BTGmoderatorLU wrote:If two of the four expressions x+y, x+5y, x-y and 5x-y are chosen random, what is the probability that their product will be of the form of x^2-(by)^2, where b is an integer?
A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6
First, notice that we are being tested on the difference of squares. We can restate the problem as: What is the probability, when selecting two expressions at random, that the product of those expressions will create a difference of two squares? Remember, the difference of two squares can be written as follows:
a^2 - b^2 = (a + b)(a - b)
So, x^2 - (by)^2 can be factored as (x + by)(x - by). Thus, we are looking for two expressions in the form of (x + by)(x - by). Although this problem is attempting to trick us with the expressions provided, the only two expressions that, when multiplied together, will give us a difference of squares are x + y and x - y. When we multiply x + y and x - y, the result is x^2 - y^2 or x^2 - (1y)^2.
We see that there is just one favorable product, namely (x + y)(x - y). In order to determine the probability of this event, we must determine the total number of possible products. Since we have a total of four expressions and we are selecting two of them to form a product, we have 4C2, which is calculated as follows:
4C2 = (4 x 3)/(2!) = 12/2 = 6 products
Of these 6 products, we have already determined that only one will be of the desired form x^2 - (by)^2. Therefore, the probability is 1/6.
Alternate Solution:
One other way to solve this problem is to use probability.
Once again, we have determined that the only two expressions that, when multiplied together, will give us a difference of squares are x + y and x - y. If we select either of those expressions first, since there are 2 favorable expressions and 4 total expressions, there is a 2/4 = 1/2 chance that either (x + y) or (x - y) will be selected. Next, since there is 1 favorable expression left and 3 total expressions, there is a 1/3 chance that the final favorable expression will be selected.
Thus, the probability of selecting x - y and x + y is 1/2 x 1/3 = 1/6.
Answer: E
