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 Your Progress

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.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Rich has 3 green, 2 red and 3 blue balls in a bag. He [#permalink]
08 Nov 2007, 07:59

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

50% (04:34) correct
50% (01:55) wrong based on 5 sessions

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?

Re: probability, colored balls [#permalink]
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.

Re: probability, colored balls [#permalink]
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%

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?

Re: probability, colored balls [#permalink]
09 Oct 2009, 22:38

12

This post received KUDOS

Expert's post

3

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.

Re: probability, colored balls [#permalink]
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... _________________

Re: probability, colored balls [#permalink]
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) _________________

Re: probability, colored balls [#permalink]
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?

Re: probability, colored balls [#permalink]
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. _________________

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

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

~~Better Burn Out... Than Fade Away~~

gmatclubot

Re: probability, colored balls
[#permalink]
17 Feb 2010, 03:01