# If two dice are rolled once, what is the probability that it will show a multiple of 3 or 4?

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The probability of an event is a ratio of the number of all favorable outcomes (rolling a multiple of 3 or 4) to the number of all possible outcomes.

There are **36** possible outcomes of rolling two dice:

(1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

The two rolled dice can possibly show any integer number between 2 and 12. From these numbers, the following are multiples of 3 or 4:

3, 4, 6, 8, and 12.

There are 2 ways to get 3: (1, 2) (2, 1)

There are **3** ways to get 4: (1, 3) (2, 2) (3, 1)

There are **5 **ways to get 6: (1, 5) (2, 4) (3, 3) (4, 2) (5, 1)

There are **5** ways to get 8: (2, 6) (3, 5) (4, 4) (5, 3) (6, 2)

There is** 1** way to get 12: (6, 6)

All together, there are 2 + 3 + 5 + 5 + 1 = 16 ways to get a multiple of 3 or 4, so there are 16 favorable outcomes.

Then, the probability of rolling a multiple of 3 or 4 is

`16/36 = 4/9`

Converted to percents and rounded, the probability would be

`4/9 * 100% = 44.4%`

**The probability that two dice will show a multiple of 3 or 4 is 4/9, or 44.4%.**