If a discrete random variable XXX has a PMF P(X=k)=k15P(X=k) = \frac{k}{15}P(X=k)=15k for k∈{1,2,3,4,5}k \in \{1, 2, 3, 4, 5\}k∈{1,2,3,4,5}, what is the probability P(X is even)P(X \text{ is even})P(X is even)?
6/15
7/15
8/15
9/15