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

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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
00:00

Difficulty:

(N/A)

Question Stats:

76% (01:53) correct 24% (00:42) wrong based on 114 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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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
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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.
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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
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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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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
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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?
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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
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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.
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25 Aug 2008, 13:40
1
KUDOS
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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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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25 Aug 2008, 16:33
Can someone give the general rules for solving probability questions please?
Manager
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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
This post was
BOOKMARKED
1- [ 3c2/ 5c2] = 7/10
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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. [#permalink]

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18 Apr 2016, 18:51
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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
Re: Five marbles are in a bag: two are red and three are blue.   [#permalink] 23 Apr 2016, 05:34
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