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by Johnrohit » Thu Oct 24, 2013 8:36 am
In a bag of marbles, there are 3 red, 2 white, 5 blue. If bob takes 2 marbles out of the bag, What is the probability that he will have one white and one blue marble?(Assuming that Bob does not replace the marbles in the bag)


Answer- 2/9

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rohit
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by Brent@GMATPrepNow » Thu Oct 24, 2013 9:18 am
Johnrohit wrote:In a bag of marbles, there are 3 red, 2 white, 5 blue. If bob takes 2 marbles out of the bag, What is the probability that he will have one white and one blue marble?(Assuming that Bob does not replace the marbles in the bag)


Answer- 2/9

Thanks!
rohit
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.

P(1 white and 1 blue) = P(1st is white and 2nd is blue OR 1st is blue and 2nd is white)
= [P(1st is white) x P(2nd is blue)] + [P(1st is blue) x P(2nd is white)]
= [2/10 x 5/9] + [5/10 x 2/9]
= [10/90] + [10/90]
= 20/90
= [spoiler]2/9[/spoiler]

Cheers,
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by GMATGuruNY » Thu Oct 24, 2013 11:23 am
Johnrohit wrote:In a bag of marbles, there are 3 red, 2 white, 5 blue. If bob takes 2 marbles out of the bag, What is the probability that he will have one white and one blue marble?(Assuming that Bob does not replace the marbles in the bag)
Alternate approach:

P = good/all.

All:
Number of ways to choose 2 marbles from 10 options = 10C2 = (10*9)/(2*1) = 45.

Good:
A good pair is composed of 1 white marble and 1 blue marble.
Number of options for the white marble = 2.
Number of options for the blue marble = 5.
To combine these options, we multiply:
2*5 = 10.

Thus:
P(1 white and 1 blue) = 10/45 = 2/9.
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by Johnrohit » Thu Oct 24, 2013 11:59 am
Sorry! new to the forum... din knw bout it! :) thanks for the info.... I was following a comprehensive method!! ur and GMAT GURUS method r way better!! Thank u!
Cheers!
Brent@GMATPrepNow wrote:
Johnrohit wrote:In a bag of marbles, there are 3 red, 2 white, 5 blue. If bob takes 2 marbles out of the bag, What is the probability that he will have one white and one blue marble?(Assuming that Bob does not replace the marbles in the bag)


Answer- 2/9

Thanks!
rohit
When posting questions, please use the spoiler function to hide the correct answer. This will allow others to attempt the question without seeing the final answer.

P(1 white and 1 blue) = P(1st is white and 2nd is blue OR 1st is blue and 2nd is white)
= [P(1st is white) x P(2nd is blue)] + [P(1st is blue) x P(2nd is white)]
= [2/10 x 5/9] + [5/10 x 2/9]
= [10/90] + [10/90]
= 20/90
= [spoiler]2/9[/spoiler]

Cheers,
Brent
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by [email protected] » Thu Oct 24, 2013 12:01 pm
Hi Johnrohit,

Both Brent and Mitch have provided great math approaches to solving the problem. You can also "brute force" if you had to:

Since there are 10 marbles, there will be 10x9 = 90 possible outcomes

We want 1 white and 1 blue (in any order), so we can quickly count up the possibilities:

W1 first and the other second
B1
B2
B3
B4
B5

5 options

W2 first and the other second
B1
B2
B3
B4
B5

5 options

B1 first and the other second
W1
W2

2 options

Then the same for B2, B3, B4, B5

2 options for each = 2 x 4 = 8 options

Grand Total = 5 + 5 + 2 + 8 = 20 options

20/90 = 2/9

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