Solution: 2022 Winter Final - 12
Author: Michiel SmidQuestion
Let $D_1$ and $D_2$ be the results of rolling two normal $6$-sided dice. What is $\Pr((D_1+D_2)\bmod 4 = 0)$?
(a)
$1/4$
(b)
$1/5$
(c)
$1/6$
(d)
$1/2$
(e)
$1/3$
Solution
Let’s make a table and see how many of the 36 outcomes is divisible by 4
4, 8, and 12 are all divisible by 4
- 4 comes up 3 times \\ $ \Pr(D_1+D_2=4) = \frac{3}{36} $
- 8 comes up 5 times \\ $ \Pr(D_1+D_2=8) = \frac{5}{36} $
- 12 comes up 1 time \\ $ \Pr(D_1+D_2=12) = \frac{1}{36} $
$ \Pr((D_1+D_2)\bmod 4 = 0) $
$ = \Pr(D_1+D_2=4) + \Pr(D_1+D_2=8) + \Pr(D_1+D_2=12) $
$ = \frac{3}{36} + \frac{5}{36} + \frac{1}{36} $
$ = \frac{9}{36} $
$ = \frac{1}{4} $