Probability question?

Question

Suppose you have the following probability mass function for the three random variables A, B, anc C:

P_(A,B,C) of (0,0,1) = 5/36
P_(A,B,C) of (0,0,2) = 1/36
P_(A,B,C) of (0,1,0) = 3/36
P_(A,B,C) of (0,1,4) = 3/36
P_(A,B,C) of (0,2,2) = 1/36
P_(A,B,C) of (0,2,3) = 5/36
P_(A,B,C) of (1,0,3) = 2/36
P_(A,B,C) of (1,0,4) = 4/36
P_(A,B,C) of (1,1,2) = 6/36
P_(A,B,C) of (1,2,0) = 4/36
P_(A,B,C) of (1,2,1) = 2/36

How would one find the expected value of B?

Answer 1

The expected value would be the sum of the B values times their probability. You can ignore the zero values, but look at the cases for B = 1 and B = 2

Add up the probability that B = 1:
3/36 + 3/36 + 6/36 = 12/36 = 1/3
Multiply this by the value of B (1) to get 1/3.

Add up the probability that B = 2:
1/36 + 5/36 + 4/36 + 2/36 = 12/36 = 1/3
Multiply this by the value of B (2) to get 2/3.

Now add these together to get your expected value:
1/3 + 2/3 = 1

Basically all values of B are equally likely 0, 1, 2. So the expected value is equal to the average of the outcome values. (0 + 1 + 2) / 3 = 1

The expected value of B is 1.

Answer 2

To find the expected value, sum over the value multiplied by the probability of the value. In your case:

E(B) = 0(5/36) + 0(1/36) + 1(3/36) +... + 2(2/36) = 36/36 = 1