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Five marbles are in a bag: two are red and three are blue. [#permalink]
03 Dec 2007, 12:12

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Difficulty:

(N/A)

Question Stats:

69% (01:43) correct
31% (00:29) wrong based on 52 sessions

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.

Re: PS probability [#permalink]
03 Dec 2007, 15:26

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

Re: PS probability [#permalink]
03 Dec 2007, 22:04

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.

Re: PS probability [#permalink]
25 Aug 2008, 12:40

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 \(p=(2C2*3C0+2C1*3C1)/ 5C2= (6+1)/10 =7/10\) _________________

Your attitude determines your altitude Smiling wins more friends than frowning

Re: PS probability [#permalink]
25 Aug 2008, 13:33

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

Re: PS probability [#permalink]
27 Sep 2009, 20:28

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

Re: PS probability [#permalink]
02 May 2011, 20:29

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

gmatclubot

Re: PS probability
[#permalink]
02 May 2011, 20:29

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