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Probability that their product (GMAT PREP 1)

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Probability that their product (GMAT PREP 1)

by alex.gellatly » Fri Aug 03, 2012 11:17 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 of the form (x^2)-[(by)^2], where b is an integer?

1/2
1/3
1/4
1/5
1/6

Thanks!
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by javzprobz » Sat Aug 04, 2012 12:47 am
Hi,

Using the combination formula, we will know how many groupings of 2 of these 4 expression we can have:4!/(4-2)!2!=6

So, now, let's see how many of these 6 combinations are of the form of (x^2)-[(by)^2], which is (x+by)(x-by)...

(x+y)(x+5y): it's not of the form of (x+by)(x-by), so NO
(x+y)(x-y): it is of the form of (x+by)(x-by), so YES
(x+y)(5x-y): not of the form of (x+by)(x-by), so NO
(x+5y)(x-y): not of the form of (x+by)(x-by), so NO
(x+5y)(5x-y): not of the form of (x+by)(x-by), so NO
(x-y)(5x-y): not of the form of (x+by)(x-by), so NO

So only one of these 6 combinations is of the form of (x+by)(x-by)...

Therefore, the correct answer is E.
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by Anurag@Gurome » Sat Aug 04, 2012 3:42 am
alex.gellatly wrote: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 of the form (x^2)-[(by)^2], where b is an integer?

1/2
1/3
1/4
1/5
1/6

Thanks!
(x + y)(x + 5y) = x^2 + 6xy + 5y^2: Not in the form x^2- (by)^2
(x + y)(5x - y) = 5x^2 + 4xy - y^2: Not in the form
(x + 5y)(x - y) = x^2 + 4xy - 5y^2: Not in the form
(x + 5y)(5x - y) = 5x^2 + 24xy - 5y^2: Not in the form
(x - y)(5x - y) = 5x^2 - 6xy + y^2: Not in the form
(x + y)(x - y) = x^2 - y^2: This is in the required form

Hence, the required probability is [spoiler]1/6[/spoiler].
Anurag Mairal, Ph.D., MBA
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