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PS: Algebra + Probability

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PS: Algebra + Probability

by haidgmat » Sun Sep 05, 2010 5:56 pm
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be the form of x^2- (by)^2, where b is an integer

a. ½
b. 1/3
c. ¼
d. 1/5
e. 1/6

Ans- E
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by Rahul@gurome » Sun Sep 05, 2010 7:15 pm
(x + y)(x + 5y) = x^2 + 6xy + 5y^2, not in the given form.
(x + y)(x - y) = x^2 - y^2, this is one of the possibilities.
(x + y)(5x - y) = 5x^2 + 4xy - y^2, not in the given form.
(x + 5y)(x - y) = x^2 + 4xy - 5y^2, not in the given form.
(x + 5y)(5x - y) = 5x^2 + 24xy - 5y^2, not in the given form.
(x - y)(5x - y) = 5x^2 - 6xy + y^2, not in the given form.
Therefore, required probability = 1/6

The correct answer is [spoiler](E)[/spoiler].
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by limestone » Sun Sep 05, 2010 7:29 pm
x^2 - (by)^2 = (x-by)(x+by)

we have 4 components here. If the first chosen is :
x+y then the second must be x-y
x+5y then the second must be x-5y ( eliminate this case as we do not have x - 5y in the set)
x-y then ........................ x+y
5x-y then .........................5x +y (eliminate too)

So only (x+y) and (x-y) can be combined to form a the satisfactory above.
The first pick: choose (x+y) or (x-y) from a pool of four items. P = 2/4 = 1/2
The second pick: choose the remain (x+y if x-y has been picked, x-y if x+y has been picked) from a pool of three items : P = 1/3
Possibility = 1/2*1/3 = 1/6
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