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crazypriya
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From a jar that contains marbles, including 7 red marbles, Jennifer Updated on: 01 Oct 2023, 05:44
Originally posted by crazypriya on 11 Dec 2013, 07:44.
Last edited by BottomJee on 01 Oct 2023, 05:44, edited 5 times in total.
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Total number of
marbles: 21 % Likelihood that
Jennifer wins: 10
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65% (hard)

Question Stats:

61% (02:17) correct 39% (02:00) wrong based on 1223 sessions

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From a jar that contains marbles, including 7 red marbles, Jennifer can select two marbles simultaneously, and if both are red she wins a prize.

From the table below, select the choices that provide a total number of marbles in the jar and a corresponding percent probability that Jennifer will win.
Total number of
marbles 		% Likelihood that
Jennifer wins 		Values
7
10
20
21
30
35
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Total number of
marbles: 21 % Likelihood that
Jennifer wins: 10
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mikemcgarry
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From a jar that contains marbles, including 7 red marbles, Jennifer 11 Dec 2013, 14:23
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crazypriya
From a jar that contains marbles, including 7 red marbles, Jennifer can select two marbles simultaneously, and if both are red she wins a prize. From the table below, select the choices that provide a total number of marbles in the jar and a corresponding percent probability that Jennifer will win.


Total number of marbles % likelihood that Jennifer wins
7

10

20

21

30

35

Show Spoiler
10 and 21
Show more
Dear crazypriya,
I'm happy to help. :-)

I don't know that there's a good way to do this other than plugging in one at at time. Let
N = total number of marbles.
If N = 7, then there's a 100% chance of picking two red marbles. That doesn't work.

If N = 10, then there's a 7/10 chance of getting red on the first pick, and 6/9 = 1/3 of getting red on the second pick. That's a (7/10)*(1/3) = 7/30 chance, which is approx. 23.33%. That doesn't work.

If N = 20, then there's a 7/20 chance of getting red on the first pick, and 6/19 of getting red on the second pick. That's a (7/20)*(6/19) = 21/190 chance, which is slightly more than 10 percent. That doesn't work.

If N = 21, then there's a 7/21 = 1/3 chance of getting red on the first pick, and 6/20 = 3/10 of getting red on the second pick. That's a (1/3)*(3/10) = 1/10, or 10% chance. This works.

I disagree with the order of answers given in the spoiler. The total number of marbles is 21, and the % probability that Jennifer wins is 10.

Does all this make sense?
Mike :-)
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From a jar that contains marbles, including 7 red marbles, Jennifer 11 Dec 2013, 19:15
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Thanks Mike....
Yes u r correct. The total number of marbles is 21, and the % probability that Jennifer wins is 10.

Actually this plugging method was given in OE....so I was wondering if there is any better method to do this....
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From a jar that contains marbles, including 7 red marbles, Jennifer 13 Dec 2013, 16:02
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crazypriya
Thanks Mike....
Yes u r correct. The total number of marbles is 21, and the % probability that Jennifer wins is 10.

Actually this plugging method was given in OE....so I was wondering if there is any better method to do this....
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Dear crazypriya,
No, I don't think there's a better way. You see, there are two variables, and only one equation relating them. If we set it up with algebra, we would get a complicated equation that we couldn't solve anyway. The format itself demands plug-in approach.
Does this make sense?
Mike :-)
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From a jar that contains marbles, including 7 red marbles, Jennifer 27 Feb 2014, 12:45
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is this question missing the other answer choices? # of marbles and % probability that Jennifer will win?
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From a jar that contains marbles, including 7 red marbles, Jennifer 27 Feb 2014, 12:54
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se7en14
is this question missing the other answer choices? # of marbles and % probability that Jennifer will win?
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Dear se7en14,
This is a Two-Part Analysis question (2PA), one of the four question types on the GMAT's IR section. In 2PA questions, the same list of possible answers applies to both questions. Here, crazypriya posted the question in plaintext format, but on the GMAT it would be part of a chart, which would make the selection process slightly clearer.
See this for an explanation of the basis of 2PA:
https://magoosh.com/gmat/2012/integrated ... -analysis/
See this for 2PA practice questions:
https://magoosh.com/gmat/2013/gmat-integ ... questions/

Please let me know if you have any further questions.
Mike :-)
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From a jar that contains marbles, including 7 red marbles, Jennifer 18 May 2024, 23:41
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mikemcgarry

crazypriya
From a jar that contains marbles, including 7 red marbles, Jennifer can select two marbles simultaneously, and if both are red she wins a prize. From the table below, select the choices that provide a total number of marbles in the jar and a corresponding percent probability that Jennifer will win.


Total number of marbles % likelihood that Jennifer wins
7

10

20

21

30

35

Show Spoiler
10 and 21
Dear crazypriya,
I'm happy to help. :-)

I don't know that there's a good way to do this other than plugging in one at at time. Let
N = total number of marbles.
If N = 7, then there's a 100% chance of picking two red marbles. That doesn't work.

If N = 10, then there's a 7/10 chance of getting red on the first pick, and 6/9 = 1/3 of getting red on the second pick. That's a (7/10)*(1/3) = 7/30 chance, which is approx. 23.33%. That doesn't work.

If N = 20, then there's a 7/20 chance of getting red on the first pick, and 6/19 of getting red on the second pick. That's a (7/20)*(6/19) = 21/190 chance, which is slightly more than 10 percent. That doesn't work.

If N = 21, then there's a 7/21 = 1/3 chance of getting red on the first pick, and 6/20 = 3/10 of getting red on the second pick. That's a (1/3)*(3/10) = 1/10, or 10% chance. This works.

I disagree with the order of answers given in the spoiler. The total number of marbles is 21, and the % probability that Jennifer wins is 10.

Does all this make sense?
Mike :-)
Show more
­Hi chetan2u KarishmaB Bunuel

Question say ''two marbles simultaneously'', but in the above explanation has not  mikemcgarry
 calculated the probablity of marbles taken one after the another?
Am i missing something?

Because, 
Picking 2 things simultaneously from 5 objects should be = 5 c 2 = 10
One after another should be = 20­
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From a jar that contains marbles, including 7 red marbles, Jennifer 19 May 2024, 00:16
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ashutosh_73

mikemcgarry

crazypriya
From a jar that contains marbles, including 7 red marbles, Jennifer can select two marbles simultaneously, and if both are red she wins a prize. From the table below, select the choices that provide a total number of marbles in the jar and a corresponding percent probability that Jennifer will win.


Total number of marbles % likelihood that Jennifer wins
7

10

20

21

30

35

Show Spoiler
10 and 21
Dear crazypriya,
I'm happy to help. :-)

I don't know that there's a good way to do this other than plugging in one at at time. Let
N = total number of marbles.
If N = 7, then there's a 100% chance of picking two red marbles. That doesn't work.

If N = 10, then there's a 7/10 chance of getting red on the first pick, and 6/9 = 1/3 of getting red on the second pick. That's a (7/10)*(1/3) = 7/30 chance, which is approx. 23.33%. That doesn't work.

If N = 20, then there's a 7/20 chance of getting red on the first pick, and 6/19 of getting red on the second pick. That's a (7/20)*(6/19) = 21/190 chance, which is slightly more than 10 percent. That doesn't work.

If N = 21, then there's a 7/21 = 1/3 chance of getting red on the first pick, and 6/20 = 3/10 of getting red on the second pick. That's a (1/3)*(3/10) = 1/10, or 10% chance. This works.

I disagree with the order of answers given in the spoiler. The total number of marbles is 21, and the % probability that Jennifer wins is 10.

Does all this make sense?
Mike :-)
­Hi chetan2u KarishmaB Bunuel

Question say ''two marbles simultaneously'', but in the above explanation has not  mikemcgarry
 calculated the probablity of marbles taken one after the another?
Am i missing something?

Because, 
Picking 2 things simultaneously from 5 objects should be = 5 c 2 = 10
One after another should be = 20­
Show more
­Yes, you are correct. The question uses word select and simultaneously, both of which tell us that we are looking at combination.

However, it didn't make a difference as it is probability question => 
Actual = \(\frac{7C2}{nC2}=\frac{7C2*2!}{nC2*2!}=\frac{7P2}{nP2}\)

Solution:
\(\frac{7C2}{nC2}=\frac{7*6}{n(n-1)}\)

Now choose from options. Start from middle.....
\(\frac{7*6}{n(n-1)}\)=\(\frac{7*6}{21(21-1)}=\frac{1}{10}\) and as a % = 10%
Thus, 21 and 10 are the answers
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From a jar that contains marbles, including 7 red marbles, Jennifer 19 May 2024, 00:31
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ashutosh_73

mikemcgarry

crazypriya
From a jar that contains marbles, including 7 red marbles, Jennifer can select two marbles simultaneously, and if both are red she wins a prize. From the table below, select the choices that provide a total number of marbles in the jar and a corresponding percent probability that Jennifer will win.


Total number of marbles % likelihood that Jennifer wins
7

10

20

21

30

35

Show Spoiler
10 and 21
Dear crazypriya,
I'm happy to help. :-)

I don't know that there's a good way to do this other than plugging in one at at time. Let
N = total number of marbles.
If N = 7, then there's a 100% chance of picking two red marbles. That doesn't work.

If N = 10, then there's a 7/10 chance of getting red on the first pick, and 6/9 = 1/3 of getting red on the second pick. That's a (7/10)*(1/3) = 7/30 chance, which is approx. 23.33%. That doesn't work.

If N = 20, then there's a 7/20 chance of getting red on the first pick, and 6/19 of getting red on the second pick. That's a (7/20)*(6/19) = 21/190 chance, which is slightly more than 10 percent. That doesn't work.

If N = 21, then there's a 7/21 = 1/3 chance of getting red on the first pick, and 6/20 = 3/10 of getting red on the second pick. That's a (1/3)*(3/10) = 1/10, or 10% chance. This works.

I disagree with the order of answers given in the spoiler. The total number of marbles is 21, and the % probability that Jennifer wins is 10.

Does all this make sense?
Mike :-)
­Hi chetan2u KarishmaB Bunuel

Question say ''two marbles simultaneously'', but in the above explanation has not  mikemcgarry
 calculated the probablity of marbles taken one after the another?
Am i missing something?

Because, 
Picking 2 things simultaneously from 5 objects should be = 5 c 2 = 10
One after another should be = 20­
Show more

Mathematically, the probability of picking items simultaneously or selecting them one by one without replacement is the same.

For example, in the case of 7 red marbles out of 21, the probability of getting two reds if two marbles are picked one by one is:

7/21 * 6/20 = 1/10

If two marbles are picked simultaneously, we get:

7C2 / 21C2 = 21/210 = 1/10.
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From a jar that contains marbles, including 7 red marbles, Jennifer 20 Jul 2025, 22:30
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One way we can think of it more smartly is that, number of red marbles are 7, they expect a probability % which is in whole number. Therefore, the total number of marbles should be such that it should get divisible by 7 and 6. This will simplify our fraction. Therefore, trying with 21 and 35 first.

With total=21, it would give us a whole number in terms of percent probability.

Hope this helps! :)
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From a jar that contains marbles, including 7 red marbles, Jennifer 23 Aug 2025, 01:59
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there is indeed a better way to solve than plugging values just like mikemcgarry did

Let total marbles be N

Probability of 2 reds simultaneously = 7*6/N*(N-1)
This probability is basically likelihood of getting 2 reds, hence it is equal to P/100 where P is likelihood %age

Hence,
P/100 = 7*6/N*(N-1)
P*N*(N-1) = 4200

Now it's easier to plug values in just 1 formula. By the looks of it, it is clearly evident that one of 3 values will be multiple of 7, because 4200 is divisible by 7.
By smart plugging, we can get P=10 and N=21
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