Chapter 2.4, Problem 19E

A business office orders paper supplies from one of three vendors, V₁, V₂, or V₃. Orders are to be placed on two successive days, one order per day. Thus, (V₂, V₃) might denote that vendor V₂ gets the order on the first day and vendor V₃ gets the order on the second day.

1. a List the sample points in this experiment of ordering paper on two successive days.
2. b Assume the vendors are selected at random each day and assign a probability to each sample point.
3. c Let A denote the event that the same vendor gets both orders and B the event that V₂ gets at least one order. Find P(A), P(B), P(A ∪ B), and P(A ∩ B) by summing the probabilities of the sample points in these events.

To determine

List out the sample points for the given experiment by ordering paper on two successive days.

Answer to Problem 19E

The sample points for the given experiment by ordering paper on two successive days is $S=\left\{\left(V_1{V}_1\right),\left(V_1{V}_2\right),\left(V_1{V}_3\right),\left(V_2{V}_1\right),\left(V_2{V}_2\right),\left(V_2{V}_3\right),\left(V_3{V}_1\right),\left(V_3{V}_2\right),\left(V_3{V}_3\right)\right\}$.

Explanation of Solution

Sample points:

When performing an experiment, it results with one or more outcomes. The possible outcomes in an experiment is called sample points. It is denoted by S.

In the given experiment, the possible outcomes by ordering paper on two successive days are $\left(V_1{V}_1\right),\left(V_1{V}_2\right),\left(V_1{V}_3\right),\left(V_2{V}_1\right),\left(V_2{V}_2\right),\left(V_2{V}_3\right),\left(V_3{V}_1\right),\left(V_3{V}_2\right),\left(V_3{V}_3\right)$

Therefore, the sample points for the given experiment by ordering paper on two successive days is $S=\left\{\left(V_1{V}_1\right),\left(V_1{V}_2\right),\left(V_1{V}_3\right),\left(V_2{V}_1\right),\left(V_2{V}_2\right),\left(V_2{V}_3\right),\left(V_3{V}_1\right),\left(V_3{V}_2\right),\left(V_3{V}_3\right)\right\}$.

To determine

Give a probability value to each of the sample points.

Answer to Problem 19E

The probability of each sample point is 0.1111.

Explanation of Solution

Probability:

For every event (say A) in sample space S, a number is called the probability of event A $\left(P\left(A\right)\right)$, if it follows the following axioms.

$\begin{array}{l}Axiom1:P\left(A\right)\ge 0\\ Axiom2:P\left(S\right)=1\\ Axiom3:\text{For a pairwise mutually exclusive events in }S,\left(\text{say }{A}_1,A_2,\dots \right),\\ P\left(A_1\cup A_2\cup A_3\cup \dots \right)={\displaystyle \sum _{i=1}^{\infty }P\left(A_i\right)}\end{array}$

It is assumed that the vendors are selected at random each day. Since it is random, the sample points are equally likely. From Part (a), it is clear that there are 9 sample points.

The probability of each sample point is calculated as follows:

$\begin{array}{c}P\left(\text{each}\text{sample}\text{point}\right)=\frac{1}{9}\\ =0.1111\end{array}$

Therefore, the probability of each sample point is 0.1111.

To determine

Compute the probability for event $A,B,A\cup B,\text{ and}A\cap B$.

Answer to Problem 19E

The required probabilities are obtained as given below:

$\begin{array}{l}P\left(A\right)=0.3333\\ P\left(B\right)=0.5556\\ P\left(A\cup B\right)=0.7778\\ P\left(A\cap B\right)=0.1111\end{array}$.

Explanation of Solution

It is clear from Part (a) that there are three favourable outcomes for event A and there are five favourable outcomes for event B. There are nine possible outcomes of this experiment.

The required probability for event A is computed as follows:

$\begin{array}{c}P\left(A\right)=\frac{3}{9}\\ =\frac{1}{3}\\ =0.3333\end{array}$

Therefore, the probability of event A is 0.3333.

The required probability for event B is computed as follows:

$\begin{array}{c}P\left(B\right)=\frac{5}{9}\\ =0.5556\end{array}$

Therefore, the probability of event B is 0.5556.

The required probability for $A\cup B$ is computed as follows:

$\begin{array}{c}A\cup B=\left\{\left(V_1{V}_1\right),\left(V_2{V}_2\right),\left(V_3{V}_3\right)\right\}\cup \left\{\left(V_1{V}_2\right),\left(V_2{V}_1\right),\left(V_2{V}_2\right),\left(V_2{V}_3\right),\left(V_3{V}_2\right)\right\}\\ =\left\{\left(V_1{V}_1\right),\left(V_1{V}_2\right),\left(V_2{V}_1\right),\left(V_2{V}_2\right),\left(V_2{V}_3\right),\left(V_3{V}_2\right),\left(V_3{V}_3\right)\right\}\\ P\left(A\cup B\right)=\frac{7}{9}\\ =0.7778\end{array}$

Therefore, the probability of $A\cup B$ is 0.7778.

The required probability for $A\cap B$ is computed as follows:

$\begin{array}{c}A\cap B=\left\{\left(V_1{V}_1\right),\left(V_2{V}_2\right),\left(V_3{V}_3\right)\right\}\cap \left\{\left(V_1{V}_2\right),\left(V_2{V}_1\right),\left(V_2{V}_2\right),\left(V_2{V}_3\right),\left(V_3{V}_2\right)\right\}\\ =\left\{\left(V_2{V}_2\right)\right\}\\ P\left(A\cap B\right)=\frac{1}{9}\\ =0.1111\end{array}$

Therefore, the probability of $A\cap B$ is 0.1111.