Expected Value of $max(X,3)$ I've seen a few of these questions of the form $max$(X,constant) but they get more math heavy than I know mine should get. Could someone help me understand how to solve this?  
What we are given is that the Distribution of our random variable X is {$1,2,3,4,5$} with probabilites {$1,\frac{39}{51},\frac{26}{50},\frac{13}{49},1$} respectively. Given this, I am told to find $E(X,3)$. How would I do this? I know there shouldn't be any integrals involved.
 A: Let us define $Z:=\max\left\{X,3\right\}.$ Now, first let us consider the values of $Z$ conditioned on the values of $X$. That is, it should be clear that $$[Z|X=x] = \left\{\begin{matrix}
3, & x = 1,\\ 
3, & x =2,\\ 
3, & x = 3,\\ 
4, & x = 4,\\ 
5, & x = 5.
\end{matrix}\right.$$
Now, because $[Z|X=x]$ is always constant, it's expectation is simply its constant value, i.e., $$[Z|X=2] = 3 \Rightarrow \mathbb{E}(Z|X=2) = 3.$$ Therefore, by conditioning we can simply compute the expectation in the normal way, i.e., $$\mathbb{E}(Z) = \sum_{x=1}^{5}\mathbb{E}(Z|X = x)\mathbb{P}(X=x) \\   = 3\left(\mathbb{P}(X=1) + \mathbb{P}(X=2) + \mathbb{P}(X=3)\right) + 4\mathbb{P}(X=4) + 5\mathbb{P}(X=5).$$
Where you can insert the PMF appropriately (it seems like in the question there is a problem with the PMF you gave as it doesn't sum to 1).
---edit---
A simpler method would be to use the following fact: for a discrete random variable $X$ and function $f$

$$\mathbb{E}[f(X)] = \sum_x f(x)\mathbb{P}(X=x).$$

In our case, $f(X)=\max \{X,3\}$, thus the solution is $$\mathbb{E}(\max \{X,3\}) =\sum_{x=2}^5\max\{x,3\}\mathbb{P}(X=x).$$ The technique in the original answer is more general though and will aid in getting the expectation of more complicated random variables, e.g., $\max\{X,Y\}$ where $X$ and $Y$ are both random variables.
