Dear emilia_3272,

Part 1)

Let
p1 = probability of rolling a 1,
p2 = probability of rolling a 2,
p3 = probability of rolling a 3,
p4 = probability of rolling a 4,
p5 = probability of rolling a 5, and
p6 = probability of rolling a 6.

We’re told that
p2 = 2 p1,
p3 = 3 p1,
p4 = 4 p1,
p5 = 5 p1, and
p6 = 6 p1.

If there are no other possibilities (such as the die balancing on an edge), then from the basic axioms of probability we know that

p1 + p2 + p3 + p4 + p5 + p6 = 1, so

p1 + (2 p1) + (3 p1) + (4 p1) + (5 p1) + (6 p1) = 1
21 p1 = 1

and thus

p1 = 1/21,
p2 = 2/21,
p3 = 3/21,
p4 = 4/21,
p5 = 5/21, and
p6 = 6/21.

Therefore, the probability mass function for this die is

f(x) = x / 21, for x in {1, 2, 3, 4, 5, 6}, otherwise f(x) = 0.

Part 2)

The mean is a weighted average of the possible outcomes, with the weights being the respective probabilities. So in this case the mean m is the sum of each x times f(x). That is,

m = (1 f(1)) + (2 f(2)) + (3 f(3)) + (4 f(4)) + (5 f(5)) + (6 f(6))
= (1 1/21) + (2 2/21) + (3 3/21) + (4 4/21) + (5 5/21) + (6 6/21)
= 1/21 + 4/21 + 9/21 + 16/21 + 25/21 + 36/21
= 91/21
= 13/3
= 4.333 (to three decimal places).

The variance v uses the same weights, but instead of applying them to the outcomes x, we apply them to (x – m)^2.

v = ((1 – 13/3)^2 f(1)) + ((2 – 13/3)^2 f(2)) + ((3 – 13/3)^2 f(3))
+ ((4 – 13/3)^2 f(4)) + ((5 – 13/3)^2 f(5)) + ((6 – 13/3)^2 f(6))
= ((10/3)^2 1/21) + ((7/3)^2 2/21) + ((4/3)^2 3/21)
+ ((1/3)^2 4/21) + ((2/3)^2 5/21) + ((5/3)^2 6/21)
= 100/189 + 98/189 + 48/189 + 4/189 + 20/189 +150/189
= 420/189
= 20/9
= 2.222 (to three decimal places).

Now you should be able to compare this with a fair die, which I think has mean 7/2 (equals 3.500) and variance 35/12 (equals 2.917 to three decimal places). You should also check to make sure my calculations are correct.