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Nine marbles are numbered one through nine and placed in a

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Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 25 May 2007, 08:17
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Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63
B. 1/6
C. 4/9
D. 809/1260
E. cannot be determined

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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 25 May 2007, 11:40
Why four marbles in the cup? 3 rounds of 3 marbles are taken from the group of 9 marbles, every time discarding 2 and putting only 1 in the cup.

Wouldn't there be only 3 marbles in the cup?
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New post 25 May 2007, 12:02
I believe the second time he puts two in the jar and throws one away so at the end we would have 4 in the jar. I'm still working on it.....my probability skills aren't that great...gotta work on that!!
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 26 May 2007, 06:48
ajluberto wrote:
Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63
B. 1/6
C. 4/9
D. 809/1260
E. cannot be determined

HELP!!!


Ajluberto, are you sure you posted correct answer choices?
There are four prime numbers between 1 and 9, namely 2,3,5 and 7, so we are asked what is the probabality that all 4 prime numbers have been put into the cup.
I tried various approaces to this Q, but all my answers do not fit to your answer choices...
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 26 May 2007, 07:23
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After hours fighting agaist my math demmons I got the solution:


make 1 2 3 4 5 6 7 8 9
x p p x p x p x x

so we have 4 primes (p)

now we want for the first round pxx, xxp or xpx
(4/9*5/8*4/7) * 3 = 30/63
but we still want 1 p and for that the chance is 1/3 from the above
30/63*1/3 = 10/63

second round xpp,ppx,pxp
(3/6*3/5*2/4) * 3=9/20
but we still want 2p and for that the chance is 2/3 from the above
9/20*2/3=3/10

finnaly, we want from the rest pxx,xxp or xpx and we get
(1/3*2/2*1/1)*3=1
but we still want 1 p and for that the chance is 1/3 from the above
1*1/3=1/3

so great people the total probability is 10/63*3/10*1/3

what equals 1/63....that´s our answer!!!!!!!!!!!!!!!
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 30 May 2007, 09:35
Great Job andrehaui! Thats the answer. This question was from an advanced Princeton Review exercise.
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New post 31 May 2007, 13:02
Thank you my friend. I can tell you that was not easy to arrive the answer, but after all it is not a monster, it is just question of time and logic. Good luck at your Gmat! :wink:
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 09 Sep 2011, 21:10
Adrehaui, I think you made a small miscalculation. In the second round, the probability you picked the two primes is 1/3 not 2/3 (equal to the probability you will not pick a composite, there is only one composite out of three).

There is a much easier way to look at the question.

By an argument of symmetry, all numbers are equally likely to end inside the cup.

Therefore the answer is
P(choose 4 from 9).
For the first number in the cup, peobability it is a prime is 4/9
For second it is 3/8 (assuming the first WAS prime).
For third it is 2/7
For fourth it is 1/6

Giving a total probability of 1/126

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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 09 Sep 2011, 22:04
ajluberto wrote:
Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63
B. 1/6
C. 4/9
D. 809/1260
E. cannot be determined
HELP!!!


\(\frac{C^5_2*C^4_1}{C^9_3}*\frac{1}{3}*\frac{C^3_2*C^3_1}{C^6_3}*\frac{2}{3}*\frac{1}{3}=\frac{1}{63}\)

Ans: "A"
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 07 Oct 2016, 18:45
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ajluberto wrote:
Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63
B. 1/6
C. 4/9
D. 809/1260
E. cannot be determined

HELP!!!

This is sooooo easy........just kidding,......but here is my try at explaining this more than solving (as two members have already solved the problem), just so people interested can understand the logic behind. This might not be a GMAT Q, but definitely worth understanding, will help in the GMAT:

4 prime and 5 non-prime numbers in the list
1st selection:
1 used and 2 dropped
Total selected = 3 of 9 = 9C3
Selecting 1 prime = 4C1 (since only one marble is placed in the cup and we want this one marble to be a prime)
Selecting 2 non prime = 5C2

P(of selecting one prime and 2 non prime from the jar) = \(\frac{(4C1 * 5C2)}{9C3}\)

Now, out of the 3 selected (i.e. one prime and 2 non-prime) only one is placed in the cup and the other 2 are dropped, hence chance of a prime being placed in the cup is 1/3

P(of selecting one prime and 2 non prime from the jar & placing the prime in the cup) = \(\frac{(4C1 * 5C2)}{9C3} * \frac{1}{3}\)

2nd selection:
2 used and 1 dropped
Total selected = 3 of 6 = 6C3
Selecting 2 prime = 3C2 (since only two marbles is placed in the cup and we want these two marbles to be prime)
Selecting 1 non prime = 3C1

P(of selecting two prime and 1 non prime from the jar) = \(\frac{(3C2 * 3C1)}{6C3}\)

Now, out of the 3 selected (i.e. two prime and 1 non-prime) two are placed in the cup and the other is dropped, hence chance of both prime being placed in the cup is 2/3

P(of selecting two prime and 1 non prime from the jar & placing both prime in the cup) = \(\frac{(3C2 * 3C1)}{6C3} * \frac{2}{3}\)

3rd selection:
1 used and 2 dropped
Total selected = 3 of 3 = 3C3 (only 3 remain)
Selecting 1 prime = 1C1 (only 1 prime remains)
Selecting 2 non prime = 2C2 (only 2 non-prime remain)

P(of selecting one prime and two non prime from the jar) = \(\frac{(1C1 * 2C2)}{3C3}\)

Now, out of the 3 selected (i.e. one prime and two non-prime) one is placed in the cup and the other two are dropped, hence chance of prime being placed in the cup is 1/3

P(of selecting one prime and two non prime from the jar & placing placing prime in the cup) = \(\frac{(1C1 * 2C2)}{3C3} * \frac{1}{3}\)

P(all 4 marbles in the cup are prime) = \(\frac{(4C1 * 5C2)}{9C3} * \frac{1}{3} * \frac{(3C2 * 3C1)}{6C3} * \frac{2}{3} * \frac{(1C1 * 1C2)}{3C3} * \frac{1}{3}\) = \(\frac{10}{21} * \frac{1}{3} * \frac{9}{20} *\frac{2}{3} * 1 * \frac{1}{3}=\frac{1}{63}\)
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 09 Oct 2016, 00:54
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If you remove all the noise from the question. It basically says that what is the probability of selecting 4 numbers from 1-9 and all of them are prime. i.e 4C4/9C4 i.e 1/63.
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 10 Oct 2016, 18:48
sahilcordeiro wrote:
If you remove all the noise from the question. It basically says that what is the probability of selecting 4 numbers from 1-9 and all of them are prime. i.e 4C4/9C4 i.e 1/63.


OK so 4C4 is 1, but then 9C4 is 9! / 4!5! = 9*8*7*6 / 4*3*2 = 9*2*7 / 1 = 126 ... so it is 1/126. What am I doing wrong?
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 16 Jul 2017, 08:22
ajluberto wrote:
Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63
B. 1/6
C. 4/9
D. 809/1260
E. cannot be determined



Bunuel IanStewart VeritasPrepKarishma

Please help with this problem

I'm getting 1/126
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 16 Jul 2017, 11:40
sahilcordeiro wrote:
If you remove all the noise from the question. It basically says that what is the probability of selecting 4 numbers from 1-9 and all of them are prime. i.e 4C4/9C4 i.e 1/63.


4C4/9C4 comes out to be 1/126
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 19 Jul 2017, 01:02
ShashankDave wrote:
ajluberto wrote:
Nine marbles are numbered one through nine and placed in a jar. Three marbles are randomly selected from the jar, one is placed in a cup, and two are discarded. Next, another three marbles are selected from the jar, two are placed in the same cup as the first, and one is discarded. Finally, out of the three that remain in the jar, one is placed in the cup, and two are discarded. What is the probability that the four marbles in the cup all have prime numbers written on them?

A. 1/63
B. 1/6
C. 4/9
D. 809/1260
E. cannot be determined



Bunuel IanStewart VeritasPrepKarishma

Please help with this problem

I'm getting 1/126


Agreed, it is 1/126. In the second round, there is a 1/3 chance of putting two primes in the cup, not 2/3 as some have calculated.
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Re: Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 23 Jul 2017, 03:26
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theprobabilotyguy wrote:
Adrehaui, I think you made a small miscalculation. In the second round, the probability you picked the two primes is 1/3 not 2/3 (equal to the probability you will not pick a composite, there is only one composite out of three).

There is a much easier way to look at the question.

By an argument of symmetry, all numbers are equally likely to end inside the cup.

Therefore the answer is
P(choose 4 from 9).
For the first number in the cup, peobability it is a prime is 4/9
For second it is 3/8 (assuming the first WAS prime).
For third it is 2/7
For fourth it is 1/6

Giving a total probability of 1/126

Posted from my mobile device





_________________________________________________________________
Probability of picking a composite out of 3 = 1/3.
Hence, Probability of not picking a composite = 1-1/3 = 2/3 = Probability of picking 2 primes out of 3
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Nine marbles are numbered one through nine and placed in a  [#permalink]

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New post 23 Jul 2017, 06:20
ShashankDave wrote:

Bunuel IanStewart VeritasPrepKarishma

Please help with this problem

I'm getting 1/126


1/126 is the correct answer. My solution would be identical to theprobabilotyguy's above. The fact that we're discarding some marbles doesn't matter - we don't know anything about those discarded marbles, so they don't make it any more or less likely that the marbles we keep display prime numbers. So the answer is just:

4/9 * 3/8 * 2/7 * 1/6 = 1/126

So the correct answer isn't among the options. This isn't what hard GMAT questions are like anyway, so better to find a more realistic set of problems.

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Nine marbles are numbered one through nine and placed in a   [#permalink] 23 Jul 2017, 06:20
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