Question

Assume that on an average one telephone number out of 15 called between 2 P.M. and 3 P.M. on week days is busy. What is the probability that if six randomly selected telephone numbers are called, at least 3 of them will be busy?

Answer 1

Let X be the number of busy calls for 6 randomly selected telephone numbers.

X follows a binomial distribution with n =6 ;$p=\mathrm{one}\mathrm{out}\mathrm{of}15=\frac{1}{15}\mathrm{and}q\mathit{}=\frac{14}{15}$
$$\begin{gathered} P(X=r)={}^{6}C_r{\left(\frac{1}{15}\right)}^r{\left(\frac{14}{15}\right)}^{6-r} \\ \mathrm{Prob}\text{ability that}\mathrm{at}\mathrm{least}3\mathrm{of}\mathrm{them}\mathrm{are}\mathrm{busy}=P(X\ge 3) \\ =1-\left\{P\right(X=0)+P(X=1)+P(X=2\left)\right\} \\ =1-\left\{{}^{\mathit{6}}C_0{\left(\frac{1}{15}\right)}^0{\left(\frac{14}{15}\right)}^{6-0}+{}^{6}C_1{\left(\frac{1}{15}\right)}^1{\left(\frac{14}{15}\right)}^{6-1}+{}^{6}C_2{\left(\frac{1}{15}\right)}^2{\left(\frac{14}{15}\right)}^{6-2}\right\} \end{gathered}$$

$=1-\left\{{\left(\frac{14}{15}\right)}^6+\frac{6}{15}{\left(\frac{14}{15}\right)}^5+\frac{1}{15}{\left(\frac{14}{15}\right)}^4\right\}$