Two standard six-sided dice are tossed. One die is red and the other die is blue. What is the probability that the number appearing on the red die is greater than the number appearing on the blue die?
(A) $\frac{18}{36}$
(B) $\frac{25}{36}$
(C) $\frac{15}{36}$
(D) $\frac{12}{36}$
(E) $\frac{17}{36}$
Solution 1

For each of the 6 possible outcomes that could appear on the red die, there are 6 possible outcomes that could appear on the blue die.

That is, the total number of possible outcomes when a standard six-sided red die and a standard six-sided blue die are rolled is $6 \times 6=36$.

These 36 outcomes are shown in the table below.

When a number that appears on the red die is greater than a number that appears on the blue die, a checkmark has been placed in the appropriate cell, corresponding to the intersection of the column and row.

For example, the table cell containing the double checkmark $\checkmark \checkmark$ represents the outcome of a 4 appearing on the red die and a 2 appearing on the blue die.

Number on the Red Die

| Blue | 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| 1 |  | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ | $\checkmark$ |
| 2 |  |  | $\checkmark$ | $\checkmark \checkmark$ | $\checkmark$ | $\checkmark$ |
| 3 |  |  |  | $\checkmark$ | $\checkmark$ | $\checkmark$ |
| 4 |  |  |  |  | $\checkmark$ | $\checkmark$ |
| 5 |  |  |  |  |  | $\checkmark$ |
| 6 |  |  |  |  |  |  |

Of the 36 possible outcomes, $1+2+3+4+5$ or 15 have a number appearing on the red die that is larger than the number appearing on the blue die.

The probability that the number appearing on the red die is greater than the number appearing on the blue die is $\frac{15}{36}$.

## Solution 2

As in Solution 1, we determine the total number of possible outcomes to be 36 .

Each of these 36 outcomes can be grouped into one of three possibilities; the number appearing on the red die is greater than the number appearing on the blue die, the number appearing on the red die is less than the number appearing on the blue die, or the numbers appearing on the two dice are equal.

There are 6 possible outcomes in which the numbers appearing on the two dice are equal (both numbers are 1 , both numbers are 2 , and so on).

Of the 36 total outcomes, this leaves $36-6=30$ outcomes in which either the number appearing on the red die is greater than the number appearing on the blue die, or the number appearing on the red die is less than the number appearing on the blue die.

These two possibilities are equally likely to happen (since both dice are identical except for colour), and so the number appearing on the red die will be greater than the number appearing on the blue die in half of the 30 outcomes, or 15 outcomes.

Thus, the probability that the number appearing on the red die is greater than the number appearing on the blue die is $\frac{15}{36}$.