A probability distribution is partially given in the followi

A probability distribution is partially given in the following table with the additional information that the even values of X are equally likely. Determine the missing entries in the table.

X

P(X)

1

.1

2

3

0

4

5

.3

6

Solution

P(1) = 0.1
P(3) = 0
P(5) = 0.3

So, P(2) + P(4) + P(6) = 1 - (0.1 + 0 + 0.3)
P(2) + P(4) + P(6) = 0.6

Since each even number outcome is equally likely, we get the answer as :

P(2) = 0.2
P(0.4) = 0.2
P(0.6) = 0.2

So, missing entries are all 0.2