Joe picks 2 distinct numbers from the set of the first 14 positive integers S = \{1,2,3,\ldots,14\}. The probability that the sum of the 2 numbers is divisible by 3 can be expressed as \frac{a}{b}, where a and b are coprime positive integers. What is the value of a+b?

122

To solve this problem, we need to find the number of favorable outcomes (where the sum of the two numbers is divisible by 3) and the total number of possible outcomes.

Let's first find the total number of possible outcomes:
Joe has to choose 2 distinct numbers from a set of 14 positive integers. The total number of ways to do this is given by the combination formula: C(14, 2) = 14! / (2! * (14-2)!) = 91.

Now, let's find the number of favorable outcomes:
For the sum of two numbers to be divisible by 3, there are a few cases we need to consider:

Case 1: Choose two numbers that are both divisible by 3.
There are 4 numbers (3, 6, 9, and 12) in the set S that are divisible by 3. Joe can choose any 2 of these numbers, so the number of favorable outcomes for this case is C(4, 2) = 6.

Case 2: Choose two numbers with a remainder of 1 when divided by 3.
There are 5 numbers (1, 4, 7, 10, and 13) in the set S that have a remainder of 1 when divided by 3. Joe can choose any 2 of these numbers, so the number of favorable outcomes for this case is C(5, 2) = 10.

Case 3: Choose two numbers with a remainder of 2 when divided by 3.
There are 5 numbers (2, 5, 8, 11, and 14) in the set S that have a remainder of 2 when divided by 3. Joe can choose any 2 of these numbers, so the number of favorable outcomes for this case is C(5, 2) = 10.

Therefore, the total number of favorable outcomes is 6 + 10 + 10 = 26.

Finally, the probability that the sum of the two numbers is divisible by 3 is given by the ratio of the number of favorable outcomes to the total number of possible outcomes: P = 26/91.

So, the value of a is 26 and b is 91, and a + b = 26 + 91 = 117.