ExplanationThink 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.
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