## Elementary Statistics (12th Edition)

a) n=12, p=$\frac{1}{6}$, therefore $n\geq100$ is not satisfied, therefore we cannot use the Poisson distribution. b) Counting by the Poisson distribution: $\frac{2^3\cdot e^{-2}}{3!}=0.18$, whilst counting by the binomial distrbution:${12\choose 3}\cdot (\frac{1}{6})^3 \cdot (\frac{5}{6})^9=0.197.$ These two results are far from each other, so the result gained by the Poisson distribution is not OK.