We will take probability space as following: $\Omega = \{ \text{Heads}, \text{Tails} \}$ with $\mathbb{P}(\text{Heads}) = \mathbb{P}(\text{Tails}) = \frac{1}{2}$ we can have a random variable $X: \Omega \to \{0, 1 \} \subset \mathbb{R}$ such that $X(\text{Tails}) = 0$ and $X(\text{Heads})=1$.

This distribution models the number of heads when tossing a coin $n$ times such that the probability of getting heads is $p$. Or, if you like it more, it models the number of heads when $n$ coins are tossed (each of the coins is again such that the probability of getting heads is $p$). Indeed, the number of ways of get exactly $k$ heads when tossing a coin $n$ times is $\binom{n}{k}$ since we simply need to pick which $k$ of the $n$ tosses should be heads (others will be tails). Once we picked the $k$ tosses for heads, we have $p^k$ as probability that these $k$ tosses are indeed all heads and $(1-p)^{n-k}$ that the rest $n-k$ tosses are indeed all tails. Thus we obtain $p_k = \binom{n}{k}p^k(1-p)^{n-k}$ as the probability that we get exactly $k$ heads.

Let's reiterate this one more time, this formula is not a definition of probability or something like that. If anything, it is a lemma coming from the definition of one of the most popular probability spaces, namely the uniform one. This formula says that if an event $A$ consists of $m$ simple events, each having probability $\frac{1}{n}$ (where $n$ is the total number of simple events), then $\mathbb{P}(A) = \frac{1}{n} + ... + \frac{1}{n} = \frac{m}{n}$. The first "=" in the this chain of equalities is the definition of $\mathbb{P}(A)$ and the second is well... obvious :)

Just work from the definitions, it is not a complicated exercise. Let's do the distributions one by one:

• Bernoulli distribution: Well, $p + (1-p) = 1$, both $p$ and $1-p$ are non-negative, so done;
• Uniform distribution: This one is also obvious, just note that $\frac{1}{n} + ... + \frac{1}{n} = 1$;
• Binomial distribution: This one is a bit more interesting. We need to show that $p_0 + p_1 + ... + p_n = 1$ the $p_k$-s are a bit ugly in general. Luckily, there is a Binomial Theorem that tells you how to expand $(a+b)^n$. And all you need to do here is to use this theorem to get that \[1 = (p + (1-p))^n = \binom{n}{0}p^n(1-p)^0 + \binom{n}{1}p^{n-1}(1-p) + ... + \binom{n}{n}p^{0}(1-p)^n = p_0 + ... + p_n ;\]
• Geometric distribution: If you know what is the sum of an infinite geometric progression, i.e that \[ 1 + q + q^2 + q^3 + ... = \frac{1}{1-q} \] for $1 \geq q \geq 0$ this one is easy too. All we need to do is to apply that formula here. If you are scared of infinite probability spaces and/or infinite sums, feel free to ignore this distribution for now.

First of all note that we are gonna work in the probability space with 36 simple events, namely all possible pairs of rolls. Each of these simple events will have probability $\frac{1}{36}$ since this is how we interpret the words "fair dice". Now, once we are clear on the probability space we are working in, let's look at all the simple events and see what is the value of $T$ we get:
\begin{array}{c  c  c  c  c  c}
(1,1), \, T=1 & (1,2), \, T=2 & (1,3), \, T=3 & (1,4), \, T=4 & (1,5), \, T=5 & (1,6), \, T=6 \\
(2,1), \, T=2 & (2,2), \, T=2 & (2,3), \, T=3 & (2,4), \, T=4 & (2,5), \, T=5 & (2,6), \, T=6 \\
(3,1), \, T=3 & (3,2), \, T=3 & (3,3), \, T=3 & (3,4), \, T=4 & (3,5), \, T=5 & (3,6), \, T=6 \\
(4,1), \, T=4 & (4,2), \, T=4 & (4,3), \, T=4 & (4,4), \, T=4 & (4,5), \, T=5 & (4,6), \, T=6 \\
(5,1), \, T=5 & (5,2), \, T=5 & (5,3), \, T=5 & (5,4), \, T=5 & (5,5), \, T=5 & (5,6), \, T=6 \\
(6,1), \, T=6 & (6,2), \, T=6 & (6,3), \, T=6 & (6,4), \, T=6 & (6,5), \, T=6 & (6,6), \, T=6 \\
\end{array} For example you could write $T((5,3)) = 5$ since $(5,3)$ is a member of $\Omega$, i.e the set of all simple events, and $T$ is a function from $\Omega$ to $\mathbb{R}$. Then, the probability distribution of $T$ is
\begin{array}{c  c  c  c  c  c}
& T=1 & T=2 & T=3 & T=4 & T=5 & T=6 \\
\text{prob: } & 1/36 & 3/36 & 5/36 & 7/36 & 9/36 & 11/36
\end{array}

Suppose that there are $n$ different fair coins, and let's say $T_i$ models whether it is heads or tails when we toss coin number $i$. Then the sum of $T_i$-s models the total number of heads when we toss all of the coins $1,2,3,...,n$ together. And we know that a random variable that models the total number of heads when we toss $n$ identical coins (each having probability $p$ as probability of getting heads) is called a Binomial random variable. So $B$ is a Binomial random variable with parameters $n$ and $p$. The distribution of such a random variable has been defined above.

First of all, let's stress that when talking about one die roll we are working in a probability space with $6$ simple events (forming $\Omega$, it was actually explicitly defined above) each having probability $\frac{1}{6}$. Now, a random variable representing a die roll is $X: \Omega \to \{ 1, 2, ..., 6 \} \subset \mathbb{R}$ such that $X(\text{roll }k) = k$". Then by the definition of mathematical expectation we get \[ \mathbb{E}[X] = \frac{1}{6} \cdot 1 + \frac{1}{6} \cdot 2 + \frac{1}{6} \cdot 3 + \frac{1}{6} \cdot 4 + \frac{1}{6} \cdot 5 + \frac{1}{6} \cdot 6 = \frac{21}{6} = 3.5 \]

Using one of the tables from the solution to the exercise 4, and using the definition of the mathematical definition only we get right away that: \[ \mathbb{E}[T] = \frac{1}{36} \cdot 1 + \frac{3}{36} \cdot 2 + \frac{5}{36} \cdot 3 + \frac{7}{36} \cdot 4 + \frac{9}{36} \cdot 5 + \frac{11}{36} \cdot 6 = \frac{161}{36} \approx 4.47 \]