Math/statistics question
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Math/statistics question

[From: ] [author: ] [Date: 12-11-10] [Hit: ]
/1!(47-1)!By definition 0! = 1 and any number to the power of 0 = 1-8/47 times 2/10, or 1/5(reduced.) It is like this because you draw 8 balls out of the total,......
Bag of 47 balls -> 10 red balls & 37 black balls.
I draw 8 balls (not putting them back).
What are the odds I draw at least 2 red balls?

Thanks in advance.

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Binomial problem
Pr (red) = 10/47
Pr (black) = 37/47
n = 47
x = 0,1
Pr of drawing at least 2 = 1 - Pr(0) + Pr(1)

Pr (0) = (47!/0!(47 - 0)!)*(10/47)^0*(37/47)^47
Pr(1) = 47!/1!(47-1)!)*(10/47)^1*(37/47)^46
By definition 0! = 1 and any number to the power of 0 = 1

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8/47 times 2/10, or 1/5(reduced.) It is like this because you draw 8 balls out of the total, which is 47, so it is the denominator. Then, you multiply, since you don't know what the odds are. You multiply 8/47 and 2/10, or 1/5 because there are a total of 10 red balls and you don't know the chance of drawing 2 red balls, since that is what you said. So, 2/10 times 8/47 is 8/235, and that is your final answer, since it cannot be simplified any further. ;D
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