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# Probability... what approach to use

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Senior Manager
Joined: 30 Nov 2010
Posts: 263
Schools: UC Berkley, UCLA
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Probability... what approach to use [#permalink]

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05 Feb 2011, 17:41
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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]

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05 Feb 2011, 17:53
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]

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05 Feb 2011, 23:44
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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Joined: 30 Nov 2010
Posts: 263
Schools: UC Berkley, UCLA
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Kudos [?]: 88 [0], given: 66

Re: Probability... what approach to use [#permalink]

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07 Feb 2011, 06:40
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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Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
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Senior Manager
Joined: 30 Nov 2010
Posts: 263
Schools: UC Berkley, UCLA
Followers: 1

Kudos [?]: 88 [0], given: 66

Re: Probability... what approach to use [#permalink]

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07 Feb 2011, 06:41
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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Thank you for your kudoses Everyone!!!

"It always seems impossible until its done."
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Re: Probability... what approach to use   [#permalink] 07 Feb 2011, 06:41
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