director  Joined: 09 Jul 2005
Posts: 542
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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For me the probability is:

P=[5!/(2!·2!)]*[(3/8)^2]*[(3/8)^2]*[(2/8)^1]=2430/16384=0.15

What is the OA?
Director  Joined: 01 Jan 2008
Posts: 581
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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probably repeating others:

(5!/2!*2!)*(2/8)*(3/8)^2*(3/8)^2=5*(3^5)/(2^13)
Manager  Joined: 08 Jul 2009
Posts: 149
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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young_gun wrote:
sportyrizwan wrote:
Here is my solution

Total no of way 5 ball can be selected from 8 = 8C5 = 56

now 1 out of 2 red balls can be selected in 2 way

2 out of 3 green ball can be selected in 3 ways

and 2 out of 3 blue ball can be selected in 3 ways

so total number of ways = 2*3*3 = 18

so the probability = 18 / 56 = 9/28

Am I right ?

No, that's without replacement.

I think the answer should be 9/28. Permutation assumes replacement. Without replacement would get really tricky.
Manager  Joined: 27 Oct 2008
Posts: 167
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

Soln:
= (2/8) * (3/8)^2 * (3/8)^2 *5!/(2! * 2!)
Intern  Joined: 25 Mar 2009
Posts: 48
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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automan wrote:
For me the probability is:

P=[5!/(2!·2!)]*[(3/8)^2]*[(3/8)^2]*[(2/8)^1]=2430/16384=0.15

What is the OA?

P=2/8*(3/8*2/8)*(3/8*2/8)
Intern  Joined: 09 Oct 2009
Posts: 1
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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The probability in this case would be as follow:

1/5+2/5+2/5 = 5/5 = 1
SVP  Joined: 29 Aug 2007
Posts: 2310
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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Bunuel wrote:
Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%

Why do you have 5!/2!2!? Does order matter? No...
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Manager  Joined: 10 Jul 2009
Posts: 102
Location: Ukraine, Kyiv
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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Thank you all, guys!
Understood it completely!

GMAT TIGER, order does not matter.
Because we have the following combination:

GGBBR

if we mark numbers G1G2B1B2R and rearrange them to have G2G1B2B1R, we will still have the same combination.
thus we eliminate repeats when we divide 5! (GGBBR) by 2! (BB) and 2! (GG)
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Manager  Joined: 11 Aug 2008
Posts: 120
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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1
I think the ans is (3/8)^2*2/8*8!/(3!*3!*2!)
Manager  Joined: 05 Jul 2009
Posts: 152
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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GMAT TIGER wrote:
Bunuel wrote:
Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%

Why do you have 5!/2!2!? Does order matter? No...

Its getting confusing. I though the order here is not important and thus the answer should be 2/8*3/8*3/8*3/8*3/8

Can someone please tell me the difference between (if there is any) the following with replacement:

1. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

2. In how many ways rich can draw five balls so that he has picked 1 red, 2 green, and 2 blue balls?
Senior Manager  Joined: 22 Dec 2009
Posts: 291
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

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2
Bunuel wrote:
Let's clear out this one:

CASE I.
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 5 from the bag WITH replacement. What is the probability that of the 5 drawn balls, Rich has picked 1 red, 2 green, and 2 blue balls?

The probability would be: 2/8*3/8*3/8*3/8*3/8*5!/2!2!=14.8%

CASE II.
If the problem was WITHOUT replacement:

Then the probability would be: 2/8*3/7*2/6*3/5*2/4*5!/2!*2!=9/28=32.14%
OR different way of counting: 2C1*3C2*3C2/ 8C5=9/28

The only difference we have that in FIRST case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.

We are then multiplying this, in BOTH CASES, by 5!/2!2!, because we want to calculate number of ways this combination is possible (1 red, 2 green, 2 blue), total combination of 5 balls is 5!, BUT as among these 5 balls we have 2 green and 2 red, we should divide 5! by 2!2! to get rid of repeated combinations.

The second way of calculating "WITHOUT replacement case" just counts the number of combinations of picking desired number of each ball from its total number (1 red out of two, 2 green out of three, and 2 blue out of three), multiplies them and divides this by the number of ways we can pick total of 5 balls out of 8.

Hope now it's clear.

Does the order matter in this question??? I dont think so.. as the question says... wat is the probability of getting 1 red, 2 green and 2 blue... and says nothing of the arrangement. Please comment.
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~~Better Burn Out... Than Fade Away~~ Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He   [#permalink] 17 Feb 2010, 04:01

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

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