, 2023

Solution 1

We begin by completing a table in which the ones digit of each
possible product is listed.

For example, when the number on the first die is 3 and the number on the
second die is 6, the entry in the table is 8 since $3\times6=18$ and the ones digit of 18 is
8.

$\times$
Number on the second die

1
2
3
4
5
6

Number on the First Die
1
1
2
3
4
5
6

2
2
4
6
8
0
2

3
3
6
9
2
5
8

4
4
8
2
6
0
4

5
5
0
5
0
5
0

6
6
2
8
4
0
6

Of the 36 possible outcomes in the table above, 6 outcomes have a
ones digit that is equal to 0.

Thus, the probability that the ones digit of the product is 0 is $\dfrac{6}{36}=\dfrac{1}{6}$.

Solution 2

Since the ones digit of the product is 0, then the product is
divisible by 5 and is even.

Since the possible numbers in the product are 1, 2, 3, 4, 5, 6, then one
of the numbers rolled must be 5.

Since the product is even (and 5 is not), then the other number rolled
must be one of the three even numbers, namely 2, 4, 6.

Thus, the possible pairs of numbers that can be rolled to give a product
whose ones digit is 0, are $(5,2)$,
$(5,4)$, $(5,6)$ or $(2,5)$, $(4,5)$, $(6,5)$. (We note that the first number in
the ordered pair represents the first number rolled, while the second
number in the pair is the second number rolled.)

Since there are 6 possible rolls for each of the two dice, then there
are $6\times6=36$ possible ordered
pairs representing all possible outcomes.

Since 6 of these ordered pairs represent a product whose ones digit is
0, then the required probability is $\dfrac{6}{36}=\dfrac{1}{6}$.