Solution
= Solution
A day's admission count is a <Poisson random variable> with mean three. “Less than three” includes zero, one and two, so
$$
\begin{aligned}
P(A<3)&=\sum_{k=0}^{2}\frac{3^ke^{-3}}{k!}\\
&=e^{-3}\left(1+3+\frac92\right)=\boxed{\frac{17}{2}e^{-3}\approx0.423190}.
\end{aligned}
$$
Thus the probability is \b[about $42.32\%$].