solution: The only way for the sum of the numbers Paul and Jesse choose to be odd is if one of them chooses 2 and the other chooses an odd prime. There are five ways for Paul to choose 2 and Jesse to choose an odd prime, and there are five ways for Jesse to choose 2 and Paul to choose an odd prime. Since there are $6\cdot 6=36$ total possible ways for Paul and Jesse to choose their numbers, the probability that the sum of the numbers Paul and Jesse choose is NOT even is $\frac{10}{36}=\frac{5}{18}$. Therefore, the probability that the sum of the numbers Paul and Jesse choose IS even is $1-\frac{5}{18}=\boxed{\frac{13}{18}}$.
prompt.content: <|im_start|>system
  You are a helpful assistant.<|im_end|>
  <|im_start|>user
  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 even?
  Please reason step by step, and put your final answer within \boxed{{}}.<|im_end|>
  <|im_start|>assistant
  
prompt.role: user
extra_info.answer: \frac{13}{18}
extra_info.question: 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 even?
extra_info.split: test