1/8 is the answer to the problem
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
Set up a proportion
EA / EB = ED/ EC
16 / (5x + 2 + 16) = 12 / (12 + 24)
16/(5x + 18) = 12 / 36 Cross multiply
16 * 36 = (4x + 18)*12 Remove the brackets
576 = 48x + 216 Subtract 216 from both sides.
360 = 48x Divide by 48
360/48 = x
7.5 = x
The absolute value measures the distance a number is away from the origin (zero) on the number line. A number can be to the left or right of zero on the number line, but the distance away from zero is always positive. Graphically:
In other words:
• If d is positive and
![\begin{gathered} |x|=d \\ \text{ then} \\ x=d\text{ or }x=-d \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%7Cx%7C%3Dd%20%5C%5C%20%5Ctext%7B%20then%7D%20%5C%5C%20x%3Dd%5Ctext%7B%20or%20%7Dx%3D-d%20%5Cend%7Bgathered%7D)
• If d is negative and
![\begin{gathered} |x|=-d \\ \text{then} \\ \text{ No solution, because distance (d) can not be negative.} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20%7Cx%7C%3D-d%20%5C%5C%20%5Ctext%7Bthen%7D%20%5C%5C%20%5Ctext%7B%20No%20solution%2C%20because%20distance%20%28d%29%20can%20not%20be%20negative.%7D%20%5Cend%7Bgathered%7D)
Therefore, the equation that has no solutions is: