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red&blue balls

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by pemdas » Sun Feb 05, 2012 4:52 pm
A box contains 5 red balls and 4 blue balls. If three balls are selected at random without replacement, what is the probability that two balls are blue and one is red?

A 1/4
B 1/3
C 2/5
D 2/3
E 5/14

made up
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by Anurag@Gurome » Sun Feb 05, 2012 5:27 pm
pemdas wrote:A box contains 5 red balls and 4 blue balls. If three balls are selected at random without replacement, what is the probability that two balls are blue and one is red?

A 1/4
B 1/35/
C 2/5
D 2/3
E 5/14

made up
Total # of balls = 9, blue balls = 4, red balls = 5
Possible combinations are: {B, B, R}, {R, B, B}, {B, R, B}
Therefore, probability that two balls are blue and one is red = (4/9 * 3/8 * 5/7) + (5/9 * 4/8 * 3/7) + (4/9 * 5/8 * 3/7) = (60 * 3)/(72 * 7) = [spoiler]5/14[/spoiler]

The correct answer is E.
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by shankar.ashwin » Sun Feb 05, 2012 8:22 pm
BBR can be arranged in 3!/2! = 3 ways.

Now probability of getting BBR = (4/9)*(3/8)*(5/7) = 5/(3*14)

Now since we gotto consider various arrangements, multiply by 3..

Therefore, we have (5*3)/(3*14) = [spoiler]5/14[/spoiler]
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by [email protected] » Sun Feb 05, 2012 8:49 pm
Another way could be,
Probability=[(2 Blue ball out of 4)*(1 Red ball out of 5)]/(3 balls out of 5 Red & 4 Blue)
=(4C2*5C1)/9C3
=6*5/84
=30/84
=5/14
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by pemdas » Sun Feb 05, 2012 11:24 pm
all answers are right, many thanks
cheers,
pemdas
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by pemdas » Mon Feb 06, 2012 3:56 am
[email protected] wrote:Another way could be,
Probability=[(2 Blue ball out of 4)*(1 Red ball out of 5)]/(3 balls out of 5 Red & 4 Blue)
=(4C2*5C1)/9C3
=6*5/84
=30/84
=5/14
I actually liked this method more, as it's more quick and myself solved this way too. Thanks Guha
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