Arrivals at a service facility occur at random (Poisson distributed) at a mean rate of $10$ per minute. Find the probability of observing more than $2$ arrivals in $10$ seconds. I've begun to answer this question using the poisson tables:
$P(\gt2$ arrivals in an interval of $10$ seconds)
$=1-P(\le2$ arrivals in $60$ seconds) $=0.9972$
I have then gone onto divide this answer by $6$ giving me the answer of $0.1662$
Really not sure if this is correct.
Note that the mean of $10$ arrivals per minute implies the mean of $\frac{5}{3}$ arrivals per $10$ seconds. So: $$P(X>2)=1-P(X\le 2)=1-P(X=0)-P(X=1)-P(X=2)=\\ 1-\frac{e^{-\frac 53}\cdot \left(\frac 53\right)^0}{0!}-\frac{e^{-\frac 53}\cdot \left(\frac 53\right)^1}{1!}-\frac{e^{-\frac 53}\cdot \left(\frac 53\right)^2}{2!}=1-e^{-\frac 53}\left(1+\frac 53+\frac{25}{18}\right)=0.234.$$ Intuitively, if the mean arrival is $\frac 53$ per $10$ seconds, the probability of more than $2$ arrivals in $10$ seconds will be less than $0.5$.