We want both digits of N to be prime. There are 4 single-digit prime numbers (2, 3, 5, 7). This means that there are 4*4=16 total options for N. We want to find how many of those 16 options are divisible by 3. So our probability will either be something/16 or - if the fraction can be reduced - something/a factor of 16. So we can eliminate A and C right off the bat because a fraction with 16 in the denominator can't be equivalent to a fraction with 3 or 25 in the denominator.
To find which of our are divisible by 3, we need to think about each number individually. We only have 16 options, and we can eliminate answers as we go, so this shouldn't take too long. Let's start with numbers that begin with 2:
22 - no
23 - no
25 - no
27 - yes, 3*9=27
Since we've found at least one option that is divisible by 3, we know that our probability is at least 1/16. This means we can eliminate answer choice E.
Now we're between B and D - a 1/4 probability means that 4 out of 16 options are divisible by 3, while a 5/16 probability means that 5 out of 16 options are. Let's go through our remaining options:
32 - no
33 - yes, 3*11=33
35 - no
37 - no
52 - no
53 - no
55 - no
57 - yes, 3*19=57
72 - yes, 3 *24=72
73 - no
75 - yes, 3*25=75
77 - no
There were 5 options in total that were divisible by 3, so our probability is 5/16, or answer choice D.
