Even sum of four primes chosen from the first sixteen

Choose $4$ different numbers at random from the first $16$ prime numbers. What is the probability that their sum is even?

$\tfrac34$. Every prime except $2$ is odd. Four odd primes sum to an even number, while including $2$ gives three odd numbers plus $2$, which is odd. So the sum is even exactly when $2$ is left out: $\binom{15}{4}/\binom{16}{4}=\tfrac{12}{16}=\tfrac34$.