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.
\(\begin{matrix}&\color{blue}2&\color{blue}3&\color{blue}5&\color{blue}7&\color{blue}11&\color{blue}13\\\color{red}2&4&5&7&\color{magenta}9&13&\color{magenta}15\\\color{red}3&5&\color{magenta}6&8&10&14&16\\\color{red}5&7&8&10&\color{magenta}12&16&\color{magenta}18\\\color{red}7&\color{magenta}9&10&\color{magenta}12&14&\color{magenta}18&20\\\color{red}11&13&14&16&\color{magenta}18&22&\color{magenta}24\\\color{red}13&\color{magenta}15&16&\color{magenta}18&20&\color{magenta}24&26\end{matrix}\)
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,
\(\text{Prob}(\text{sum divisible by 3}) = \dfrac{13}{36}\)
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.
\(\begin{matrix}&\color{blue}2&\color{blue}3&\color{blue}5&\color{blue}7&\color{blue}11&\color{blue}13\\\color{red}2&4&5&7&\color{magenta}9&13&\color{magenta}15\\\color{red}3&5&\color{magenta}6&8&10&14&16\\\color{red}5&7&8&10&\color{magenta}12&16&\color{magenta}18\\\color{red}7&\color{magenta}9&10&\color{magenta}12&14&\color{magenta}18&20\\\color{red}11&13&14&16&\color{magenta}18&22&\color{magenta}24\\\color{red}13&\color{magenta}15&16&\color{magenta}18&20&\color{magenta}24&26\end{matrix}\)
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,
\(\text{Prob}(\text{sum divisible by 3}) = \dfrac{13}{36}\)