Answer
$P(0)=0.3936$
$P(1)=0.3670$
$P(2)=0.1711$
$P(3)=0.0532$
$P(4~or~more~hits)=0.0151$
Work Step by Step
$P(x)=\frac{μ^x}{x!}e^{-μ}$
$μ=0.9323$
$P(0)=\frac{0.9323^0}{0!}e^{-0.9323}=0.3936$
$P(1)=\frac{0.9323^1}{1!}e^{-0.9323}=0.3670$
$P(2)=\frac{0.9323^2}{2!}e^{-0.9323}=0.1711$
$P(3)=\frac{0.9323^3}{3!}e^{-0.9323}=0.0532$
The event "4 or more hits" is the complement of the event "0 hits or 1 hit or 2 hits or 3 hits"
Using the Complement Rule (see page 275):
$P(4~or~more~hits)=P(X\geq4)=1-[P(0)+P(1)+P(2)+P(3)]=1-(0.3936+0.3670+0.1711+0.0532)=0.0151$
