A cube with its sides numbered 1 through 6 is rolled twice

Total # of outcomes is 6*6=36;

Favorable outcomes:

a-b --> prime
1-1 --> 2;

1-2 --> 3;
2-1 --> 3;

1-4 --> 5;
4-1 --> 5;
2-3 --> 5;
3-2 --> 5;

1-6 --> 7;
6-1 --> 7;
2-5 --> 7;
5-2 --> 7;
3-4 --> 7;
4-3 --> 7;

6-5 --> 11;
5-6 --> 11.

Total of 15 favorable outcomes.

P=15/36.

Answer: C.

The question says "first landing on a and then landing on b" so I assumed that the result in both rolls of the dice had to be different, so I didn't count 1,1. So for me answer was D. Was my assumption wrong?

Question says the roll landed on a and b, a and b can be different or same. The question does not give any idea about that, so assuming they are different is not correct in my opinion.

Unless it is explicitly stated otherwise, different variables CAN represent the same number.

I don't know if there is a short way to solve this, but I put down everything that can be obtained from the values of a and b:
1+1 = 2 -prime
1+2=3-prime
1+3=4-not
1+4=5-prime
1+5=5-not
1+6=prime

so when a is 1, the other must be 1,2, 4, or 6. the probability that a is 1 is 1/6. the probability that b is 1,2,4,6 is 4/6 or 2/3
the first scenario when a is 1, the overall probability that the a+b is prime is 1/6 * 2/3 or 1/9

when a is 2, in order for a+b to be prime, b must be: 1,3, or 5. the probability for a to be 2 is 1/6 and the probability of b to be 1,3, or 5 is 1/2 so overall probability for a+b to be prime when a is 2 is 1/6*1/2 or 1/12

when a is 3, b must be 2 or 4 and the overall probability is 1/6 * 1/3 = 1/18

when a is 4, b must be 1 or 3 -> overall probability 1/18

when a is 5, b must be 2 or 6 -> 1/18

when a is 6, b must be 1 or 5 -> 1/18

now, the sum of all probabilities:

1/9 + 1/12 + 4(1/18)

find LCM of 9, 12, and 18 -> 36
multiply first fraction by 4, second by 3, and the last by 2.

(4+3+8)/36 or 15/36. divide by 3 both numerator and denominator -> 5/12

Rephrased, the question is asking how many ways we can get 2,3,5,7,11 from choosing 1 through 6 twice.

First, 2 ways to choose 6 numbers = 6*6 = 36 in denominator.

In numerator:
1 way, 1 choice to get a 2 {1,1}
2 ways, 1 choice to 3 {1,2} {2,1}
2 ways, 2 choices to get 5 {1,4} {4,1} & {2,3} {3,2}
2 ways, 3 choices to get 7 {3,4} {4,3} & {5,2} {2,5} & {6,1} {1,6}
2 ways, 1 choice to get a 11 {5+6} {6,5}

1+2+4+6+2 = 15

15/36 = 5/12

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