Chapter 2, Problem 144SE

Let S contain four sample points, E₁, E₂, E₃, and E₄.

1. a List all possible events in S (include the null event).
2. b In Exercise 2.68(d), you showed that ${\displaystyle {\sum }_{i=1}^n\left(\begin{array}{l}n\\ i\end{array}\right)}=2^n$. Use this result to give the total number of events in S.
3. c Let A and B be the events {E₁, E₂, E₃} and {E₂, E₄}, respectively. Give the sample points in the following events: $A\cup B$, A ∩ B, $\bar{A}\cap \bar{B}$, and $\bar{A}\cup B$.

To determine

Find the list of all possible events in S.

Explanation of Solution

Sample space:

The set of all possible sample points of the experiment is termed as sample space.

The sample points of the sample space are $E_1,E_2,E_3\text{ and }{E}_4$.

The list of all possible events in S is shown below:

$S=\left\{\begin{array}{l}\left(\varphi \right),\left(E_1\right),\left(E_2\right),\left(E_3\right),\left(E_4\right),\left(E_1,E_2\right),\left(E_1,E_3\right),\left(E_1,E_4\right),\\ \left(E_2,E_3\right),\left(E_2,E_4\right),\left(E_3,E_4\right),\left(E_1,E_2,E_3\right),\left(E_1,E_2,E_4\right),\\ \left(E_1,E_3,E_4\right),\left(E_2,E_3,E_4\right),\left(E_1,E_2,E_3,E_4\right)\end{array}\right\}$

To determine

Find the total number of events in S.

Explanation of Solution

Combination rule:

If the number of unordered subsets of size r selected from n objects then it can be termed as combination.

The combination formula is shown below.

$\left(\begin{array}{l}n\\ r\end{array}\right)=\frac{n!}{r!\left(n-r\right)!}$

Here, n represents the total number of objects and r represents the number of unordered subsets.

The total number of events in S is obtained as follows:

$\begin{array}{c}\text{Total number of events}=\left(\begin{array}{l}\left(\text{Number of ways of getting zero sample point}\right)+\\ \left(\text{Number of ways of getting one sample point}\right)+\\ \left(\text{Number of ways of getting two sample points}\right)+\\ \left(\text{Number of ways of getting three sample points}\right)+\\ \left(\text{Number of ways of getting four sample points}\right)\end{array}\right)\\ =\left(\begin{array}{l}4\\ 0\end{array}\right)+\left(\begin{array}{l}4\\ 1\end{array}\right)+\left(\begin{array}{l}4\\ 2\end{array}\right)+\left(\begin{array}{l}4\\ 3\end{array}\right)+\left(\begin{array}{l}4\\ 4\end{array}\right)\\ =\frac{4!}{0!\left(4-0\right)!}+\frac{4!}{1!\left(4-1\right)!}+\frac{4!}{2!\left(4-2\right)!}+\frac{4!}{3!\left(4-3\right)!}+\frac{4!}{4!\left(4-4\right)!}\\ =1+4+6+4+1\\ =16\\ =2^4\end{array}$

Thus, the total number of events in S is 16.

To determine

Find the sample points in the given events.

Explanation of Solution

From the given information, $A=\left\{E_1,E_2,E_3\right\}\text{ and }B=\left\{E_2,E_4\right\}$.

For $A\cup B$:

$\begin{array}{c}A\cup B=\text{Sample points in A and B}\\ =\left\{E_1,E_2,E_3,E_4\right\}\end{array}$

For $A\cap B$:

$\begin{array}{c}A\cup B=\text{Common sample points in A and B}\\ =\left\{E_2,E_4\right\}\end{array}$

For $\overline{A}\cap \overline{B}$:

$\begin{array}{c}\overline{A}\cap \overline{B}=\text{Sample points neither in A nor in B}\\ =\varphi \end{array}$

For $\overline{A}\cup B$:

$\begin{array}{c}\overline{A}\cup B=\text{Sample points in }B\text{ and not in }B\\ =\left\{E_2,E_4\right\}\end{array}$