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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] New post 08 Nov 2007, 07:59
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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?
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 [#permalink] New post 08 Nov 2007, 08:12
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.

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Re: probability, colored balls [#permalink] New post 08 Nov 2007, 08:28
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

Last edited by Damager on 08 Nov 2007, 09:22, edited 1 time in total.
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 [#permalink] New post 08 Nov 2007, 09:10
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 ?
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Re: [#permalink] New post 08 Jan 2008, 22:30
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?
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Re: probability, colored balls [#permalink] New post 08 Jan 2008, 23:40
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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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Re: probability, colored balls [#permalink] New post 09 Jan 2008, 00:23
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?
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Re: probability, colored balls [#permalink] New post 09 Jan 2008, 23:39
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.
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Re: probability, colored balls [#permalink] New post 10 Jan 2008, 00:51
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?
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Re: probability, colored balls [#permalink] New post 08 Sep 2009, 13:01
probably repeating others:

(5!/2!*2!)*(2/8)*(3/8)^2*(3/8)^2=5*(3^5)/(2^13)
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Re: [#permalink] New post 24 Sep 2009, 18:02
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.
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Re: probability, colored balls [#permalink] New post 28 Sep 2009, 10:03
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!)
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Re: probability, colored balls [#permalink] New post 28 Sep 2009, 10:10
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)
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Re: probability, colored balls [#permalink] New post 09 Oct 2009, 13:55
The probability in this case would be as follow:

1/5+2/5+2/5 = 5/5 = 1
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Re: probability, colored balls [#permalink] New post 09 Oct 2009, 22:38
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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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Re: probability, colored balls [#permalink] New post 10 Oct 2009, 21:39
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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Re: probability, colored balls [#permalink] New post 11 Oct 2009, 02:39
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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Re: probability, colored balls [#permalink] New post 24 Oct 2009, 06:21
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I think the ans is (3/8)^2*2/8*8!/(3!*3!*2!)
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Re: probability, colored balls [#permalink] New post 24 Oct 2009, 20:51
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?
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Re: probability, colored balls [#permalink] New post 17 Feb 2010, 03:01
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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Re: probability, colored balls   [#permalink] 17 Feb 2010, 03:01
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