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# Marbles

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Manager
Joined: 05 Nov 2005
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18 Dec 2005, 08:40
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Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.

1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20

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Director
Joined: 17 Dec 2005
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18 Dec 2005, 09:35
Same procedure as with PS 1-29 I think,

We must add the propability that,

one marble is red: 2/5 * 3/4= 6/20
with
two marbles are red: 2/5 * 1/4= 2/20

Which gives 2/5, so it is A.

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Manager
Joined: 02 Jun 2005
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18 Dec 2005, 09:38
sandalphon wrote:

Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.

1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20

Vote for 3)

Andy pick 2 out of 5 without replacing the marbles back into the bag. The chances of picking up 2 blue marbles are 3/5*1/2=3/10, thus the probability pikcing up at least 1 red marble is 1-3/10=7/10.

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Director
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18 Dec 2005, 09:52
@tinyseal,

can you see why we have different results, while our approaches are directly reverse? You come from behind and I'm from the front. Both correct. Can't see the problem.

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Manager
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18 Dec 2005, 10:15
two marbles are red: 2/5 * 1/4= 2/20

allabout, I think that the probility of picking up two red marbles is: 2/5.

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Director
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19 Dec 2005, 00:35

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Intern
Joined: 06 Dec 2005
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19 Dec 2005, 01:38
I think 2/5 ..as said by allabout

To elucidate

The event can be seen as

1.atleast 1 red marble + no red marble .

prob of drawing atleast 1 red marble + probablity of no red marble(meaning that the two marbles drawn are blue) = 1 ....(a)

probablity of drawing 2 blue marbles = 3/5
hence prob of atleast on red marble = 1-3/5 = 2/5

thanx
_________________

Any one can know the seeds in an apple ,BUT only GOD know the number of apples in the seed

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Senior Manager
Joined: 15 Apr 2005
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Location: India, Chennai

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19 Dec 2005, 01:44
sandalphon wrote:

Five marbles are in a bag: two are red and three are blue. If Andy randomly picks two of the marbles, what is the probability that at least one of the marbles he chooses will be red.

1.) 2/5
2.) 6/10
3.) 7/10
4.) 15/20
5.) 19/20

P(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5

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Intern
Joined: 16 Oct 2005
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19 Dec 2005, 15:14
krisrini wrote:
P(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5

Krisrini, it's a smart way of doing it..

I calculated this as
Probability that both are red + Probability that one is read = 2/5*1/4 + 2/5 * 3/4 = 2/20 + 6/20 = 2/5

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Senior Manager
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19 Dec 2005, 16:17
strange wrote:
krisrini wrote:
P(Atleast one red) = 1- probablity (no red)
= 1 - 3/5
= 2/5

Krisrini, it's a smart way of doing it..

I calculated this as
Probability that both are red + Probability that one is read = 2/5*1/4 + 2/5 * 3/4 = 2/20 + 6/20 = 2/5

It should be 7/10.

Probability of no reds is not 3/5. It is 3/5 *2/4 = 3/10
1 - Prob of red = 1 - 3/10 = 7/10

Different approach:
- There are 5 balls, thus there are 5C2 ways to pick 2, or (5*4)/2! = 10.
- There are 3C2 ways to pick only blues (3*2)/2! = 3.
- (Total combinations) - (combinations with no reds) = 10 - 3 = 7
- Therefore, 7/10 is the prob of picking at least 1 red.

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GMAT Club Legend
Joined: 07 Jul 2004
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19 Dec 2005, 19:41
P(at least one is red) = 1-P(pick 2 blue)
P(pick 2 blue) = P(none is red)

P(pick 2 blue) = 3/5*2/4 = 3/10
P(at least 1 red) = 1-3/10 = 7/10

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Senior Manager
Joined: 09 Aug 2005
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19 Dec 2005, 20:43
total combos = 5C2 = 10

good results

R1R2

R1B1 R1B2 R1B3

R2B1, R2B2, R2B3

7 in all

so 7/10

in GMAT usually numbers are small - I find this approach more confidance building and convincing.

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SVP
Joined: 28 May 2005
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19 Dec 2005, 22:28
it can be 1R1B or 1B1R or 2R

= 2/5*1/4 + 3/5*2/4 + 2/5*3/4
= 14/20 or 7/10
_________________

hey ya......

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Director
Joined: 17 Dec 2005
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20 Dec 2005, 01:40
nakib77 wrote:
it can be 1R1B or 1B1R or 2R

= 2/5*1/4 + 3/5*2/4 + 2/5*3/4
= 14/20 or 7/10

That's the point why I think 2/5 is right.

the order in which we choose the two marbles doesn't matter, so that we have not to consider RB and BR as different possibilities.

Hope sandalphon will clear it up soon, I'm curious about the result.

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Director
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20 Dec 2005, 03:03
P(at least 1 red)= 1 - P(no red marble) = 1- (3/5)*(2/4)=14/20=7/10

C.

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Manager
Joined: 05 Nov 2005
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21 Dec 2005, 13:44
OA is A, altought I have great problems deciphering that question. I have no OE since it is ffrom the Princeton CD, sorry guys.

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CEO
Joined: 21 Jan 2007
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23 Feb 2008, 21:11
sandalphon wrote:
OA is A, altought I have great problems deciphering that question. I have no OE since it is ffrom the Princeton CD, sorry guys.

there is no way that the OA is A.

A is only possible if the question asked what is the probability of choosing 1 red marble out of the total.

at least one red = 1 - (3/5)(2/4)
OR
at least one red = (5C2 - 3C2)/(5C2)
_________________

You tried your best and you failed miserably. The lesson is 'never try'. -Homer Simpson

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Manager
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23 Feb 2008, 21:19
Yeah there is something wrong with the question or the answer. For probability for atleast one red from two should be 7/10

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Re: Marbles   [#permalink] 23 Feb 2008, 21:19
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