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Probability

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Manager
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Probability [#permalink]  16 Dec 2006, 06:51
Q)
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Intern
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Can you explain how you got 1/6. I got 1/4.
Director
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I am getting 1/6 also. The only 2 expressions which will give that result are: (x-y) and (x+y)

For x+y to be selected first and x-y second, the probability is: 1/4*1/3
It could also happen vice versa with x-y selected first.

So the total probability is: 2*1/4*1/3 = 1/6
Director
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Two equations can be selected in 4c2 ways which is n(s)
For the product to be x^2-by^2 we can select first eq in 2c1 ways and second equation in 1 way

probability =2c1*1/4c2=2/6=1/3
Senior Manager
Joined: 20 Feb 2006
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We need to pick (x+y)(x-y). Other combinations will result in an xy term.

# of ways to pick any two terms = 4C2 = 6
# of ways to pick (x+y)(x-y) = 1

P = 1/6
VP
Joined: 28 Mar 2006
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yogeshsheth wrote:
Two equations can be selected in 4c2 ways which is n(s)
For the product to be x^2-by^2 we can select first eq in 2c1 ways and second equation in 1 way
probability =2c1*1/4c2=2/6=1/3

Yogesh: Can you explain the bolded text please?
Thanks!
Director
Joined: 01 Oct 2006
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trivikram wrote:
yogeshsheth wrote:
Two equations can be selected in 4c2 ways which is n(s)
For the product to be x^2-by^2 we can select first eq in 2c1 ways and second equation in 1 way
probability =2c1*1/4c2=2/6=1/3

Yogesh: Can you explain the bolded text please?
Thanks!

Got this one wrong ..forgot the equation is x-5y..
so only one combination..
Manager
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1/6 it is.......thanks for the explanation....
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