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?

Guest Nov 26, 2020

#1**0 **

*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**

_{.}

Guest Nov 26, 2020

#2**0 **

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**

Guest Nov 26, 2020