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Five marbles are in a bag: two are red and three are blue.

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Current Student
Joined: 31 Aug 2007
Posts: 365
Five marbles are in a bag: two are red and three are blue.  [#permalink]

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03 Dec 2007, 13:12
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Question Stats:

77% (00:52) correct 23% (00:42) wrong based on 145 sessions

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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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CEO
Joined: 17 Nov 2007
Posts: 3458
Concentration: Entrepreneurship, Other
Schools: Chicago (Booth) - Class of 2011
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03 Dec 2007, 13:35
C

1-3C2/5C2 = 1-3/10=7/10

3C2 - 2 blue of three
5C2 - the total number of combination
Current Student
Joined: 31 Aug 2007
Posts: 365

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03 Dec 2007, 14:04
walker wrote:
C

1-3C2/5C2 = 1-3/10=7/10

3C2 - 2 blue of tree
5C2 - total number of combination

agreed, supposedly the OA is 2/5...must be wrong.
CEO
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03 Dec 2007, 14:14
young_gun wrote:
agreed, supposedly the OA is 2/5...must be wrong.

hm... other way:

1-st marble:

1. [Red;2/5]
2. [Blue;3/5]

2-nd marble:

1. [Red;2/5][Red or Blue;1] = 2/5
2. [Blue;3/5][Red;2/4] = 6/20=3/10

p=2/5+3/10=7/10
SVP
Joined: 29 Aug 2007
Posts: 2410

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03 Dec 2007, 16:26
young_gun 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

at least 1 is red = 1 - no red
at least 1 is red = 1 - all blue
at least 1 is red = 1 - 3c2/5c2
at least 1 is red = 1 - 3/10 = 7/10
SVP
Joined: 29 Mar 2007
Posts: 2465

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03 Dec 2007, 23:04
young_gun 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

How bout none will be red. Very fast calculation then.

3/5*2/4 = 6/20 --> 14/20 --> 7/10. C
Director
Joined: 09 Aug 2006
Posts: 742

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03 Dec 2007, 23:06
I suck at these but I did the following and got the same answer as OA:

You can either pick both red OR you can pick 1 red and 1 blue.

p(1st is red) * p(2nd is red) = 2/5 * 1/4 = 2/20

p(1st is red) * p(2nd is blue) = 2/5 * 3/4 = 6/20

Total p = 2/20 + 6/20 = 2/5

What's the source?
CEO
Joined: 17 Nov 2007
Posts: 3458
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03 Dec 2007, 23:48
GK_Gmat wrote:
I suck at these but I did the following and got the same answer as OA:

You can either pick both red OR you can pick 1 red and 1 blue.

p(1st is red) * p(2nd is red) = 2/5 * 1/4 = 2/20

p(1st is red) * p(2nd is blue) = 2/5 * 3/4 = 6/20

Total p = 2/20 + 6/20 = 2/5

What's the source?

You should add p(1st is blue) * p(2nd is red) = 3/5*2/4=6/20

2/5+6/20=7/10
Director
Joined: 09 Aug 2006
Posts: 742

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04 Dec 2007, 00:00
walker wrote:
GK_Gmat wrote:
I suck at these but I did the following and got the same answer as OA:

You can either pick both red OR you can pick 1 red and 1 blue.

p(1st is red) * p(2nd is red) = 2/5 * 1/4 = 2/20

p(1st is red) * p(2nd is blue) = 2/5 * 3/4 = 6/20

Total p = 2/20 + 6/20 = 2/5

What's the source?

You should add p(1st is blue) * p(2nd is red) = 3/5*2/4=6/20

2/5+6/20=7/10

Ok; didn't consider that. Thanks a lot! Agree on 7/10.
SVP
Joined: 07 Nov 2007
Posts: 1707
Location: New York

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25 Aug 2008, 13:40
1
young_gun 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

$$p=1- (3C2/ 5C2)=7/10$$
or
$$p=(2C2*3C0+2C1*3C1)/ 5C2= (6+1)/10 =7/10$$
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Manager
Joined: 22 Jul 2008
Posts: 138

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25 Aug 2008, 14:33
You cannot do 2 C 1 and 3 C 1 in this case because they are identical marbles. Therefore, to choose 1 Red or 1 blue marble, there is only 1 way from amongst the Red or Blue marbles. However, the method used is very simple.

1 R + 1 B--- 2/5 * 3/4 = 6/20
2 R -----2/5 * 1/4 = 2/20
Therefore, total Probability = 6/20 + 2/20 = 8/20 = 2/5
Intern
Joined: 23 May 2008
Posts: 47

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25 Aug 2008, 16:33
Can someone give the general rules for solving probability questions please?
Manager
Joined: 27 Oct 2008
Posts: 180

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27 Sep 2009, 21:28
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

Soln:
Probability that atleast one will be red is
= Probability that exactly one is red + probability that both are red
= (2C1 * 3C1/5C2) + 2C2/5C2
= 7/10
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02 May 2011, 20:02
1
1- [ 3c2/ 5c2] = 7/10
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Manager
Joined: 09 Aug 2010
Posts: 97

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02 May 2011, 21:29
Another approach:

R R B B B

Using Anagram Method:

How many ways can we select 2 of any color from 5? 5!/(2!3!) = 10
How many ways can we select (R,B) from R R B B B? 2!/1!1! x 3!/1!2! = 2 x 3 = 6
How many ways can we select (R,R) from R R B B B? 2!/2! = 1

Probability(of at least one Red) = (6 + 1)/10 = 7/10 OR C
Intern
Joined: 28 May 2014
Posts: 6
Schools: HBS '19 (WD)
Re: Five marbles are in a bag: two are red and three are blue.  [#permalink]

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23 Apr 2016, 05:34

Total ways of getting 2 balls = 5C2
Total ways of getting atleast one ball = 2C1*3C1 + 2C2

Probability = 2C1*3C1+2C2 /5C2 = 4/10=2/5

--== Message from the GMAT Club Team ==--

THERE IS LIKELY A BETTER DISCUSSION OF THIS EXACT QUESTION.
This discussion does not meet community quality standards. It has been retired.

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Joined: 09 Sep 2013
Posts: 8176
Re: Five marbles are in a bag: two are red and three are blue.  [#permalink]

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27 Aug 2018, 14:15
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Re: Five marbles are in a bag: two are red and three are blue. &nbs [#permalink] 27 Aug 2018, 14:15
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