Probability ( Random Variable ). 
Let $$p_X(x) = \begin{cases}\frac{x}{15},& x\in\{1,2,3,4,5\}\\ 0,& \text{otherwise}\end{cases}$$ be the probability mass function of $X$. We need to find $$\mathbb P\left(\frac{1}{2}<X<\frac{5}{2}\right)$$ and $$\mathbb P\left(1\leq X \leq2\right).$$

What I figured : For the first one , I found out the cumulative distribution function and divided the interval $\left(\frac{1}{2},\frac{5}{2}\right)$ and wrote the probability as: 
$$\mathbb P\left(\frac{1}{2}<X<1\right) + \mathbb P(X=1) + \mathbb P(1<X<2) + \mathbb P(X=2) + \mathbb P\left(2<X<\frac{5}{2}\right),$$ which comes out to be $\frac{8}{15}$. Am I correct ?
If yes , do I need to carry on like this for the second part or $\mathbb P(1\leq X \leq2)$ is simply $\mathbb P(X=1) + \mathbb P(X=2)$? 
Help!

Keep it simple!
$$X\in\left\{\frac{1}{2}<X<\frac{5}{2}\right\} = \left\{1,2\right\}$$
Then:
$$P\left(\frac{1}{2}<X<\frac{5}{2}\right) = P(X=1) + P(X=2) = \frac{3}{15}$$

$$X\in\left\{1 \leq X \leq2\right\} = \left\{1,2\right\}$$
Then:
$$P\left(1 \leq X \leq2\right) = P(X=1) + P(X=2) = \frac{3}{15}$$
