Answer
$P(more~than~one)=0.00121$
In about 121 of next one hundred thousand 100 million miles of flight, there will be more than one fatal accident.
Work Step by Step
$P(fewer~than~2)=P(x\lt2)=P(0)+P(1)=\frac{(0.05\times1)^0}{0!}e^{-0.05\times1}+\frac{(0.05\times1)^1}{1!}e^{-0.05\times1}=\frac{1}{1}e^{-0.05}+\frac{0.05}{1}e^{-0.05}=0.99879$
The probability that there will be more than one fatal accident during the next 100 million miles of flight is the complement of the probability that there will be fewer than 2 fatal accidents during the next 100 million miles of flight.
Using the Complement Rule (see page 275):
$P(more~than~one)=P(x\gt1)=1-P(fewer~than~2)=1-0.99879=0.00121$
