For calls between 2pm and 3pm on week days assume that, on average, 1 telephone number out of 15 is busy.
What is the probability, if 6 randomly selected telephone numbers called, that:
1.) No more than 3 are busy.
2.) At least 3 of them will be busy.
Please, help me to solve this example.
Let $X$ be the number of busy calls. The calls will have a binomial distribution.
$$X\sim\mathcal{B}(n, p) \iff \mathbb{P}(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$
So, given that $n=6$, $p=\frac1{15}$
$$X\sim\mathcal{B}(6, \frac 1{15}) \iff \mathbb{P}(X=x)=\frac{6!}{x!(6-x)!}\cdot\frac{14^{6-x}}{15^6}$$
Use this to find:
(1) $\mathbb{P}(X\leq 3) = \mathbb{P}(X=0)+ \mathbb{P}(X=1)+ \mathbb{P}(X=2)+\mathbb{P}(X=3)$
(2) $\mathbb{P}(X\geq 3) = \mathbb{P}(X=3)+ \mathbb{P}(X=4)+ \mathbb{P}(X=5)+\mathbb{P}(X=6)$
Alternatively, you can use:
(1) $\mathbb{P}(X\leq 3) = 1 - \mathbb{P}(X=4) - \mathbb{P}(X=5) - \mathbb{P}(X=6)$
(2) $\mathbb{P}(X\geq 3) = 1 - \mathbb{P}(X=0)- \mathbb{P}(X=1)- \mathbb{P}(X=2)$