Last visit was: 19 Nov 2025, 14:02 It is currently 19 Nov 2025, 14:02
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
Close
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
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
655-705 Level|   Probability|      
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,355
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,355
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 17 Feb 2010, 14:01
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jeeteshsingh
Bunuel
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case we are always drawing from 8 balls and the initial number of balls in bag remains the same, as we are putting withdrawn balls back (we have replacement).

In the SECOND case after each withdrawal we are reducing total number of balls by 1 AND reducing the withdrawn balls number by 1.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

No, order does not matter here and nowhere I mention in my solution that it does. The point is that, in both cases, combination of RGGBB can occur in 5!/2!2! # of ways (RGGBB, GGRBB, BRBGG, ... total 30 combinations), so that's why we are multiplying by it.
User avatar
claytont
User avatar
Joined: 18 Oct 2006
Last visit: 28 Feb 2010
Posts: 13
Own Kudos:
6
Posts: 13
Kudos: 6
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 28 Feb 2010, 15:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMAT TIGER
Bunuel
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...
Show more

It matters in the sense that you have to sum all the different ways in which to choose the balls. BUNUEL's answer is correct.

You have to multiply by 5!/(2!2!1!) because that is the number of ways to select 5 balls of which 2 are green (i.e. are similar), 2 are blue (i.e. are similar) and 1 is red.


look at it this way... Try to find the probability of all the different ways in which you can draw 5 balls in the manner suggested.
P(R,G,G,B,B) = 2/8*3/8*3/8*3/8*3/8
or
P(G,R,G,B,B) = 2/8*3/8*3/8*3/8*3/8
or
P(G,G,R,B,B) = 2/8*3/8*3/8*3/8*3/8
or
.
.
.

Thus the desired probability is = P(R,G,G,B,B) + P(G,R,G,B,B) + ....

Since there are 5!/(2!2!) similar addends, the desired probability is 5!/(2!2!1!) * 2/8*3/8*3/8*3/8*3/8
= 5!/(2!2!) * (1/4) * (3/8)^2 * (3/8)^2
User avatar
mainhoon
User avatar
Joined: 18 Jul 2010
Last visit: 10 Oct 2013
Posts: 534
Own Kudos:
392
Given Kudos: 15
Status:Apply - Last Chance
Affiliations: IIT, Purdue, PhD, TauBetaPi
Concentration: $ Finance $
Schools:Wharton, Sloan, Chicago, Haas
GPA: 4.0
WE 1: 8 years in Oil&Gas
Posts: 534
Kudos: 392
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 10 Aug 2010, 18:45
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show more

"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."

Bunuel,
About the above statement, can you explain the way of calculating Case I this way - using the counting approach. How would the calculation be different with the WITH case.

Also as a general doubt when one says 5 balls are drawn randomly WITH replacement - drawing 5 balls randomly suggests I grab 5 directly, however with the WITH replacement clause, it seems to indicate that I draw one then put it back, then draw another and then replace it, essentially drawing balls 1 by 1.. Is this understanding correct?
User avatar
Bunuel
User avatar
Math Expert
Joined: 02 Sep 2009
Last visit: 19 Nov 2025
Posts: 105,390
Own Kudos:
778,355
Given Kudos: 99,977
Products:
GMAT Club Tests Forum Quiz
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 105,390
Kudos: 778,355
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 11 Aug 2010, 03:28
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mainhoon
Bunuel
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.

"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."

Bunuel,
About the above statement, can you explain the way of calculating Case I this way - using the counting approach. How would the calculation be different with the WITH case.

Also as a general doubt when one says 5 balls are drawn randomly WITH replacement - drawing 5 balls randomly suggests I grab 5 directly, however with the WITH replacement clause, it seems to indicate that I draw one then put it back, then draw another and then replace it, essentially drawing balls 1 by 1.. Is this understanding correct?
Show more

When we are told that 5 balls are drawn WITH replacement it means that we draw 1 ball and put it back and then another one and put it back and so on. GMAT wording would make it clear and unambiguous.

So in the case WITH replacement we pick one ball at a time, so it's no need to use C here, as you wold have 1 from desired number divided 1 from all. For example instead of 2/8 you would have \(C^1_2/C^1_8\), which is complicated way of writing the same 2/8.
User avatar
feruz77
User avatar
Joined: 08 Oct 2010
Last visit: 22 Dec 2021
Posts: 169
Own Kudos:
2,565
Given Kudos: 974
Location: Uzbekistan
Concentration: Finance and accounting
Schools:Johnson, Fuqua, Simon, Mendoza
GPA: 4.0
WE 3: 10
Posts: 169
Kudos: 2,565
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 18 Jan 2011, 10:54
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Total ways of picking 5 out of 8 is 8C5=8!/5!3!=56
For what does follow next, we have to take into account that each time picked balls will be replaced:
Probability of picking 2 out of 3 green balls is 3C2/56=3/56
Probability of picking 2 out of 3 blue balls is 3C2/56=3/56
Probability of picking 1 out of 2 red balls is 2C1/56=2/56

Inasmuch as picking green, blue and red balls are independent of each other because of replacement, we can add individual probabilities:
(3C2/56)+( 3C2/56)+(2/56)=(3/56)+(3/56)+(2/56)=8/56=1/7≈14%

Here I would request participants of this forum to advice me whether my approach does work in all the cases with replacement when we have to do with probability?
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
 [4]
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
 [4]
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 27 Sep 2013, 20:26
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Bunuel
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.
Show more

Responding to a pm:
@Himanshu

Why are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR
Here we are picking 5 balls RGGBB. How many different cases can we have?
RGGBB
RGBGB
RBBGG
BBGGR
BGBGR
and many more...
The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.)

NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before.
avatar
jegf1987
avatar
Joined: 26 Sep 2015
Last visit: 25 Oct 2016
Posts: 2
Given Kudos: 454
Posts: 2
Kudos: 0
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 08 Dec 2015, 17:39
Kudos
Add Kudos
Bookmarks
Bookmark this Post
jeeteshsingh
Bunuel
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.
Show more

For the counting method on Case II by Bunnuel, why are we not accounting for all the different combinations, just like in the probability example?

What I mean is that for 2C1*3C2*3C2/ 8C5, unlike Bunnuel, I think that you need to multiply the numerator and the denominator by the number of combinations that are possible (since the order matters). For the probability method on Case II, we multiplied it by 5!/2!2!. Having said this, wouldn't we need to account for the number of combinations in the combination method? Thanks.
User avatar
KarishmaB
User avatar
Joined: 16 Oct 2010
Last visit: 19 Nov 2025
Posts: 16,267
Own Kudos:
77,000
Given Kudos: 482
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,267
Kudos: 77,000
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 22 Feb 2017, 03:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma


Responding to a pm:
himanshu

Why are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR
Here we are picking 5 balls RGGBB. How many different cases can we have?
RGGBB
RGBGB
RBBGG
BBGGR
BGBGR
and many more...
The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.)

NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before.
Show more

Quote:

I couldn figure out yet why dont we find simply the probability to select. For example selectin 2 green balls of 10 balls just 2/10. what is the difference here? dont it is asked simply to find prob of selecting these balls?
Show more

This is not correct. The probability of selecting 2 green balls out of 10 would not be 2/10.
2/10 would be the probability of selecting 1 green ball if out of 10 balls, exactly 2 were green.

Probability = Favourable cases / Total cases

Probability of selecting a green ball = 2/10 (if out of 10 balls, exactly 2 are green)

Probability of selecting two green balls without replacement = 2/10 * 1/9 (for the first ball, the probability of it being green is 2/10 and the for the second ball, the probability of it being green is 1/9 because now we have only 9 balls left and 1 of them is green)

Probability of selecting two green balls with replacement = 2/10 * 2/10 (each time the probability of picking a green ball is 2/10)

Does this help?
avatar
Taikiqi
avatar
Joined: 29 Dec 2020
Last visit: 21 Feb 2021
Posts: 13
Own Kudos:
6
Given Kudos: 14
Posts: 13
Kudos: 6
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 17 Jan 2021, 14:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Bunuel
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.
Show more

Bunuel, I am confused, can you please explain a bit more on your answer? I got (2/8)x(3/8)^2 x (3/8)^2.

Thanks.

Posted from my mobile device
User avatar
DanTheGMATMan
User avatar
Joined: 02 Oct 2015
Last visit: 18 Nov 2025
Posts: 378
Own Kudos:
226
Given Kudos: 9
Expert
Expert reply
Posts: 378
Kudos: 226
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 31 May 2024, 05:22
Kudos
Add Kudos
Bookmarks
Bookmark this Post
­Have to remember the orders for each case:

­
User avatar
kanwar08
User avatar
Joined: 18 Apr 2023
Last visit: 17 Nov 2025
Posts: 23
Own Kudos:
6
Given Kudos: 44
Location: India
Schools: ISB '26 IIMA '26
GMAT Focus 1: 655 Q88 V70 DI82
GMAT 1: 580 Q47 V26
Schools: ISB '26 IIMA '26
GMAT Focus 1: 655 Q88 V70 DI82
GMAT 1: 580 Q47 V26
Posts: 23
Kudos: 6
Rich has 3 green, 2 red and 3 blue balls in a bag. He randomly picks 01 Jul 2025, 19:55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Thanks KarishmaB for your response to this question.

I had one doubt. In probability, we are mainly concerned with Mutually Exclusive Events i.e. why we use Combination instead of permutation since order won't affect the probability.

Based on this understanding, I always approach questions by considering all MEEs in the question.

In this example, I understood the WITH replacement part but when considering for MEE, I got confused because I was typically used to applying this formula i.e. nCr(1/p)n-r (1-1/p)^r

Now, in this question I was thinking from the lens of nCr and was unable to arrive at it to consider MEE events

The one doubt I have in your answers is when we consider 5!/2!2!, isn't it simply considering all the possible ARRANGEMENTS?

Eg. If I have to find the number of ways to arrange AGGRRB = 6!/2!2!. So, this must consider that he order does matter and AB or BA will both be unique.

Can you please help me understand how I can fill this Gap?

KarishmaB
Bunuel
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.

Responding to a pm:
@Himanshu

Why are we multiplying by \(\frac{5!}{2!2!}\)? This is the equivalent of 2 in the previous question. There we picked 2 balls so we had 2 different cases RW and WR
Here we are picking 5 balls RGGBB. How many different cases can we have?
RGGBB
RGBGB
RBBGG
BBGGR
BGBGR
and many more...
The total number of cases is 5!/2!*2! (We divide by 2!s because there are 2 blues and 2 greens.)

NOte that we multiply by this number in both with and without replacement. The only thing that changes in without replacement case is the probability as discussed before.
Show more
   1   2 
Moderators:
Bunuel
Math Expert
105390 posts
Krunaal
Tuck School Moderator
805 posts