Question

A particular telephone number is used to receive both voice calls and fax messages. Suppose that 25% of the incoming calls involve faxmessages, and consider a sample of 25 incoming calls. (Round your answers to three decimal places.)

(a) What is the probability that at most 4 of the calls involve a fax message?

(b) What is the probability that exactly 4 of the calls involve a fax message?

(c) What is the probability that at least 4 of the calls involve a fax message?

(d) What is the probability that more than 4 of the calls involve a fax message?

You may need to use the appropriate table in the Appendix ofTables to answer this question.

Answer 1

Concepts and reason

The concept of binomial distribution is used to solve this problem.

A binomial experiment is a discrete probability experiment which is repeated for a fixed number of trials, and each of the trials is independent of the other trials. The possible outcomes for each trial are two, and they are defined as a success (S) or as a failure (F).

The probability of success p for each trial remains the same. The random variable X in the experiment represents the number of successes in n independent trials.

Fundamentals

The experiments where each trial has only two possible outcomes can be explained with the help of binomial distribution. One of the outcomes is termed as success with $p$ being the probability of success, while the other is termed as a failure with $1 - p$ being the probability of failure. The probability of x successes out of n trials is defined as,

$P\left( {X = x} \right) = \left( \begin{array}{c}\\n\\\\x\\\end{array} \right){\mkern 1mu} {p^x}{(1 - p)^{n - x}}$

Where n is the number of trials, p is the probability of success x is the number of successes and q is the probability of failures.

(a)

Consider the variable X as the number of fax messages in the incoming calls. Then the success is defined as the event that the incoming call is a voice call. The proportion of voice calls is 25 percent. Therefore, the parameters for the binomial distribution can be listed as:

$\begin{array}{l}\\n = 25\\\\p = 0.25\\\end{array}$

The probability that at most 5 of the calls involve a fax message is calculated as,

$\begin{array}{c}\\P\left( {x \le 5} \right) = \sum\limits_{x = 0}^5 {P\left( {X = x} \right)} \\\\ = \sum\limits_{x = 0}^5 {\frac{{25!}}{{x! \times \left( {25 - x} \right)!}} \times {{\left( {0.25} \right)}^x}{{(1 - 0.25)}^{25 - x}}} \\\\ = \left( \begin{array}{c}\\\frac{{25!}}{{0! \times \left( {25 - 0} \right)!}} \times {\left( {0.25} \right)^0}{(1 - 0.25)^{25 - 0}}\\\\ + \frac{{25!}}{{1! \times \left( {25 - 1} \right)!}} \times {\left( {0.25} \right)^1}{(1 - 0.25)^{25 - 1}}\\\\ + \frac{{25!}}{{2! \times \left( {25 - 2} \right)!}} \times {\left( {0.25} \right)^2}{(1 - 0.25)^{25 - 2}}\\\\ + \frac{{25!}}{{3! \times \left( {25 - 3} \right)!}} \times {\left( {0.25} \right)^3}{(1 - 0.25)^{25 - 3}}\\\\ + \frac{{25!}}{{4! \times \left( {25 - 4} \right)!}} \times {\left( {0.25} \right)^4}{(1 - 0.25)^{25 - 4}}\\\\ + \frac{{25!}}{{5! \times \left( {25 - 5} \right)!}} \times {\left( {0.25} \right)^5}{(1 - 0.25)^{25 - 5}}\\\end{array} \right)\\\\ = \left( \begin{array}{l}\\0.0008 + 0.0063 + 0.0251 + 0.0641\\\\ + 0.1175 + 0.1645\\\end{array} \right)\\\end{array}$

$= 0.3783$

(b)

The probability that exactly five of the 25 calls are fax messages is calculated as:

$\begin{array}{c}\\P\left( {x = 5} \right) = \left( \begin{array}{c}\\n\\\\x\\\end{array} \right){\mkern 1mu} {p^x}{(1 - p)^{n - x}}\\\\ = \frac{{25!}}{{5! \times \left( {25 - 5} \right)!}} \times {0.25^5}{(1 - 0.25)^{25 - 5}}\\\\ = 0.1645\\\end{array}$

(c)

The probability that at least five of the incoming calls are fax messages is calculated as:

$\begin{array}{c}\\P\left( {x \ge 5} \right) = 1 - P\left( {x < 5} \right)\\\\ = 1 - P\left( {x \le 4} \right)\\\\ = 1 - \sum\limits_{x = 0}^4 {P\left( {X = x} \right)} \\\\ = 1 - \left\{ {\sum\limits_{x = 0}^4 {\frac{{25!}}{{x! \times \left( {25 - x} \right)!}} \times {{\left( {0.25} \right)}^x}{{(1 - 0.25)}^{25 - x}}} } \right\}\\\end{array}$

$\begin{array}{c}\\ = 1 - \left( \begin{array}{c}\\\frac{{25!}}{{0! \times \left( {25 - 0} \right)!}} \times {0.25^0}{(1 - 0.25)^{25 - 0}}\\\\ + \frac{{25!}}{{1! \times \left( {25 - 1} \right)!}} \times {0.25^1}{(1 - 0.25)^{25 - 1}}\\\\ + \frac{{25!}}{{2! \times \left( {25 - 2} \right)!}} \times {0.25^2}{(1 - 0.25)^{25 - 2}}\\\\ + \frac{{25!}}{{3! \times \left( {25 - 3} \right)!}} \times {0.25^3}{(1 - 0.25)^{25 - 3}}\\\\ + \frac{{25!}}{{4! \times \left( {25 - 4} \right)!}} \times {0.25^4}{(1 - 0.25)^{25 - 4}}\\\end{array} \right)\\\\ = 1 - \left( \begin{array}{l}\\0.0008 + 0.0063 + 0.0251\\\\ + 0.0641 + 0.1175\\\end{array} \right)\\\\ = 1 - 0.2138\\\\ = 0.7862\\\end{array}$

(d)

The probability that more than five out of the 25 incoming calls are fax messages is calculated as,

$\begin{array}{c}\\P\left( {x > 5} \right) = 1 - P\left( {x \le 5} \right)\\\\ = 1 - \left\{ {\sum\limits_{x = 0}^4 {\frac{{25!}}{{x! \times \left( {25 - x} \right)!}} \times {{0.25}^x}{{(1 - 0.25)}^{25 - x}}} } \right\}\\\\ = 1 - 0.3783\\\\ = 0.6217\\\end{array}$

Ans: Part a

The probability that at most 5 of the 25 calls are fax messages is 0.3783.

Part b

The probability of exactly five fax messages out of the 25 incoming calls is 0.1645.

Part c

The probability that at least 5 of the incoming 25 calls are fax messages is 0.7862.

Part d

The probability that more than 5 of the incoming calls are fax messages is 0.6217.