Let X denote the number on the top side of the unbiased die.
The probability mass function is given by the following table.
X = x |
1 |
2 |
3 |
4 |
5 |
6 |
P(x) |
1/6 |
1/6 |
1/6 |
1/6 |
1/6 |
1/6 |
The expected value for the random variable X is
E(X) = ∑xxPx(x)
= (1 x 1/6) + (2 x 1/6) + (3 x 1/6) + (4 x 1/6) + (5 x 1/6) + (6 x 1/6)
= 1/6(1 + 2 + 3 + 4 + 5 + 6)
= 21/6 = 7/2 = 3.5