GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 03 Aug 2020, 22:03 ### 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

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # Rich has 3 green, 2 red and 3 blue balls in a bag. He

Author Message
TAGS:

### Hide Tags

Current Student Joined: 31 Aug 2007
Posts: 339
Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

5
36 00:00

Difficulty:   65% (hard)

Question Stats: 59% (02:09) correct 41% (02:08) wrong based on 396 sessions

### HideShow timer Statistics

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) $$(\frac{3}{8})^2 * (\frac{3}{8})^2$$
Math Expert V
Joined: 02 Sep 2009
Posts: 65768
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

23
43
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.
_________________
##### General Discussion
Senior Manager  Joined: 13 Dec 2006
Posts: 297
Location: Indonesia
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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
Manager  Joined: 01 Sep 2006
Posts: 201
Location: Phoenix, AZ, USA
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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, 08:28.
Last edited by Damager on 08 Nov 2007, 09:22, edited 1 time in total.
Intern  Joined: 01 Nov 2007
Posts: 49
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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 ?
Director  Joined: 22 Nov 2007
Posts: 802
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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: 2913
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]

### Show Tags

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
_________________
HOT! GMAT Club Forum 2020 | GMAT ToolKit 2 (iOS) - The OFFICIAL GMAT CLUB PREP APPs, must-have apps especially if you aim at 700+
Director  Joined: 22 Nov 2007
Posts: 802
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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

SVP  Joined: 29 Mar 2007
Posts: 1648
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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.
Senior Manager  Joined: 09 Jul 2005
Posts: 411
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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?
Senior Manager  Joined: 01 Jan 2008
Posts: 418
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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: 121
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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: 125
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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: 35
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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]

### Show Tags

The probability in this case would be as follow:

1/5+2/5+2/5 = 5/5 = 1
SVP  Joined: 29 Aug 2007
Posts: 1712
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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...
Manager  Joined: 10 Jul 2009
Posts: 83
Location: Ukraine, Kyiv
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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)
Manager  Joined: 11 Aug 2008
Posts: 91
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

1
I think the ans is (3/8)^2*2/8*8!/(3!*3!*2!)
Manager  Joined: 05 Jul 2009
Posts: 124
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

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?
Manager  Joined: 22 Dec 2009
Posts: 225
Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He  [#permalink]

### Show Tags

3
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. Re: Rich has 3 green, 2 red and 3 blue balls in a bag. He   [#permalink] 17 Feb 2010, 03:01

Go to page    1   2    Next  [ 31 posts ]

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