How many outcomes there are depends on what you’re counting as different outcomes. If each possible choice of dice and each possible way the two chosen dice can come up is counted as a separate outcome, there are $2840$ possible outcomes; here’s the reasoning.
If you pick the $20$- and $12$-sided dice, there are $20\cdot12=240$ possible outcomes; if you pick the $20$-and $10$-sided dice, there are $20\cdot10=200$ possible outcomes; and so on. The total possible number of outcomes is therefore the sum of all possible products of two of the numbers $20,12,10,8,6$, and $4$. You can compute this by brute force, or you can take advantage of the fact that $$\begin{align*}&(a_1+a_2+a_3+a_4+a_5+a_6)^2=\\
&a_1^2+a_2^2+a_3^2+a_4^2+a_5^2+a_6^2+2(\text{the sum that you want})\;,
\end{align*}$$ so that $$\begin{align*}&\text{the sum that you want}=\\
&(20+12+10+8+6+4)^2-(20^2+12^2+10^2+8^2+6^2+4^2)=\\
&60^2-(400+144+100+64+36+16)=\\
&3600-760=\\
&2840\;.
\end{align*}$$
These $2840$ outcomes are not all equally likely, however. Since there are $\binom62=15$ possible pairs of dice, and all pairs are equally likely, the probability of drawing any particular pair is $\frac1{15}$. If you draw the $20$- and $12$-sided dice, each of the $240$ possible rolls is equally likely, so the overall probability of drawing those two dice and getting a particular roll is $\frac1{15}\cdot\frac1{240}=\frac1{3600}$. If you draw the $6$- and $4$-sided dice, on the other hand, there are only $24$ possible rolls, so the probability that you draw those two dice and get a particular roll is $\frac1{15}\cdot\frac1{24}=\frac1{360}$, or ten times as large.
However, you might decide to count as different outcomes only the different possible totals on the two dice. The minimum possible total is $1+1=2$, the maximum is $20+12=32$, and every integer total between these extremes is possible, so there are $32-1=31$ possible totals, as you said. Of course these $31$ outcomes also have very different probabilities: to take the easiest example, there is only one way to get a total of $32$, but there are $15$ ways to get a total of $2$, one for every possible pair of dice.
This is relevant to the second question: because the different totals have different probabilities, you can’t simply say that there are $16$ even and $15$ odd totals, and therefore the probability of getting an even total is $\frac{16}{31}$. You’re going to have to analyze the possibilities in more detail.
Let’s look at a particular pair of dice, say the $10$- and the $6$-sided dice. The $10$-sided die has $5$ even and $5$ odd numbers, and the $6$-sided die has $3$ even and $3$ odd numbers. The $10\cdot 6=60$ possible rolls of these two dice can therefore be broken down as follows: $5\cdot 3=15$ rolls with an even number on both dice; $5\cdot 3=15$ rolls with an even number on the $10$-sided die and an odd number on the $6$-sided die; $5\cdot 3=15$ rolls with an odd number on the $10$-sided die and an even number on the $6$-sided die; and $5\cdot 3=15$ rolls with an odd number on both dice. When do you get an even total? In the first and last cases, which amount to $15+15=30$ out of the $60$, or half of them. Thus, if you happen to draw the $10$- and the $6$-sided dice, your probability of rolling an even total is $\frac12$.
We could have got this with much less work by realizing right away that in order to get an even total, we must get either two even or two odd numbers. Each die has half its faces even and half of them odd, so no matter which two dice we pick, the probability of getting even + even is $\frac12\cdot\frac12=\frac14$, and so is the probability of getting odd + odd, which means that the probability of getting an even total is $\frac14+\frac14=\frac12$. Since the probability of getting an even total is $\frac12$ no matter which pair of dice we draw, the overall probability of getting an even total must also be $\frac12$.
For the last question you must also look a bit more closely at the possible outcomes. First, you can’t roll an $8$ or higher on both dice unless both dice have at least $8$ faces, so you must begin by drawing two of the four dice with the most faces; there are $\binom42=6$ ways to do this, so $\frac9{15}=\frac35$ of the time you won’t even draw dice on which it’s possible to get $8$ or higher on both. What you should do is look at these six pairs individually, counting the number of rolls that give you two numbers that are at least $8$. For instance, if you’re rolling the $12$- and $10$-sided dice, you need to get one of the $5$ largest numbers on the $12$-sided die and one of the $3$ largest on the $10$-sided die; the probability of doing so is $\frac5{12}\cdot\frac3{10}=\frac18$. Of course the probability of drawing these two dice in the first place is only $\frac1{15}$, so the probability of drawing them and getting a successful roll is $\frac1{15}\cdot\frac18=\frac1{120}$. Similarly, the probability of success when you roll the $12$- and the $8$-sided dice is $\frac5{12}\cdot\frac18=\frac5{96}$, and the probability of getting those two dice in the first place is $\frac1{15}$, so the probability of drawing them and getting a successful roll is $\frac1{15}\cdot\frac5{96}=\frac1{288}$. If you perform similar calculations for each of the other four pairs of dice that can give you successful outcomes, you can then add the results to get the final probability of rolling two numbers greater than or equal to $8$.
You can’t simply count successful outcomes and divide by $2840$, since the $2840$ possible rolls aren’t equally likely, as noted at the beginning.
Your answer to 1 is correct.
Your answer to 2
$A:$ the die rolled is a loaded
$\lnot A$: not A or the die rolled is fair
$B$ = the sum of 3 values is 4.
$P(B|A) = 3(\frac {2}{6})(\frac {2}{6})(\frac {1}{6})=\frac {12}{216}$
The only way to do it is if you roll $2$ two's and a one, and the one could show up on any of $3$ rolls.
$P(B|\lnot A) = 3(\frac {1}{6})(\frac {1}{6})(\frac {1}{6})=\frac {3}{216}$
$P(B) = P(A)P(B|A) + P(\lnot A)P(B|\lnot A) = (\frac 12)\frac {12}{216} + (\frac 12)(\frac {3}{216})$
And that gives you all of the necessary components to plug into Bayes law.
$\frac {12}{15} = \frac 45$
Best Answer
Your first and most critical error is that the set of faces on the four dice are not equiprobable. You enumerated the outcomes as $$\{1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,6,6,6,6,6,6,6\}$$ but this is not representative of how likely each face is to occur, because the sampling process is not uniform on the faces, but rather, on the dice. In other words, there are two random processes at work here: the first is you pick a die uniformly and at random from the bag. The second is that you roll the die that you picked.
Because of this two-step process, it is not the case that the probability of rolling a $6$ is simply $7/22$. Instead, you must consider the fact that the tetrahedral die has a $1/4$ chance of being selected, thus the six that it will invariably roll if selected are more likely to occur than would be the case if all faces are equiprobable. The correct probability of rolling a $6$ is given as follows. Let $T$ be the event that the tetrahedral (unfair) die is selected, and $X \in \{1, 2, 3, 4, 5, 6\}$ be a random variable describing the outcome of the die roll. Then by the law of total probability,
$$\Pr[X = 6] = \Pr[X = 6 \mid T] \Pr[T] + \Pr[X = 6 \mid \bar T]\Pr[\bar T],$$
where $\bar T$ is the complementary event of selecting a cubic (fair) die. Then since $\Pr[T] = 1/4$ and $\Pr[\bar T] = 1 - 1/4 = 3/4$, we have
$$\Pr[X = 6] = 1 \cdot \frac{1}{4} + \frac{1}{6}\cdot \frac{3}{4} = \frac{3}{8}.$$
In the second part of the question, we are asked for $$\Pr[T \mid X = 6],$$ the probability of having selected the tetrahedral die given that the outcome of the roll was $6$. We simply apply Bayes' rule here:
$$\Pr[T \mid X = 6] = \frac{\Pr[X = 6 \mid T]\Pr[T]}{\Pr[X = 6]}.$$ We already have all of the quantities on the right hand side: $$\Pr[T \mid X = 6] = \frac{1 \cdot \frac{1}{4}}{\frac{3}{8}} = \frac{2}{3}.$$
For the last part, we are asked for $$\Pr[X_{\text{new}} = 6 \mid X = 6],$$ that is to say, the posterior predictive probability that the next roll is $6$ on the same die that was selected after the first roll observed was $6$. Unlike the previous calculation, we cannot blindly use Bayes' rule, since the temporal order of events is that $X$ is first observed, then $X_{\text{new}}$ is observed afterwards. Instead, we must again condition on the fairness of the die that was selected.
Think for a moment what we need. Given we rolled a $6$, the posterior probability that we had selected the unfair die is $2/3$, which is higher than the prior probability $1/4$ of selecting the unfair die. So our intuition suggests that rolling a second $6$ on the same die should also be more likely than $3/8$. In particular,
$$\Pr[X_{\text{new}} = 6 \mid X = 6] = \Pr[X_{\text{new}} = 6 \mid T]\Pr[T \mid X = 6] + \Pr[X_{\text{new}} \mid \bar T]\Pr[\bar T \mid X = 6].$$
This is again an application of the law of total probability, except we are conditioning on the fairness of the die after we observed a $6$ on the first roll. This works because once a die is selected, the outcome of the next roll does not depend on the outcome of previous rolls, only on the distribution of the face values that are on that die. This is why it is unnecessary to write on the right-hand side instead $$\Pr[X_{\text{new}} = 6 \mid T \cap (X = 6)]\Pr[T \mid X = 6] + \Pr[X_{\text{new}} \mid \bar T \cap (X = 6)]\Pr[\bar T \mid X = 6],$$ because the independence of rolls given a particular die implies $$\Pr[X_{\text{new}} = 6 \mid T \cap (X = 6)] = \Pr[X_{\text{new}} = 6 \mid T]$$ and $$\Pr[X_{\text{new}} = 6 \mid \bar T \cap (X = 6)] = \Pr[X_{\text{new}} = 6 \mid \bar T].$$ Now all that is left is to use the previous two parts to calculate the last part. We already established $\Pr[T \mid X = 6] = 2/3$, so the complementary event is $$\Pr[\bar T \mid X = 6] = 1/3,$$ i.e. given that you rolled a $6$, the probability the die selected is fair is $1/3$. So $$\Pr[X_{\text{new}} = 6 \mid X = 6] = 1 \cdot \frac{2}{3} + \frac{1}{6} \cdot \frac{1}{3} = \frac{13}{18},$$ which is greater than $3/8$ as our earlier intuition suggested.