Answer
a) $$p\left(X_{1}=1\right)=0.14$$
b)
$$
p\left(X_{2}=2\right)= 0.32 \\$$
c)
$$p\left(X_{2} \geq 1\right)=0.96
$$
Work Step by Step
We can calculate directly from the table given in exercise $3-33$ as follows.
a) We model the number of people as a binomial random variable $X_{1}$ with the parameters $n=5$ and
$$
p_{1}=0.038+0.102+0.172+0.204=0.516
$$
and calculate the probability:
$$
p\left(X_{1}=1\right)=5 \times 0.516 \times(1-0.516)^{4}=0.14
$$
For b) and $c$ ) We model the number of people as a binomial random variable
$X_{2}$ with the parameters $n=5$ and
$$
p_{2}=1-p_{1}=0.484
$$
and calculate the probability:
$$\begin{array}{l}
p\left(X_{2}=2\right)=\left(\begin{array}{l}
5 \\
2
\end{array}\right) \times 0.484^{2} \times(1-0.484)^{3}= 0.32 \\
p\left(X_{2} \geq 1\right)=1-p\left(X_{2}=0\right)=1-0.516^{5}=0.96
\end{array}
$$
