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 ;

\[p = \text{ one out of } 15 = \frac{1}{15}\text{ and }  q = \frac{14}{15}\]

\[P(X = r) = ^{6}{}{C}_r \left( \frac{1}{15} \right)^r \left( \frac{14}{15} \right)^{6 - r} \]
\[\text{ Probability that at least 3 of them are busy}  = P(X \geq 3) \]
\[ = 1 - {P(X = 0) + P(X = 1) + P(X = 2)}\]
\[ = 1 - \left\{ ^{6}{}{C}_0 \left( \frac{1}{15} \right)^0 \left( \frac{14}{15} \right)^{6 - 0} + ^{6}{}{C}_1 \left( \frac{1}{15} \right)^1 \left( \frac{14}{15} \right)^{6 - 1} + ^{6}{}{C}_2 \left( \frac{1}{15} \right)^2 \left( \frac{14}{15} \right)^{6 - 2} \right\}\]

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