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Probability [#permalink] New post 16 Dec 2006, 06:51
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 [#permalink] New post 16 Dec 2006, 13:40
Can you explain how you got 1/6. I got 1/4.
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 [#permalink] New post 16 Dec 2006, 13:51
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
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 [#permalink] New post 16 Dec 2006, 14:09
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
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 [#permalink] New post 16 Dec 2006, 17:04
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
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 [#permalink] New post 16 Dec 2006, 17:40
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!
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 [#permalink] New post 17 Dec 2006, 02:33
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..
so the answer must 1/6
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 [#permalink] New post 17 Dec 2006, 05:58
1/6 it is.......thanks for the explanation....
  [#permalink] 17 Dec 2006, 05:58
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