It is currently 21 Oct 2017, 15:02

### 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

# There are 10 solid colored balls in a box, including 1 Green

Author Message
TAGS:

### Hide Tags

Senior Manager
Joined: 07 Sep 2010
Posts: 329

Kudos [?]: 1032 [3], given: 136

There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

29 Jun 2012, 23:17
3
KUDOS
25
This post was
BOOKMARKED
00:00

Difficulty:

25% (medium)

Question Stats:

80% (01:33) correct 20% (08:06) wrong based on 426 sessions

### HideShow timer Statistics

There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

Experts, request you to please explain this using both Combiantion approach as well as probability approach.

Also, one more doubt, can we solve this question using 1- P(Green, Yellow and Other ball) approach
Thanks
H
[Reveal] Spoiler: OA

_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Kudos [?]: 1032 [3], given: 136

Senior Manager
Joined: 23 Oct 2010
Posts: 381

Kudos [?]: 396 [3], given: 73

Location: Azerbaijan
Concentration: Finance
Schools: HEC '15 (A)
GMAT 1: 690 Q47 V38
Re: There are 10 solid colored balls in a box - Probability [#permalink]

### Show Tags

30 Jun 2012, 00:21
3
KUDOS
the worst case is

Some color-some another color- Green

ok, we have 10 balls , 1 of them is yellow, and the another one is green.
need to find that worst case

3*(8/10)*(7/9)*(1/8)=7/30

here the integer 3 means that u can order Some color-some another color-green in 3 ways (green -some c-some an.c.; some c-green-some an.c.-some c.-some an.c.-green)

8/10 means that u can select 8 (10 minus 1 yellow minus 1 green color) out of 10 colors
7/9 means that u can select 7 (9 minus 1 yellow minus 1 green color) out of 9 remaining colors
1/8 means that u have only one green out of 8 remaining colors

hope it helps
_________________

Happy are those who dream dreams and are ready to pay the price to make them come true

I am still on all gmat forums. msg me if you want to ask me smth

Kudos [?]: 396 [3], given: 73

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [8], given: 12194

Re: There are 10 solid colored balls in a box,including 1 Green [#permalink]

### Show Tags

30 Jun 2012, 03:09
8
KUDOS
Expert's post
17
This post was
BOOKMARKED
imhimanshu wrote:
There are 10 solid colored balls in a box,including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

Experts, request you to please explain this using both Combiantion approach as well as probability approach.

Also, one more doubt, can we solve this question using 1- P(Green, Yellow and Other ball) approach
Thanks
H

We have that there are 1 Green (G), 1 Yellow (Y) and 8 some other colors (X) balls in the box.

Approach #1:

We need to find the probability of GXX. $$P(GXX)=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}*\frac{3!}{2!}=\frac{7}{30}$$, we are multiplying by $$\frac{3!}{2!}$$ since GXX scenario could occur in 3 ways: GXX, XGX, or XXG (the number of permutations of 3 letters GXX out of which 2 X's are identical).

Approach #2:

$$P=\frac{C^2_8*C^1_1}{C^3_{10}}=\frac{7}{30}$$, where $$C^2_8$$ is ways to select 2 other color balls out of 8, $$C^1_1$$ is ways to select 1 green ball, and $$C^3_{10}$$ is total ways to select 3 balls out of 10.

Hope it's clear.
_________________

Kudos [?]: 129160 [8], given: 12194

Director
Joined: 22 Mar 2011
Posts: 610

Kudos [?]: 1059 [0], given: 43

WE: Science (Education)
Re: There are 10 solid colored balls in a box,including 1 Green [#permalink]

### Show Tags

30 Jun 2012, 05:06
1
This post was
BOOKMARKED
imhimanshu wrote:
There are 10 solid colored balls in a box, including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

Experts, request you to please explain this using both Combiantion approach as well as probability approach.

Also, one more doubt, can we solve this question using 1- P(Green, Yellow and Other ball) approach
Thanks
H

Probability approach, direct not for the complementary event (it is more complicated, not worth trying in this case: it can be one G and also one Y or neither G nor Y, plus an additional different color):
P(G) * P(noG & noY) * P( another noG & noY) * 3 = 1/10 * 8/9 * 7/8 * 3 = 7/30.
We need the factor of 3 because the Green ball can be either the first, second or third.

_________________

PhD in Applied Mathematics
Love GMAT Quant questions and running.

Kudos [?]: 1059 [0], given: 43

Senior Manager
Joined: 07 Sep 2010
Posts: 329

Kudos [?]: 1032 [0], given: 136

Re: There are 10 solid colored balls in a box,including 1 Green [#permalink]

### Show Tags

30 Jun 2012, 05:17
Hi Bunuel,

Thanks for the reply. I was able to solve it using Approach # 2. But I have doubt in Approach # 1 . Request you to please provide your comments.

Isn't it an assumption that remaining 8 balls are of same color. If the remaining 8 balls each of them are of different color, then the number of arrangements will get changed and hence probability.

Please tell me what I am missing
Thanks.
H

Bunuel wrote:

We have that there are 1 Green (G), 1 Yellow (Y) and 8 some other colors (X) balls in the box.

_________________

+1 Kudos me, Help me unlocking GMAT Club Tests

Kudos [?]: 1032 [0], given: 136

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [1], given: 12194

Re: There are 10 solid colored balls in a box,including 1 Green [#permalink]

### Show Tags

30 Jun 2012, 05:50
1
KUDOS
Expert's post
imhimanshu wrote:
Hi Bunuel,

Thanks for the reply. I was able to solve it using Approach # 2. But I have doubt in Approach # 1 . Request you to please provide your comments.

Isn't it an assumption that remaining 8 balls are of same color. If the remaining 8 balls each of them are of different color, then the number of arrangements will get changed and hence probability.

Please tell me what I am missing
Thanks.
H

Bunuel wrote:

We have that there are 1 Green (G), 1 Yellow (Y) and 8 some other colors (X) balls in the box.

We are not assuming that 8 other balls are necessarily of some one particular color: they can be of one color or 8 different colors, but for us the only thing which is important that they are of different color than green and yellow.

Consider this: if the first ball selected is green then we are left with 9 balls out of which 1 is yellow and 8 other balls are not yellow (that's the only thing we care), so the probability of selecting non-yellow ball is 8/9.

Hope it's clear.
_________________

Kudos [?]: 129160 [1], given: 12194

Intern
Status: Trying to crack GMAT
Joined: 17 May 2012
Posts: 39

Kudos [?]: 9 [0], given: 4

Location: India
Concentration: Operations, Technology
GMAT Date: 07-11-2012
GPA: 3.82
WE: Engineering (Computer Software)
Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

01 Jul 2012, 21:52
Hi Bunuel,

I would like to apologize in advance for asking such a "weird" question but i would still like to know:

Why did u consider that the remaining 8 balls did not have a green or yellow or both in them as well. I mean the question didn't state that the 1 G and 1 Y were the ONLY balls present, right?

But, I guess in PS questions we DO need an answer, so i can understand your logic. In that case, what would happen if this were a DS question? (I mean this IS a 700 level question, right?)
_________________

“When I was young I observed that nine out of ten things I did were failures, so I did ten times more work.” ~ Bernard Shaw

Kudos me if I helped you in any way.

Kudos [?]: 9 [0], given: 4

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [0], given: 12194

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

02 Jul 2012, 01:21
Aki wrote:
Hi Bunuel,

I would like to apologize in advance for asking such a "weird" question but i would still like to know:

Why did u consider that the remaining 8 balls did not have a green or yellow or both in them as well. I mean the question didn't state that the 1 G and 1 Y were the ONLY balls present, right?

But, I guess in PS questions we DO need an answer, so i can understand your logic. In that case, what would happen if this were a DS question? (I mean this IS a 700 level question, right?)

But we ARE told that there is is only 1 green ball and only 1 yellow ball in the box, because "there are 10 solid colored balls in a box, including 1 Green and 1 Yellow" means exactly that. How else?

Would it make ANY sense if there were for example 2 green balls and we were told that "there are 10 solid colored balls in a box, including 1 Green and 1 Yellow"?
_________________

Kudos [?]: 129160 [0], given: 12194

Intern
Status: Trying to crack GMAT
Joined: 17 May 2012
Posts: 39

Kudos [?]: 9 [0], given: 4

Location: India
Concentration: Operations, Technology
GMAT Date: 07-11-2012
GPA: 3.82
WE: Engineering (Computer Software)
Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

02 Jul 2012, 02:16
Bunuel wrote:
Aki wrote:
Hi Bunuel,

I would like to apologize in advance for asking such a "weird" question but i would still like to know:

Why did u consider that the remaining 8 balls did not have a green or yellow or both in them as well. I mean the question didn't state that the 1 G and 1 Y were the ONLY balls present, right?

But, I guess in PS questions we DO need an answer, so i can understand your logic. In that case, what would happen if this were a DS question? (I mean this IS a 700 level question, right?)

But we ARE told that there is is only 1 green ball and only 1 yellow ball in the box, because "there are 10 solid colored balls in a box, including 1 Green and 1 Yellow" means exactly that. How else?

Would it make ANY sense if there were for example 2 green balls and we were told that "there are 10 solid colored balls in a box, including 1 Green and 1 Yellow"?

Hmm.. yeah, that does make sense. I'm a non-native English speaker so I tend to over-analyze simple things. In this case, since the word only was missing i.e. there are 10 solid colored balls in a box, including exactly/only 1 Green and 1 Yellow . But yeah, in retrospect this is a Quant question not a Verbal SC Thanks for the explanation
_________________

“When I was young I observed that nine out of ten things I did were failures, so I did ten times more work.” ~ Bernard Shaw

Kudos me if I helped you in any way.

Kudos [?]: 9 [0], given: 4

Intern
Joined: 20 Sep 2011
Posts: 23

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

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

02 Jul 2012, 19:06
Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

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

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [2], given: 12194

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

03 Jul 2012, 03:56
2
KUDOS
Expert's post
3
This post was
BOOKMARKED
plock3vr wrote:
Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

The probability of each case will be the same:
{GXX} - $$P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XXG} - $$P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}$$;

The sum: $$\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}$$.

Hope it helps.
_________________

Kudos [?]: 129160 [2], given: 12194

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16587

Kudos [?]: 273 [0], given: 0

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

12 Oct 2013, 06:55
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Manager
Joined: 13 Jul 2013
Posts: 70

Kudos [?]: 13 [0], given: 21

GMAT 1: 570 Q46 V24
Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

02 Jan 2014, 01:59
Bunuel wrote:
plock3vr wrote:
Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

The probability of each case will be the same:
{GXX} - $$P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XXG} - $$P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}$$;

The sum: $$\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}$$.

Hope it helps.

In the second case {XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$,

how did we get 8/10? If we are considering any ball other than green, why is it not 9/10?

Kudos [?]: 13 [0], given: 21

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [1], given: 12194

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

02 Jan 2014, 05:59
1
KUDOS
Expert's post
theGame001 wrote:
Bunuel wrote:
plock3vr wrote:
Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

The probability of each case will be the same:
{GXX} - $$P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XXG} - $$P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}$$;

The sum: $$\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}$$.

Hope it helps.

In the second case {XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$,

how did we get 8/10? If we are considering any ball other than green, why is it not 9/10?

Because we don't need the yellow ball, X is any but green and yellow.
_________________

Kudos [?]: 129160 [1], given: 12194

Manager
Joined: 14 Jan 2013
Posts: 152

Kudos [?]: 378 [1], given: 30

Concentration: Strategy, Technology
GMAT Date: 08-01-2013
GPA: 3.7
WE: Consulting (Consulting)
Re: There are 10 solid colored balls in a box,including 1 Green [#permalink]

### Show Tags

14 Jan 2014, 16:54
1
KUDOS
Bunuel wrote:
imhimanshu wrote:
There are 10 solid colored balls in a box,including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

Experts, request you to please explain this using both Combiantion approach as well as probability approach.

Also, one more doubt, can we solve this question using 1- P(Green, Yellow and Other ball) approach
Thanks
H

We have that there are 1 Green (G), 1 Yellow (Y) and 8 some other colors (X) balls in the box.

Approach #1:

We need to find the probability of GXX. $$P(GXX)=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}*\frac{3!}{2!}=\frac{7}{30}$$, we are multiplying by $$\frac{3!}{2!}$$ since GXX scenario could occur in 3 ways: GXX, GXG, or XGG (the number of permutations of 3 letters GXX out of which 2 X's are identical).

Approach #2:

$$P=\frac{C^2_8*C^1_1}{C^3_{10}}=\frac{7}{30}$$, where $$C^2_8$$ is ways to select 2 other color balls out of 8, $$C^1_1$$ is ways to select 1 green ball, and $$C^3_{10}$$ is total ways to select 3 balls out of 10.

Hope it's clear.

Hi Bunuel,

I am confused with the red part.

I feel it should be - GXX. XGX, XXG. Please suggest. Thanks
_________________

"Where are my Kudos" ............ Good Question = kudos

"Start enjoying all phases" & all Sections

__________________________________________________________________
http://gmatclub.com/forum/collection-of-articles-on-critical-reasoning-159959.html

http://gmatclub.com/forum/percentages-700-800-level-questions-130588.html

http://gmatclub.com/forum/700-to-800-level-quant-question-with-detail-soluition-143321.html

Kudos [?]: 378 [1], given: 30

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [0], given: 12194

Re: There are 10 solid colored balls in a box,including 1 Green [#permalink]

### Show Tags

15 Jan 2014, 00:58
Mountain14 wrote:
Bunuel wrote:
imhimanshu wrote:
There are 10 solid colored balls in a box,including 1 Green and 1 Yellow. If 3 of the balls in the box are chosen at random, without replacement, what is the probability that the 3 balls chosen will include the Green ball but not the yellow ball.

A. 1/6
B. 7/30
C. 1/4
D. 3/10
E. 4/15

Experts, request you to please explain this using both Combiantion approach as well as probability approach.

Also, one more doubt, can we solve this question using 1- P(Green, Yellow and Other ball) approach
Thanks
H

We have that there are 1 Green (G), 1 Yellow (Y) and 8 some other colors (X) balls in the box.

Approach #1:

We need to find the probability of GXX. $$P(GXX)=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}*\frac{3!}{2!}=\frac{7}{30}$$, we are multiplying by $$\frac{3!}{2!}$$ since GXX scenario could occur in 3 ways: GXX, GXG, or XGG (the number of permutations of 3 letters GXX out of which 2 X's are identical).

Approach #2:

$$P=\frac{C^2_8*C^1_1}{C^3_{10}}=\frac{7}{30}$$, where $$C^2_8$$ is ways to select 2 other color balls out of 8, $$C^1_1$$ is ways to select 1 green ball, and $$C^3_{10}$$ is total ways to select 3 balls out of 10.

Hope it's clear.

Hi Bunuel,

I am confused with the red part.

I feel it should be - GXX. XGX, XXG. Please suggest. Thanks

Correct, It was a typo. Edited. Thank you.
_________________

Kudos [?]: 129160 [0], given: 12194

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16587

Kudos [?]: 273 [0], given: 0

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

18 Jan 2015, 14:26
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Chat Moderator
Joined: 19 Apr 2013
Posts: 689

Kudos [?]: 170 [0], given: 537

Concentration: Strategy, Healthcare
Schools: Sloan '18 (A)
GMAT 1: 730 Q48 V41
GPA: 4
Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

11 May 2015, 23:13
Bunuel wrote:
plock3vr wrote:
Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

The probability of each case will be the same:
{GXX} - $$P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XXG} - $$P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}$$;

The sum: $$\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}$$.

Hope it helps.

Bunuel, is there any way to know beforehand that all scenarios will be the same? I always write them out to see whether they are the same or not. Thanks.
_________________

If my post was helpful, press Kudos. If not, then just press Kudos !!!

Kudos [?]: 170 [0], given: 537

Math Expert
Joined: 02 Sep 2009
Posts: 41892

Kudos [?]: 129160 [0], given: 12194

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

12 May 2015, 03:44
Ergenekon wrote:
Bunuel wrote:
plock3vr wrote:
Bunuel question for you--and it may be dumb. When I tried the problem I knew that there were three possibilities for getting the green ball. However, I ended up calculating the probabilities for each scenario. Do we know that the probably is the same for each scenario or could it be different?

The probability of each case will be the same:
{GXX} - $$P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XGX} - $$P=\frac{8}{10}*\frac{1}{9}*\frac{7}{8}=\frac{7}{90}$$;

{XXG} - $$P=\frac{8}{10}*\frac{7}{9}*\frac{1}{8}=\frac{7}{90}$$;

The sum: $$\frac{7}{90}+\frac{7}{90}+\frac{7}{90}=\frac{7}{30}$$.

Hope it helps.

Bunuel, is there any way to know beforehand that all scenarios will be the same? I always write them out to see whether they are the same or not. Thanks.

1 G and 2 X's has the same probability no matter the order.
_________________

Kudos [?]: 129160 [0], given: 12194

GMAT Club Legend
Joined: 09 Sep 2013
Posts: 16587

Kudos [?]: 273 [0], given: 0

Re: There are 10 solid colored balls in a box, including 1 Green [#permalink]

### Show Tags

15 May 2016, 10:03
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________

Kudos [?]: 273 [0], given: 0

Re: There are 10 solid colored balls in a box, including 1 Green   [#permalink] 15 May 2016, 10:03

Go to page    1   2    Next  [ 22 posts ]

Display posts from previous: Sort by