Record the result of the tossing as an ordered pair $(a,b)$, where $a$ and $b$ are integers between $1$ and $4$. Here the first coordinate records the result of the first toss, and the second coordinate records the result of the second toss.
(a) Let the number of such ordered pairs be $N$. What is the value of $N$?
There is a fast way of finding $N$. But one can do it slowly, by listing and counting. Here is a start: $(1,1)$, $(1,2)$, $(1,3)$, $(1,4)$.
(b) All these ordered pairs are equally likely. How many ordered pairs give you a sum of $4$ or less? List and count them carefully. Let the number be $S$.
(c) Your probability is $\dfrac{S}{N}$.
It looks as if we are expected to assume the dice have numbers $1$, $2$, $3$ written on their faces. For the probability mass function of $X$, we need to calculate $\Pr(X=k)$ for all possible values of $k$. There values are $2$ to $6$.
Suppose one die is red and the other green. Record the result of a toss as an ordered pair $(u,v)$ where $u$ is the number on the green, and $v$ the number on the red. There are $9$ equally likely possibilities,
We get $X=2$ if we get $1$ on red and $1$ on gree. So $\Pr(X=2)=\frac{1}{9}$. We get $X=3$ if we have $1$ on green and $2$ on red, or the other way around. So $\Pr(X=3)=\frac{2}{9}$. Continue.
For $E(X)$, the mean of $X$, we need to calculate
$$2\Pr(X=2)+3\Pr(X=3)+\cdots+6\Pr(X=6).$$
But there is a simpler way. Let random variable $U$ be the number on the green, and $V$ the number on the red. Then $X=U+V$ and therefore $E(X)=E(U)+E(V)$.
It is easy to see that $E(U)=E(V)=2$, so $E(X)=4$. We could also show that $E(X)=4$ by a symmetry argument.
For the variance of $X$, we can proceed from the definition of variance, or, somewhat more simply, from the fact that $\text{Var}(X)=E(X^2)-(E(X))^2$. Note that
$$E(X^2)=2^2\Pr(X=2)+3^2\Pr(X=3)+\cdots+6^2\Pr(X=6).$$
A much nicer way is to note that since $U$b and $V$ are independent, the variance of $U+V$ is the sum of the variances of $U$ and $V$. These can be computed quickly.
Best Answer
We throw a fair 6-sided die independently four times and let $X$ denote the minimal value rolled.
For each roll, the value must be greater than three. This event has probability $\frac{3}{6}=\frac{1}{2}$. As this must occur on each of $4$ rolls, we have: $$P(X\ge4)=\left(\frac{1}{2}\right)^4=\frac{1}{16}$$
$$P(X=1)=P(X>0)-P(X>1)=1-\left(\frac{5}{6}\right)^4=1-\frac{625}{1296}=\frac{671}{1296}$$
$$P(X=2)=P(X>1)-P(X>2)=\left(\frac{5}{6}\right)^4-\left(\frac{4}{6}\right)^4=\frac{625}{1296}-\frac{256}{1296}=\frac{369}{1296}$$
$$P(X=3)=P(X>2)-P(X>3)=\left(\frac{4}{6}\right)^4-\left(\frac{3}{6}\right)^4=\frac{256}{1296}-\frac{81}{1296}=\frac{175}{1296}$$
$$P(X=4)=P(X>3)-P(X>4)=\left(\frac{3}{6}\right)^4-\left(\frac{2}{6}\right)^4=\frac{81}{1296}-\frac{16}{1296}=\frac{65}{1296}$$
$$P(X=5)=P(X>4)-P(X>5)=\left(\frac{2}{6}\right)^4-\left(\frac{1}{6}\right)^4=\frac{16}{1296}-\frac{1}{1296}=\frac{15}{1296}$$
$$P(X=6)=P(X>5)-P(X>6)=\left(\frac{1}{6}\right)^4-0=\frac{1}{1296}$$