Chapter 13.6, Problem 40E

To determine

To find: the probability that Luis will not get bonus.

Answer to Problem 40E

The probability of not getting a bonus is $\frac{15}{16}$ .

Explanation of Solution

Given:

  $P\left(\text{selling }1\text{ policy}\right)=\frac{1}{2}.$ .

Concept used:

Binomial probability refers to the probability of exactly $x$ successes on $n$ repeated trials in an Experiment which has two possible outcomes common called the binomial experiment.

If the probability of success on an individual trial is $p$ and $q$ where $p+q=1$ or $q=1-p$ .

Then the binomial probability is:

  $C{}_x{p}^x{q}^{n-x}\text{ or }C{}_x{p}^x{\left(1-p\right)}^{n-x}$ .

The probability is minimum $0$ and maximum $1$ .

  $0\le p\left(A\right)\le 1.$

If $p\left(A\right)+p\left(B\right)=1$ then $p\left(B\right)=1-p\left(A\right)$ .

At least $x$ in the probability means considering the $x$ and greater than $x$ .

At most $x$ in the probability means considering the $x$ and lower than $x$ .

Calculation:

According to the given:

  $\begin{array}{l}P\left(\text{selling }1\text{ policy}\right)=\frac{1}{2}.\\ P\left(\text{not selling }1\text{ policy}\right)=\frac{1}{2}.\end{array}$

The probability of not getting a bonus can be solved through the binomial expression as:

  $\begin{array}{l}P\left(\text{not getting a bonus}\right)=P\left(\text{less than or equal to }3\text{}\text{policies}\right).\\ \Rightarrow 1-P\left(4\text{ policies}\right).\\ \Rightarrow 1-C{}_4\times {\left(\frac{1}{2}\right)}^4\times {\left(\frac{1}{2}\right)}^0.\\ \Rightarrow 1-1\times \frac{1}{16}\times 1.\\ \Rightarrow \frac{15}{16}.\end{array}$

Hence, the probability of not getting a bonus is $\frac{15}{16}$ .