Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 29 Jun 2016, 05:44

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

Events & Promotions

Events & Promotions in June
Open Detailed Calendar

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: 369
Followers: 1

Kudos [?]: 102 [4] , given: 1

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

Show Tags

08 Nov 2007, 08:59
4
KUDOS
16
This post was
BOOKMARKED
00:00

Difficulty:

55% (hard)

Question Stats:

60% (02:19) correct 40% (01:41) wrong based on 126 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) $$(\frac{3}{8})^2 * (\frac{3}{8})^2$$
[Reveal] Spoiler: OA
Director
Joined: 13 Dec 2006
Posts: 518
Location: Indonesia
Followers: 6

Kudos [?]: 167 [0], given: 0

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

Show Tags

08 Nov 2007, 09:12
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.

Amardeep
Senior Manager
Joined: 01 Sep 2006
Posts: 301
Location: Phoenix, AZ, USA
Followers: 1

Kudos [?]: 19 [0], given: 0

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

Show Tags

08 Nov 2007, 09:28
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, 10:22, edited 1 time in total.
Manager
Joined: 01 Nov 2007
Posts: 69
Followers: 1

Kudos [?]: 7 [0], given: 0

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

Show Tags

08 Nov 2007, 10:10
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 ?
VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 392 [0], given: 0

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

Show Tags

08 Jan 2008, 23:30
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?
CEO
Joined: 17 Nov 2007
Posts: 3580
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
GMAT 1: 750 Q50 V40
Followers: 485

Kudos [?]: 2802 [1] , given: 359

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

Show Tags

09 Jan 2008, 00:40
1
KUDOS
Expert's post
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 TOOLKIT 2 (iOS) / GMAT TOOLKIT (Android) - The OFFICIAL GMAT CLUB PREP APP, a must-have app especially if you aim at 700+ | PrepGame

VP
Joined: 22 Nov 2007
Posts: 1092
Followers: 9

Kudos [?]: 392 [0], given: 0

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

Show Tags

09 Jan 2008, 01:23
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

CEO
Joined: 29 Mar 2007
Posts: 2583
Followers: 19

Kudos [?]: 341 [0], given: 0

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

Show Tags

10 Jan 2008, 00:39
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.
Director
Joined: 09 Jul 2005
Posts: 595
Followers: 2

Kudos [?]: 42 [0], given: 0

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

Show Tags

10 Jan 2008, 01:51
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?
Director
Joined: 01 Jan 2008
Posts: 629
Followers: 5

Kudos [?]: 162 [0], given: 1

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

Show Tags

08 Sep 2009, 14:01
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: 174
Followers: 3

Kudos [?]: 59 [0], given: 13

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

Show Tags

24 Sep 2009, 19:02
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.
Manager
Joined: 27 Oct 2008
Posts: 185
Followers: 1

Kudos [?]: 127 [0], given: 3

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

Show Tags

28 Sep 2009, 11:03
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!)
Manager
Joined: 25 Mar 2009
Posts: 55
Followers: 1

Kudos [?]: 12 [0], given: 9

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

Show Tags

28 Sep 2009, 11:10
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)
Intern
Joined: 09 Oct 2009
Posts: 1
Followers: 0

Kudos [?]: 0 [0], given: 0

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

Show Tags

09 Oct 2009, 14:55
The probability in this case would be as follow:

1/5+2/5+2/5 = 5/5 = 1
Math Expert
Joined: 02 Sep 2009
Posts: 33552
Followers: 5945

Kudos [?]: 73799 [15] , given: 9903

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

Show Tags

09 Oct 2009, 23:38
15
KUDOS
Expert's post
31
This post was
BOOKMARKED
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.
_________________
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 65

Kudos [?]: 669 [0], given: 19

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

Show Tags

10 Oct 2009, 22:39
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...
_________________
Manager
Joined: 10 Jul 2009
Posts: 129
Location: Ukraine, Kyiv
Followers: 3

Kudos [?]: 111 [0], given: 60

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

Show Tags

11 Oct 2009, 03:39
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)
_________________

Never, never, never give up

Manager
Joined: 11 Aug 2008
Posts: 161
Followers: 1

Kudos [?]: 46 [1] , given: 8

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

Show Tags

24 Oct 2009, 07:21
1
KUDOS
I think the ans is (3/8)^2*2/8*8!/(3!*3!*2!)
Manager
Joined: 05 Jul 2009
Posts: 182
Followers: 1

Kudos [?]: 37 [0], given: 5

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

Show Tags

24 Oct 2009, 21:51
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?
Senior Manager
Joined: 22 Dec 2009
Posts: 362
Followers: 11

Kudos [?]: 319 [2] , given: 47

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

Show Tags

17 Feb 2010, 04:01
2
KUDOS
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.
_________________

Cheers!
JT...........
If u like my post..... payback in Kudos!!

|For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~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

Go to page    1   2    Next  [ 29 posts ]

Similar topics Replies Last post
Similar
Topics:
4 In a jar there are 3 red balls and 2 blue balls. What is the 8 31 Jul 2014, 06:47
11 A bag has 4 blue, 3 yellow and 2 green balls. The balls of 7 30 Sep 2012, 09:13
36 A bag contains 3 white balls, 3 black balls & 2 red balls 21 29 Aug 2010, 09:06
2 A bag contains 3 white balls, 3 black balls & 2 red balls. 13 10 Aug 2010, 17:41
2 A has to pick a ball from a bag with 3 red & 3 blue balls. 11 16 Apr 2010, 09:04
Display posts from previous: Sort by