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pep4 =probabilty tuf1

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by Azntycoon » Sun Oct 05, 2008 12:01 pm
I think it's D.

The only way you can get x^2 - (by)^2 is if you get x+y and x-y, which means 2 out of the 4 options. Using the formula for combinations, we find that there are 6 possible combinations.

n!/r!(n-r)!
4!/2!(4-2)! = 6

Therefore, there is 1/6 chance.


What's the OA?
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by afterhours » Sun Oct 05, 2008 12:23 pm
4 possible "right" combination, 12 total combination's

I have the answer as 1/3
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by vittalgmat » Sun Oct 05, 2008 12:26 pm
Hi,
I solved it differently and got the same answer. 1/6. Can someone let me know if this approach is correct.

P(getting x^2 - (by)^2 ) =
P(choosing x-y) * P (choosing x+y given that x -y is chosen) +
P(choosing x+y) * P (choosing x -y given that x +y is chosen)
= 1/4 * 1/3 + 1/4 + 1/3.
= 2/12 = 1/6.

Thanks
-V
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by cramya » Sun Oct 05, 2008 3:38 pm
Vittal,
I think thats correct.

Other way using combinatiosn would be

1c1*1c1 / 4c2 = 1/6
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by stop@800 » Mon Oct 06, 2008 3:36 am
4 nos so products possible = 6 (4c2)

we have one possible case ie (x-y)(x+y)

so 1/6
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by amitansu » Mon Oct 06, 2008 3:50 am
Yes 1/6 should be the ans.

x^2-(by)^2=(x+by)(x-by) where b is an integer.

there are 2 items out of 4 whose product is in the aforesaid form.

Choosing one is 2/4 (4 is total no. of items )
choosing the other one is 1/3 (since repitition is not cosidered as not stated)

P=2/4 * 1/3=1/6


Amit
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