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Probability... what approach to use

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Probability... what approach to use [#permalink] New post 05 Feb 2011, 16:41
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There are 5 red marbles, 3 blue marbles and 2 greeen marbles. If a user chooses to use two marbles,what is the probability that the two marbles will be a different color?

The approach that I used was:

5/10*4/9=2/9
3/10*2/9=1/15
2/10*1/9=1/45

I added the results and came up with 9/45 then when that is simplified I get 1/5.
Then
1-1/5=4/5 this is the result I got but the answer says 31/45 after using the combination approach.

Shouldn't the two approaches bring about the same result? Or did I do something wrong?

Mari
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Re: Probability... what approach to use [#permalink] New post 05 Feb 2011, 16:53
Expert's post
mariyea wrote:
There are 5 red marbles, 3 blue marbles and 2 greeen marbles. If a user chooses to use two marbles,what is the probability that the two marbles will be a different color?

The approach that I used was:

5/10*4/9=2/9
3/10*2/9=1/15
2/10*1/9=1/45

I added the results and came up with 9/45 then when that is simplified I get 1/5.
Then
1-1/5=4/5 this is the result I got but the answer says 31/45 after using the combination approach.

Shouldn't the two approaches bring about the same result? Or did I do something wrong?

Mari


You did everything right except the addition: 2/9+1/15+1/45=10/45+3/45+1/45=14/45 (not 9/45) --> 1-14/45=31/45.
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Re: Probability... what approach to use [#permalink] New post 05 Feb 2011, 22:44
Mariyea did it by finding out probability of selecting two marbles of same colors and then taking the complement.

The other way using combination;

The total number of ways of choosing any 2 marbles out of 10 is
C^{10}_2 = \frac{10*9}{2} = 45

5R(red marbles), 3B(blue marbles) and 2G(greeen marbles)

Favorable ways of selecting 2 distinct colored marbles:
1R out of 5R AND 1B out of 3B
OR
1R out of 5R AND 1G out of 2G
OR
1B out of 3B AND 1G out of 2G

C^5_1*C^3_1+C^5_1*C^2_1+C^3_1*C^2_1
=5*3+5*2+3*2
=15+10+6=31

Ans: P=\frac{31}{45}
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Re: Probability... what approach to use [#permalink] New post 07 Feb 2011, 05:40
Bunuel wrote:
mariyea wrote:
There are 5 red marbles, 3 blue marbles and 2 greeen marbles. If a user chooses to use two marbles,what is the probability that the two marbles will be a different color?

The approach that I used was:

5/10*4/9=2/9
3/10*2/9=1/15
2/10*1/9=1/45

I added the results and came up with 9/45 then when that is simplified I get 1/5.
Then
1-1/5=4/5 this is the result I got but the answer says 31/45 after using the combination approach.

Shouldn't the two approaches bring about the same result? Or did I do something wrong?

Mari


You did everything right except the addition: 2/9+1/15+1/45=10/45+3/45+1/45=14/45 (not 9/45) --> 1-14/45=31/45.

Unbelievable! I can't believe I did that... Thanks Bunuel! I don't know what's wrong with me..
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Re: Probability... what approach to use [#permalink] New post 07 Feb 2011, 05:41
fluke wrote:
Mariyea did it by finding out probability of selecting two marbles of same colors and then taking the complement.

The other way using combination;

The total number of ways of choosing any 2 marbles out of 10 is
C^{10}_2 = \frac{10*9}{2} = 45

5R(red marbles), 3B(blue marbles) and 2G(greeen marbles)

Favorable ways of selecting 2 distinct colored marbles:
1R out of 5R AND 1B out of 3B
OR
1R out of 5R AND 1G out of 2G
OR
1B out of 3B AND 1G out of 2G

C^5_1*C^3_1+C^5_1*C^2_1+C^3_1*C^2_1
=5*3+5*2+3*2
=15+10+6=31

Ans: P=\frac{31}{45}

Yes this is the approach that I saw was being used...
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Re: Probability... what approach to use   [#permalink] 07 Feb 2011, 05:41
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