Answer
$P(fewer~than~4)=0.414$
On about 414 of every 1000 days, there will be fewer than 4 cars arriving at the bank between 4:00 p.m. and 4:10 p.m.
Work Step by Step
$P(fewer~than~4)=P(x\lt4)=P(0)+P(1)+P(2)+P(3)=\frac{(0.41\times10)^0}{0!}e^{-0.41\times10}+\frac{(0.41\times10)^1}{1!}e^{-0.41\times10}+\frac{(0.41\times10)^2}{2!}e^{-0.41\times10}+\frac{(0.41\times10)^3}{3!}e^{-0.41\times10}=\frac{1}{1}e^{-4.1}+\frac{4.1}{1}e^{-4.1}+\frac{16.81}{2}e^{-4.1}+\frac{68.921}{6}e^{-4.1}=0.414$
