Answer
$P(at~least~4)=0.586$
On about 586 of every 1000 days, there will be at least 4 cars arriving at the bank between 4:00 p.m. and 4:10 p.m.
Work Step by Step
The probability that there will be at least 4 cars arriving at the bank between 4:00 p.m. and 4:10 p.m. is the complement of the probability that there will be fewer than 4 cars arriving at the bank between 4:00 p.m. and 4:10 p.m.
$P(at~least~4)=P(x\geq4)1-P(x\lt4)=1-0.414=0.586$
