Answer:
a) 
We can multiply both sides by 3 and we got:



And if we replace we got:
X 1 2 3
P(X) 0.383 0.367 0.25
b) 

![Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.101-%20%5B1.867%5D%5E2%20%3D0.615)
c) X 1 2 3
F(X) 0.383 0.367+0.383=0.75 0.25+0.75 = 1
Step-by-step explanation:
For this case we have the following probability mass function given:
X 1 2 3
P(X) (1+3k)/3 (1+2k)/3 (0.5+5k)/3
Part a
In order to satisfy the definition of probability distribution the sum of all the probabilities needs to be 1 and each of the individual probabilities needs to be higher or equal than 0, using this we can do that:

We can multiply both sides by 3 and we got:



And if we replace we got:
X 1 2 3
P(X) 0.383 0.367 0.25
And we satisfy all the conditions.
Part b
For this case we can find the mean using this formula:

For the variance we need to find the second moment first like this:

And we can find the variance with this formula:
![Var(X) = E(X^2) -[E(X)]^2 = 4.101- [1.867]^2 =0.615](https://tex.z-dn.net/?f=%20Var%28X%29%20%3D%20E%28X%5E2%29%20-%5BE%28X%29%5D%5E2%20%3D%204.101-%20%5B1.867%5D%5E2%20%3D0.615)
Part c
The cumulative distribution on this case is given by:
X 1 2 3
F(X) 0.383 0.367+0.383=0.75 0.25+0.75 = 1