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Current Student
Joined: 31 Aug 2007
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Five marbles are in a bag: two are red and three are blue.
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03 Dec 2007, 13:12
Question Stats:
77% (00:52) correct 23% (00:42) wrong based on 158 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 == 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. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



CEO
Joined: 17 Nov 2007
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Concentration: Entrepreneurship, Other
Schools: Chicago (Booth)  Class of 2011

C
13C2/5C2 = 13/10=7/10
3C2  2 blue of three 5C2  the total number of combination



Current Student
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walker wrote: C
13C2/5C2 = 13/10=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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young_gun wrote: agreed, supposedly the OA is 2/5...must be wrong.
hm... other way:
1st marble:
1. [Red;2/5]
2. [Blue;3/5]
2nd 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



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Joined: 29 Aug 2007
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Re: PS probability
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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



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Re: PS probability
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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: 693

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
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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
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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.



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Re: PS probability
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25 Aug 2008, 13:40
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
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Re: PS probability
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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



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Re: PS probability
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25 Aug 2008, 16:33
Can someone give the general rules for solving probability questions please?



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Re: PS probability
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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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Re: PS probability
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02 May 2011, 20:02
1 [ 3c2/ 5c2] = 7/10



Manager
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Re: PS probability
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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



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Re: Five marbles are in a bag: two are red and three are blue.
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23 Apr 2016, 05:34
A is the answer 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. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.



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