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Probability

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by arbiter » Sat Aug 22, 2009 8:38 am
If two of the foue expressions x+y,x+5y,x-y,5x-y are chosen at random,what is the probability taht their product will be of the form of x2-(by)2, where b is an interger?

a>1/2
b>1/3
c>1/4
d>1/5
e>1/6

OA is E
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by kaulnikhil » Sat Aug 22, 2009 8:52 am
only x2-y2 , x2 + y2
satisfy the condition therefore
2c2/4c2
= 1/6
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by arbiter » Sat Aug 22, 2009 9:14 am
only x-y and x+y pair wil give result as x2-by2

pl explain how u got 2c2
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by kaulnikhil » Sat Aug 22, 2009 10:36 am
two numbers satisty the condition x^2 - Y^2
selecting two things out of a total of 2 things (2 pairs )is 1 = 2c2
Last edited by kaulnikhil on Sun Aug 23, 2009 9:03 am, edited 1 time in total.
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by arbiter » Sat Aug 22, 2009 11:57 am
which are those two pairs?
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Let's see

by enniguy » Sat Aug 22, 2009 6:19 pm
1st pair: x+y and x-y Here b = 1
2nd pair: x+5y and x-5y. Here b = 5
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Is it 1/3?

by enniguy » Sat Aug 22, 2009 6:25 pm
I am just confused. Please help me understand.
I understood Total selection = 4C2 = 6
But, we have a choice between 2 pairs. We can select them in 2C1 ways.

1st pair: x+y and x-y Here b = 1
2nd pair: x+5y and x-5y. Here b = 5

So isn't the answer 2/6 = 1/3?

Am I doing something wrong/?
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by kaulnikhil » Sat Aug 22, 2009 9:48 pm
read the question
2nd pair: x+5y and x-5y. Here b = 5
the two pairs u talking about are 5x-y and x+5y
:D
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Oh darn

by enniguy » Sun Aug 23, 2009 3:16 am
kaulnikhil wrote:read the question
2nd pair: x+5y and x-5y. Here b = 5
the two pairs u talking about are 5x-y and x+5y
:D
Oh damn. Thanks myte. So the answer is 1/6 then.
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by umaa » Sun Aug 23, 2009 8:16 am
I still don't understand how you got 2C2. Can you pls explain me?
What we think, we become
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by kaulnikhil » Sun Aug 23, 2009 9:02 am
x+y and x-y
are the only two pairs which give us a solution in the form mentioned in the question
we have 2 no and we have to select a pair from them ...this can be done in 2c2 ways = only one way ....
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