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05 Feb 2011, 17:41
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There are 5 red marbles, 3 blue marbles and 2 green 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
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
05 Feb 2011, 17:53
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mariyea
You did everything right except the addition:
\(\frac{2}{9}+\frac{1}{15}+\frac{1}{45}=\frac{10}{45}+\frac{3}{45}+\frac{1}{45}=\frac{14}{45}\) (not 9/45);
--> \(1-\frac{14}{45}=\frac{31}{45}\).
05 Feb 2011, 23:44
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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: \(P=\frac{31}{45}\)
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: \(P=\frac{31}{45}\)
