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?
Here are the first 6 primes: {2,3,5,7,11,13}
We can take a look at all the possibilities by making a table.
235711132457913153568101416578101216187910121418201113141618222413151618202426
Blue numbers and red numbers represent the number that Paul chose and the number that Jesse chose respectively.
Pink numbers are those divisible by 3.
By trial-and-error, out of all 36 possible cases, 13 cases has a sum divisible by 3.
Therefore,
Prob(sum divisible by 3)=1336
Assuming they can pick the same number
36 choices
2 7
2 13
5 7
5 13
7 11
3 3 the red ones can be reversed 11 choices out of 36
OOps...I see from Max answer I missed 11 13 13 11 so the answer would be the same 13 /36 (Thanx, max!)
Here are the first 6 primes: {2,3,5,7,11,13}
We can take a look at all the possibilities by making a table.
235711132457913153568101416578101216187910121418201113141618222413151618202426
Blue numbers and red numbers represent the number that Paul chose and the number that Jesse chose respectively.
Pink numbers are those divisible by 3.
By trial-and-error, out of all 36 possible cases, 13 cases has a sum divisible by 3.
Therefore,
Prob(sum divisible by 3)=1336