here is a probability question:

if two of the four expressions (x+y), (x+5y), (x-y) and (5x-y) are selected at random, what is the probability of their product being of the form x^2-(by)^2, where b is an integer? thanks in advance

Results 1 to 2 of 2

- Oct 20th 2006, 04:54 PM #1

- Joined
- Oct 2006
- Posts
- 5

- Oct 20th 2006, 06:17 PM #2
Probability is not my forte so someone please correct me if I'm wrong.

1. $\displaystyle (x+y)(x+5y) = x^2 + 6xy + 5y^2$

2. $\displaystyle (x+y)(x-y) = x^2 - y^2$

3. $\displaystyle (x+y)(5x-y) = 5x^2 + 4xy - y^2$

4. $\displaystyle (x+5y)(x-y) = x^2 + 4xy - 5y^2$

5. $\displaystyle (x+5y)(5x-y) = 5x^2 -5y^2$

6. $\displaystyle (x-y)(5x-y) = 5x^2 - 6xy +y^2$

These are all the*different*possibilities. Of these only one pair produces the required form, #2. So we need to pick two choices, of which there are only two elements we can choose. So by the counting principle we have 2 choices for the first factor, then we are left with only 1 possibility for the second choice. There are 4*3 ways to make a product of any two of the factors. So:

$\displaystyle P = \frac{2*1}{4*3} = \frac{1}{6}$.

So we have a 1 in 6 chance of getting the required product.

-Dan