Chapter 11.4, Problem 28PPS

To determine

What is the probability that a person takes notes less than twice in $10$ meetings?

Answer to Problem 28PPS

  $\boxed{0.37}$

Explanation of Solution

Given information:

A committee of five people randomly selects someone to take the notes of each meeting.

What is the probability that a person takes notes less than twice in $10$ meetings?

Calculation:

Generally, the probability of $k$ successes in $n$ trials of a binomial experiment is,

  $P(k){=}^n{C}_k{(p)}^k{(q)}^{n-k}$

Here $p$ represents the probability of success and $q$ represents the probability of failure.

Examine that a committee of $5$ people selects someone to take notes in each meeting.

Since one person has to be selected from $5$ people,

The probability of success is,

  $\begin{array}{l}p=\frac{1}{5}\\ =0.2\end{array}$

The probability of failure is,

  $\begin{array}{l}q=1-p\\ =1-0.2\\ =0.8\end{array}$

So, the probability that a person takes notes less than twice in $10$ meetings is,

  $\text{P}\left(\text{ Less than 2}\right)=\text{ P}\left(0\right)+\text{P}\left(\text{1}\right)$

Substitute the above values in $k$ in the above formula,

  $\text{P}\left(\text{ Less than 2}\right)$ = ${}^{10}{C}_0{(0.2)}^0{(0.8)}^{10-0}{+}^{10}{C}_1{(0.2)}^1{(0.8)}^{10-1}$

  $\begin{array}{l}=1{(}^{0.8}+10(0.2){(}^{0.8}\\ =0.10+0.27\\ =0.37\end{array}$

Hence, the probability that a person takes notes less than twice in $10$ meetings is $\boxed{0.37}$ .