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Where did I go wrong?

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by winniethepooh » Wed Jun 15, 2011 9:57 pm
Q.In a bag of marbles, there are 3 red, 2 white, and 5 blue marbles. If Maddy takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble?


What I have done is calculated probability of taking out 1 white marble that is 2/10 and multiplied that with the probability of taking out the blue marble that is 5/9..Multiplying 2/10 and 5/9 I get 1/9, which is not the answer!!
Can I know, whether is my approach wrong or did I go wrong somewhere?
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Source: — Problem Solving |

by Geva@EconomistGMAT » Wed Jun 15, 2011 10:06 pm
winniethepooh wrote:Q.In a bag of marbles, there are 3 red, 2 white, and 5 blue marbles. If Maddy takes 2 marbles out of the bag, what is the probability that he will have one white and one blue marble?


What I have done is calculated probability of taking out 1 white marble that is 2/10 and multiplied that with the probability of taking out the blue marble that is 5/9..Multiplying 2/10 and 5/9 I get 1/9, which is not the answer!!
Can I know, whether is my approach wrong or did I go wrong somewhere?
I'm guessing that the right answer is twice that, or 2/9: your calculation assumes that Maddy draws white first, blue second, but the question asks for "one white and one blue" - which doesn't really imply order. Thus, you need to bring both scenarios into account:

white, then blue: 2/10 * 5/9 = 1/9
Blue, then white: 5/10 * 2/9 = 1/9

total probability = 1/9+1/9=2/9.
Geva
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by Frankenstein » Wed Jun 15, 2011 10:08 pm
Hi,
You have taken white marble first and blue marble next. He has actually taken both the balls at the same. So, you have to add the case of picking blue marble first and then white marble as well.
So, it will be (2/10)(5/9) + (5/10)(2/9) = 2/9
(or)
Selecting 1 white and 1 blue can be done in 2C1*5C1
Number of ways of selecting 2 balls from 10 is 10C2.
So, probability is 2C1*5C1/10C2 = 2/9
Cheers!

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