Here
In the number from 1 to 18,
number of even numbers = 9

number of odd numbers = 9

Therefore probability of picking two even numbers = \(\frac{9}{18} * \frac{8}{17}\) = \(\frac{4}{17}\)
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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.
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there are 9 even numbers between 1 to 18(inclusive).
total outcomes=18C2
total favourable outcomes=9C2
probability=9C2/18C2 =4/17

P(both even) = P(1st ticket is even AND 2nd ticket is even)
= P(1st ticket is even) x P(2nd ticket is even)
= 9/18 x 8/17
= 4/17

Answer: B

Cheers,
Brent
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