You cannot find the correct answer this way.

The answer should be B.

The sample space is not the sum of two dice but it is [ {1,1} , {1,2} , {1,3} , .................., {6,6} ]

so we have total number of cases 36.

the number of cases with prime sum will be {1,1} , {2,1} , {1,2} , .............. so on.

Thus the answer is 15/36 = 5/12.

I initially thought that as well - it was my first answer. Thinking about it now, you and Carcass are correct and the answer given is incorrect. Should've stuck with my gut

When I looked at the answer I figured the sample space only cared about the result of the dice being thrown. So getting (2,1) is the same as getting (1,2), and so on.

Thanks for the correction.

The problem here is that you are giving each of the possible sums an equal weight. That is, you are giving them an equal probability of occurring which is not the case.
The probability of getting a sum of 7 for example is (6/36), whereas the probability of getting a sum of 2 is (1/36).
So you don't have a simple sample space.

If you want to do it this way you have to first find the probability of the possible sums.

P(Sum=2)=1/36
P(Sum=3)=2/36
P(Sum=5)=4/36
P(Sum=7)=6/36
P(Sum=11)=2/36

Then we are looking for P(Sum=2 or Sum=3 or Sum=5 or Sum=7 or Sum=11)
Since each event is disjoint
P(Sum=2 or Sum=3 or Sum=5 or Sum=7 or Sum=11)
=P(Sum=2)+ P(Sum=3)+ P(Sum=5)+ P(Sum=7)+ P(Sum=11)
= (1/36)+ (2/36)+ (4/36)+ (6/36)+ (2/36)
= 15/36
= 5/12

Final Answer: B

May you explain your shortcut ?

Given that Two fair dice with sides numbered 1 to 6 are tossed and We need to find What is the probability that the sum of the exposed faces on the dice is a prime number?

As we are rolling two dice => Number of cases = \(6^2\) = 36

We know that out of the numbers from 2(1+1) to 12 (6+6) we have only five prime numbers and they are 2, 3, 5, 7 and 11
(Watch this video if you are not aware of Prime Numbers)

Sum of the exposed faces has to be prime. So, following are the possible cases
(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)

=> 15 possibilities

=> Probability that the sum of the exposed faces on the dice is a prime number = \(\frac{15}{36}\) = \(\frac{5}{12}\)

So, Answer will be B
Hope it helps!

Watch the following video to learn How to Solve Dice Rolling Probability Problems

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