The problem doesn't say whether you are looking for the probablity that the * sum* of the number cubes will be even and prime, or if one cube is even and the other cube is prime. Therefore, I will answer both.

If it is the probability of the sum:

- There are...

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The problem doesn't say whether you are looking for the probablity that the * sum* of the number cubes will be even and prime, or if one cube is even and the other cube is prime. Therefore, I will answer both.

If it is the probability of the sum:

- There are 36 possible outcomes when rolling 2 number cubes.
- Only 1 outcome is both even and prime (# 2)
- Therefore, the probability that the sum will be even and prime is 2/36 = 1/18 or about 6%

If it is the probability of rolling an even number and rolling a prime number:

- Again, there are 36 possible outcomes.
- There are 3 even numbers, so P(even) = 3/36
- There are 3 prime numbers, so P(prime) = 3/36
- The probability of both events occuring is equal to the product of each separate event. Therefore P(even and prime) = 3/36 * 3/36 = 9/1296 = 1/144 or about 0.7%

If your number cubes have 6 faces on them,

To get primes, 1+1, 1+2, 1+4, 1+6, 2+1, 2+3, 2+5, 3+2, 3+4, 4+1, 4+3, 5+2, 5+6, 6+1, 6+5

To get evens 1+1, 1+3, 1+5, 2+2, 2+4, 2+6, 3+1, 3+3, 3+5, 4+2, 4+4, 4+6, 5+1, 5+3, 5+5, 6+2, 6+4, 6+6

So P(prime) = 15/36, P(even) = 18/36

As you stated it, the P(prime and even) = 2/36 1+1 is the only even prime, but I suspect you meant P(prime or even)

P(prime or even) = P(prime) + P(even) - P(prime and even) so we get

P(prime or even) = 15/36 + 18/36 - 2/36 = 31/36