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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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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.

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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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

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.

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

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

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

Re: Five marbles are in a bag: two are red and three are blue. [#permalink]

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18 Apr 2016, 18:51

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