Note that if you want to figure out if a function has a slant asymptote, you have to realize that the equation of a slant asymptote is linear, so the exponent of the leading coefficient of the numerator of the function must be one greater than that of the denominator.

When doing long division (denominator into numerator), it might not divide evenly. Do not concern yourself with the remainder. That slant or oblique asymptote is just the linear portion.

Suppose you roll two 6 sides dice and we want to see the outcomes and probability distribution for the difference between the two dice. 3 on the first and 2 on the second would be 3-2=1 and 2 on the first and 3 on the second would be 2-3 = -1

To get a 0 when subtracting the numbers on the dice, the numbers must be the same. This happens 6 ways.

(1,1), (2,2), (3,3), (4,4), (5,5), (6,6).. since there are 36 possibilities, we have 6/36 = 1/6

For problem 6, to subtract numbers to get -2, the second number must be two larger than the first number.

(1,3), (2,4), (3,5), (4,6) . So we have 4 out of 36 outcomes that will subtract to -2. The probability is 4/36 = 1/9

For problem 7, It's basically the same as problem 6, except now the first number is two larger than the second number.

This gives us (3,1), (4,2), (5,3), (6,4). Again this is 4/36, simplified to 1/9

For the last problem, we want the possible outcomes with corresponding probabilities

We can get 0 if the numbers are the same, which I showed in problem 5, that is probability 1/6

We can get 1 when subtracting the numbers if we have (2,1), (3,2), (4,3), (5,4), (6,5) that is 5/36

We can get -1 when subtracting if we have (1,2), (2,3), (3,4), (4,5), (5,6), again with probability 5/36

We can get 2 as shown in problem 7 with probability of 1/9
We can get -2 as shown in problem 6 with probability of 1/9
We can get 3 with rolls of (4,1), (5,2), (6,3) with probability of 3/36 = 1/12
We can get -3 with rolls (1,4), (2,5), (3,6) with probability of 3/36 = 1/12
We can get 4 with rolls (5,1), (6,2) with probability of 2/36 = 1/18
We can get -4 with rolls (1,5), (2,6) with probability of 2/36 = 1/18
We can get 5 with a roll of (6,1) with probability 1/36
We can get -5 with a roll of (1,6) with probability of 1/36

Putting it all together we get this model

outcome Probability
5 1/36
4 1/18
3 1/12
2 1/9
1 5/36
0 1/6
-1 5/36
-2 1/9
-3 1/12
-4 1/18
-5 1/36

Notice that the probabilities will add to 1. That must always be the case for a legitimate probability model and the probabilities for each event must be between 0 and 1 inclusive

Suppose you have the following distribution

x = 0, 2, 5

p(x) = 1/4, 1/4, 1/2

find the mean and variance.

To get the mean you take the sum of x(P(x))

so for x = 0, 1,5 with P(x) = 1/4, 1/4, and 1/2

You get 0(1/4) + 1(1/4) + 5(1/2) = 0 + 1/4 + 5/2 = 2.75

To calculate sigma squared (variance)

It's the [sum (x- mean)^2P(x)]/n

So we have (0 - 2.75)^2 + (2-2.75)^2 + (5-2.75)^2

The equals 7.5625 + 0.5625 + 5.0625 = 13.1875
now take 13.1875/3 = 4.396

For a sample size of two you can have these possibilities (0,0), (0,2), (0,5), (2,0), (2,2), (2,5), (5,0), (5,2), (5,5)

The means are the two numbers added and divided by two. That gives us
(0 + 0)/2 = 0
(0 + 2)/2 = 1
(0 + 5)/2 = 2.5
(2 + 0)/2 = 1
(2 + 2)/2 = 2
(2 + 5)/2 = 3.5
(5 + 0)/2 = 2.5
(5 +2)/2 = 3.5
(5 + 5)/2 + 5

So you can have mean of 0 with (0, 0) with probability (1/4)(1/4) = 1/16
mean of 1 with (0, 2) and (2,0) with probability 2(1/4)(1/4) = 1/8
mean of 2 with (2, 2) with probability of (1/4)(1/4) = 1/16
mean of 2.5 with (0,5) and (5,0) with probability of 2(1/4)(1/2) = 1/4
mean of 3.5 with (2,5) and (5,2) with probability of 2(1/4)(1/2) = 1/4
mean of 5 with (5,5) with probability of (1/2)(1/2) = 1/4