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Marbles

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Manager
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Marbles [#permalink] New post 18 Dec 2005, 07:40
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Please explain:

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
Director
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 [#permalink] New post 18 Dec 2005, 08:35
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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Re: Marbles [#permalink] New post 18 Dec 2005, 08:38
sandalphon wrote:
Please explain:

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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 [#permalink] New post 18 Dec 2005, 08:52
@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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 [#permalink] New post 18 Dec 2005, 09:15
allabout wrote:
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.
Thus answer is 3/10+2/5=7/10.
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 [#permalink] New post 18 Dec 2005, 23:35
can you give the answer sandalphon, please?
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 [#permalink] New post 19 Dec 2005, 00:38
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
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Re: Marbles [#permalink] New post 19 Dec 2005, 00:44
sandalphon wrote:
Please explain:

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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Re: Marbles [#permalink] New post 19 Dec 2005, 14:14
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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Re: Marbles [#permalink] New post 19 Dec 2005, 15:17
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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 [#permalink] New post 19 Dec 2005, 18:41
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
Senior Manager
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 [#permalink] New post 19 Dec 2005, 19:43
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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 [#permalink] New post 19 Dec 2005, 21:28
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
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 [#permalink] New post 20 Dec 2005, 00:40
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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 [#permalink] New post 20 Dec 2005, 02:03
P(at least 1 red)= 1 - P(no red marble) = 1- (3/5)*(2/4)=14/20=7/10

C.
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 [#permalink] New post 21 Dec 2005, 12:44
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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Re: [#permalink] New post 23 Feb 2008, 20:11
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)
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Re: Marbles [#permalink] New post 23 Feb 2008, 20:19
Yeah there is something wrong with the question or the answer. For probability for atleast one red from two should be 7/10
Re: Marbles   [#permalink] 23 Feb 2008, 20:19
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