How to find the Var(X) for a sum of a rolled die and a 3 sided spinner?

Question

A fair 6-sided die has the numbers −1, −1, 0, 0, 1, 2 on its faces. A fair 3-sided spinner has edges numbered −1, 0, 1. The die is thrown and the spinner is spun. The number on the uppermost face of the die and the number on the edge on which the spinner comes to rest are noted. The sum of these two numbers is denoted by X . What is the Var(X)? Please explain each step. Thank you!

Answer 1

You can either get the distribution of the possible sums and calculate the variance or make use of the fact that the variance of the sums is the sum of the variances when variables are independent.

The sum of squares of integers from 1 to n is n(n+1)(2n+1)/6

The sum of integers from 1 to n is n(n+1)/2

variance may be calculated as E(x^2) - (E(x))^2

For the die, then, this is 1/6 ∑x^2 - ((1/6)∑x)^2 = 1/6(6)(6+1)(2*6+1)/6 - [(1/6)6(6+1)/2]^2 =

91/6 - 49/4 = 35/12

For the spinner, this is 1/3 ∑x^2 - ((1/3)∑x)^2 = 1/3(3)(3+1)(2*3+1)/6 - [(1/3)3(3+1)/2]^2 =

28/6 - 4 = 14/3 - 4 = 2/3

Then, the sum of the variances is 35/12 + 2/3 = 43/12

The other way, accumulating the 18 combinations of 1-6 and 1-3, each with probability 1/18:

P(2) = P(9) = 1/18

P(3) = P(8) = 2/18 = 1/9

P(4)=P(5) = P(6) = P(7) = 3/18 = 1/6

Then, from symmetry, E(x) = 11/2

Thus, ∑x^2P(x) - (E(x))^2 = 203/6 - 121/4 = 43/12