Answer
a) $0.4096$
b) $0.218$
c) $0.37$
Work Step by Step
Let $X$ be the random variable of number(amount to) of days (out of total of $n$ days) when the green light is observed.
$X$ has the binomial distribution with parameters $n, p=0.2 .$ The (chance)probability mass function of $X$ is stated by:
$$\mathbb{P}(X=k)=\left(\begin{array}{c}{n} \\ {k}\end{array}\right) 0.2^{k} \times 0.8^{n-k}, k=0,1, \ldots, n$$
Calculate using this formula:
(a)
$\operatorname{Set} n=5 :$
$\mathbb{P}(X=1)=\left(\begin{array}{l}{5} \\ {1}\end{array}\right) 0.2^{1} \times 0.8^{4}=[0.4096]$
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(b)
$\operatorname{Set} n=20 :$
$\mathbb{P}(X=4)=\left(\begin{array}{c}{20} \\ {4}\end{array}\right) 0.2^{4} \times 0.8^{16}=0.218$
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(c)
$\operatorname{Set} n=20 :$
$\mathbb{P}(X>4)=1-\mathbb{P}(X \leq 4)=$
$\quad \quad =1-\mathbb{P}(X=0)-\mathbb{P}(X=1)-\mathbb{P}(X=2)-\mathbb{P}(X=3)-\mathbb{P}(X=4)=$
$\quad\quad =0.37$
