# 2 number cubes are rolled. Find P(an even muber and a prime number) 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%

Approved by eNotes Editorial Team 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

Approved by eNotes Editorial Team