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 a multiple of 3?
Primes[2, 3, 5, 7, 11, 13]
(5, 7, 8, 9, 10, 12, 13, 14, 15, 16, 16, 18, 18, 20, 24) >>Total Sums= 15
[9, 12, 15, 18, 18, 24]>>Total = 6 - numbers that are divisible by 3.
Probability is: 6 / 15 =2 / 5
I think this is more accurate:
(4, 5, 6, 7, 8, 9, 10, 10, 12, 13, 14, 14, 15, 16, 16, 18, 18, 20, 22, 24, 26)>>Total Sums =21
[6, 9, 12, 15, 18, 18, 24] = 7 Numbers divisible by 3
Probability is: 7 / 21 = 1 / 3
