Answer
0.06, not unlikely.
Work Step by Step
By using the formula: $P(12)=\frac{\mu^k\cdot e^{-\mu}}{x!}=\frac{8.5^{12}\cdot e^{-8.5}}{12!}=0.06.$
If a value is unusual, then it is more than two standard deviations far from the mean. $Minimum \ usual \ value=mean-2\cdot(standard \ deviation)=8.5-2\cdot\sqrt{8.5}=2.67$
$Maximum \ usual \ value=mean+2\cdot(standard \ deviation)=8.5+2\cdot\sqrt{8.5}=14.33$.
12 is between the two bounds therefore it is not unlikely.