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

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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

Re: There are 10 solid colored balls in a box - Probability [#permalink]

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30 Jun 2012, 00:21

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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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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: D.

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.

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

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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 [#permalink]

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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.

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.

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

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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

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 [#permalink]

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

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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?

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} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

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

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12 Oct 2013, 06:55

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

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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} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

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} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

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

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14 Jan 2014, 16:54

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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).

Answer: D.

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: D.

Hope it's clear.

Hi Bunuel,

I am confused with the red part.

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

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).

Answer: D.

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: D.

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

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18 Jan 2015, 14:26

Hello from the GMAT Club BumpBot!

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

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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} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

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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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} - \(P=\frac{1}{10}*\frac{8}{9}*\frac{7}{8}=\frac{7}{90}\);

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 [#permalink]

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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).

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