Explanation

Think of this problem as if you’re pulling out an even ticket and then another even ticket.

So, for the first ticket there are 9 possible evens out of 18 total, so the probability that the first ticket is even is $\frac{9}{18}$. Now you have one fewer even ticket in the hat. So there are 8 evens out of 17 total tickets for the second ticket, thus, the probability is $\frac{8}{17}$. You want an even AND aneven, so multiply: $\frac{9}{18}\times\frac{8}{17}=\frac{4}{17}$.

The answer is choice B.