1. Let $p(x)=\frac{x}{15}, x=1,2,3,4,5$ be the pmf of $X$. Find the following probabilities:
a. $P(X=1$ or 2$) \quad \frac{1}{15}$

Step 1 ：Let $p(x)=\frac{x}{15}, x=1,2,3,4,5$ be the probability mass function (pmf) of $X$.

Step 2 ：We are asked to find the probability that the random variable $X$ takes on the value 1 or 2. Since $X$ can only take on one value at a time, these are mutually exclusive events, so the probability of either occurring is the sum of their individual probabilities.

Step 3 ：We substitute $x=1$ and $x=2$ into the pmf to find the probabilities of these events: $p(1)=\frac{1}{15}$ and $p(2)=\frac{2}{15}$.

Step 4 ：We add these probabilities together to find the total probability: $p(1 \text{ or } 2) = p(1) + p(2) = \frac{1}{15} + \frac{2}{15} = \frac{3}{15} = 0.2$.

Step 5 ：Final Answer: The probability that $X$ equals 1 or 2 is \(\boxed{0.2}\).