College Algebra, Chapter 10, 10.4, Section 10.4, Problem 34

A mortgage company advertises its rates by making unsolicited telephone calls to random numbers. About $2 \%$ if the calls reach consumers who are interested in the company's services. A telephone consultant can make 100 calls per evening shift.

a.) What is the probability that two or more calls will reach an interested party in one shift?

b.) How many calls does a consultant need to make to ensure at least a $0.5$ probability of reaching one or more interested parties?

Recall that the formula for the binomial probability is given by

$C(n,r) p^r q^{n-r}$

a.) In this case, the probability of success is $p=0.02$ while the probability of failure is $q=1-p=0.98$. To solve this more easily, we can use the complement of the probability that none of 1 call reach an interested party. Thus, the probability that two or more calls will reach an interested party is


$\begin{equation} \begin{aligned} =& 1 - [P(0) + P(1)] \\ \\ =& 1 - \left[C(100,0)(0.02)^0 (0.98)^{100-0} + C(100,1)(0.02)^1 (0.98)^{100-1}\right] \\ \\ =& 1 - [0.1326 + 0.2707] \\ \\ =& 0.5967 \end{aligned} \end{equation}$



b.) In this case, we are required to solve for $n$. Again, by applying complement of the probability that none has reached the interested party, we get


$\begin{equation} \begin{aligned} 0.50 =&1 - \left[C(n,0) (0.02)^0 (0.98)^{n-0}\right] && \\ \\ 0.50 =& \frac{n!}{0! (n-0)!} (0.98)^{n-0} && \\ \\ 0.50 =& \frac{n!}{n!} (0.98)^n && \\ \\ 0.50 =& (0.98)^n && \text{Take $\ln$ of both sides} \\ \\ \ln (0.50) =& n \ln (0.98) && \text{Solve for } n \\ \\ n =& 34.34 && \\ \\ & \text{or} && \\ \\ n \approx & 35 && \end{aligned} \end{equation}$



It shows that the consultant must make 35 calls in order to ensure a $0.5$ probability of reaching one or more interested party.