Gmat Prep: Probability Question

The first thing I did when solving this problem was finding out all of the possible combinations between the 4 terms.

To do this:
4C2 = 4!/(2!*2!) = 6

With this information you can immediately eliminate 1/4 as a possibility.

The only possible answers are 1/6,2/6,3/6,4/6, 5/6, or 1

(x+y)(x-y)= x^2 - y^2

There is no other possible combination that fits the requirement so the answer is 1/6

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Vemuri: Legendary Member; Posts: 682; Joined: Fri Jan 16, 2009 2:40 am; Thanked: 32 times; Followed by:1 members

Re: Gmat Prep: Probability Question

by Vemuri » Sun May 31, 2009 5:36 am

Selecting 2 out of 4 expressions ==> 4c2 ===> 6ways. These are the available outcomes.

The product of the 2 expressions should be in the form x^2-(by)^2. This form is only possible when (x+y) & (x-y) are multiplied, i.e. x^2-y^2, where b=1. For all other products, this result is not possible. So, the probability = 1/6

Hope this helps.

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dtweah: Master | Next Rank: 500 Posts; Posts: 487; Joined: Fri Mar 27, 2009 5:49 am; Thanked: 36 times

Re: Gmat Prep: Probability Question

by dtweah » Sun May 31, 2009 6:42 am

exhilaration wrote:Hi,

Can someone please solve the question below.

Thanks!

One way to see this problem is to figure out the number of terms involved in the product.
(X+Y)(X-Y) has 2 terms when multiplied (difference of sqaures)
(X+5y)(5X-Y) has 3 terms when muiltipled (trinomial: 4 terms actually but they are not distinct so one will disappear)

By FCP Total terms = 2 x 3=6

If the product must be in the form of a difference of squares, then the problem is asking you for the probability of choosing (X+Y)(X-Y) and since there is only one (X+Y)(X-Y), the required probabilty is:

1/6.