 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2013/discrete-mathematics-probability-and-continuous-mathematics/8/c/iv/solution

Count the ordered roll pairs producing each value. The probability mass functions are

P (X = i) = \frac{6 - ∣ i - 7∣}{36} (i = 2, \dots, 12), P (Y = j) = \frac{6 - ∣ j ∣}{36} (j = - 5, \dots, 5),

and are zero outside these ranges. Both counts have a unique peak:

i = 7, j = 0; P (X = 7) = P (Y = 0) = \frac{1}{6} .

However,

Y = 0

forces equal rolls, whose sum is even. Hence

P (X = 7, Y = 0) = 0

, whereas the product of the two marginal probabilities is

1/36 > 0

X and Y are not independent .

 New to topics? Read the docs here!