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

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

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New post 29 Jun 2012, 23:17
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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
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 30 Jun 2012, 03:09
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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).

Answer: B.

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.

Answer: B.

Hope it's clear.
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 30 Jun 2012, 00:21
7
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 :)
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 30 Jun 2012, 05:06
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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.

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

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New post 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.

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

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New post 30 Jun 2012, 05:50
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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.
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 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?)
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 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"?
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 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 :lol: Thanks for the explanation
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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

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New post 03 Jul 2012, 03:56
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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.
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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

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New post 02 Jan 2014, 05:59
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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.
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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

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

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New post 17 Jun 2017, 21:30
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


Let the other color be X.

First pick \(= \frac{1}{10}\) --------------- (Picking green ball or X but Not yellow)

Second pick \(= \frac{8}{9}\) -------------- (Picking green ball or X but Not yellow)

Third pick \(= \frac{7}{8}\) ----------------- (Picking green ball or X but Not yellow)

There are 3 ways to pick the required combination. GXX, XGX, XXG

Therefore required probability \(= 3 * \frac{1}{10}*\frac{8}{9}*\frac{7}{8} = \frac{7}{30}\) . Answer (B)...
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 28 Jun 2018, 18:03
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


The number of ways to select 3 balls that include the green ball but not the yellow ball is:

1C1 x 1C0 x 8C2 = 1 x 1 x (8 x 7)/2! = 28 ways

(Note: 1C1 is the number of ways to select 1 green ball from the 1 green ball, 1C0 is the number of ways to select 0 yellow ball from the 1 yellow ball, and 8C2 is the number of ways to select 2 balls from the 8 remaining balls.)

The number of ways to select 3 balls from 10 is:

10C3 = (10 x 9 x 8)/3! = (10 x 9 x 8)/(3 x 2) = 120

Thus, the probability that the 3 balls chosen will include the Green ball but not the yellow ball is 28/120 = 7/30.

Alternate Solution:

Let’s first determine the probability of selecting the one green ball on the first draw and any of the remaining balls (except the yellow ball) on the succeeding two draws: 1/10 x 8/9 x 7/8 = 56/720.

Now, we see that we could also pick the green ball on the second draw, with any ball (except the yellow one) on the first and third draws, with probability: 8/10 x 1/9 x ⅞ = 56/720.

Similarly, we could instead pick the green ball on the third draw, with any ball (except the yellow one) on the first and second draws, with probability: 8/10 x 7/9 x 1/8 = 56/720.

Thus the total probability for the event where the green ball is drawn but the yellow one is not is:
56/720 + 56/720 + 56/720 = 168/720 = 21/90 = 7/30.

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

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New post 16 May 2019, 06:30
I am wondering if my method here is correct as I arrived at the correct answer:

Total ways to choose 3 balls out of 10 = 10C3 = 120

Then I go to how many correct selections there are: 1 Green Selection * 8 Other Selections * 7 Other Selections

So my answer is (1*8*7)/120=7/30

Is this a proper way to do it?
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 16 May 2019, 06:41
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



Let the 10 balls be G,Y and a1...a8.

Probability= (Given Case)/(Total case)

Total Case= Choosing any three balls out of the lot-> 10C3

Now given case-> One of the ball must be green and other two from the lot a1...a8= 1C1 * 8C2

Therefore answer= (1C1 * 8C2)/(10C3)
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Re: There are 10 solid colored balls in a box, including 1 Green and 1  [#permalink]

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New post 26 Jun 2019, 12:44
Probability:
Green as 1st pick, Other as 2nd and 3rd = 1/10 * 8/9 * 7/8 = 7/90
Green as 2nd pick, Other as 1st and 3rd = 8/10 * 1/9 * 7/8 = 7/90
Green as 3rd pick, Other as 1st & 2nd = 8/10 * 7/9 * 1/8 = 7/90
21/90 = 3/70

Bunuel IanStewart EgmatQuantExpert or other experts:
I see that (E) is a trap answer if you don't fulfil the requirement of picking 3 balls (e.g. if you pick green first 1/10 and leave it like that instead of also picking non-yellow for other 2... made this mistake initially. But, (E) would be the correct answer if the question stated "If up to 3 balls are chosen at random, without replacement," yes?
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Re: There are 10 solid colored balls in a box, including 1 Green and 1   [#permalink] 26 Jun 2019, 12:44
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