Easy peasy, this is a representation of differnce of 2 perfect squares
a^2-b^2=(a-b)(a+b)
-4x +12 -12 = -4 -12
-4x = -16
x = 4
No . but she can buy 2 of the books but not all 3
Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose
![1\leq x\leq 3,\quad 1\leq y \leq 5](https://tex.z-dn.net/?f=1%5Cleq%20x%5Cleq%203%2C%5Cquad%201%5Cleq%20y%20%5Cleq%205)
We have
and
, because the dice are fair.
Now we use the assumption of independence to claim that
![P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}](https://tex.z-dn.net/?f=%20P%28X%3Dx%2C%20Y%3Dy%29%20%3D%20P%28X%3Dx%29%5Ccdot%20P%28Y%3Dy%29%20%3D%5Cdfrac%7B1%7D%7B3%7D%5Ccdot%5Cdfrac%7B1%7D%7B5%7D%20%3D%20%5Cdfrac%7B1%7D%7B15%7D)
Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:
- 2 in a unique way (1+1)
- 3 in two possible ways (1+2, 2+1)
- 4 in three possible ways
- 5 in three possible ways
- 6 in three possible ways
- 7 in two possible ways
- 8 in a unique way
This implies that the probabilities of the outcomes of
are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5