probability

 

Paul and Jesse each choose a number at random from the first six primes. What is the probability that the sum of the numbers they choose is divisible by 3?  

The first six primes are 2, 3, 5, 7, 11, 13  

 

I see nothing in the problem that would prohibit Paul and Jesse  

from choosing the same number, so we have the following possible 

 

2+2, 2+3, 2+5, 2+7, 2+11, 2+13 

3+3, 3+5, 3+7, 3+11, 3+13  

5+5, 5+7, 5+11, 5+13  

7+7, 7+11, 7+13  

11+11, 11+13  

13+13 

 

The pairs that total a multiple of 3 (marked in purple above) are 

2+7,  2+13,  3+3,  5+7,  5+13,  7+11,  and  11+13  

 

  7 pairs that total a multiple of 3  

21 pairs possible 

 

So the probability is 7 out of 21   

 

               7               1  

             –––    =     –––  

              21              3  

_.

If you choose 2 numbers out of the first 6 primes {2, 3, 5, 7, 11, 13}, with repetitions allowed, you should have this many sums: [6 + 2 - 1] C 2 =7 C 2 =21 sums. You should get the following sums:

 

(4, 5, 6, 7, 8, 9, 10, 10, 12, 13, 14, 14, 15, 16, 16, 18, 18, 20, 22, 24, 26)>Total = 21 Sums

 

(6, 9, 12, 15, 18, 18, 24) = 7 sums that are divisible by 3.

 

Therefore, the probability is: 7 / 21 = 1 / 3