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probability

by nikhilgupta » Sat Jan 28, 2012 11:47 am
Suppose I have 2red balls, 3 white balls and 4 black balls. If I choose 2 balls without replacement, what is the probability that both the balls have same colour?
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by MakeUrTimeCount » Sat Jan 28, 2012 12:49 pm
Total number of balls = 2 + 3 + 4 = 9
Probability to get 2 balls = 9C2

Probability to get 2 Red balls = 2C2
Probability to get 2 White balls = 3C2
Probability to get 2 Black balls = 4C2

Probability to get 2 balls of same color = (2C2 + 3C2 + 4C2)/ 9C2 = 5/18
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by tomada » Sat Jan 28, 2012 12:52 pm
I would think that you want the sum of the following probabilities:

a) P (2 red balls)
b) P (2 white balls)
c) P (2 black balls)

If I'm correct about that, then...

a) P (2 red balls) = (2/9) * (1/8) = 2/72
b) P (2 white balls) = (3/9) * (2/8) = 6/72
c) P (2 black balls) = (4/9) * (3/8) = 12/72

Summing, (2/72) + (6/72) + (12/72) = 20/72 = 5/18

Of course, if my initial assumption is incorrect, then ignore everything. :-)
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by pemdas » Sat Jan 28, 2012 1:50 pm
do you count probability or the number of subsets?
below 9C2 should return the total number of subsets, it's not probability
if the problem asked not *without replacement* the method of Tomada would be the best one. However, if we approach with the method of MakeYourTimeCount here we must consider this only for *the same color* balls selected *without replacement*.

@nikhilgupta, to avoid confusion in test use method of Tomada
use the method of MakeYourTimeCount to improve your strategy with combinatorics question types
MakeUrTimeCount wrote:Total number of balls = 2 + 3 + 4 = 9
Probability to get 2 balls = 9C2

Probability to get 2 Red balls = 2C2
Probability to get 2 White balls = 3C2
Probability to get 2 Black balls = 4C2

Probability to get 2 balls of same color = (2C2 + 3C2 + 4C2)/ 9C2 = 5/18
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