Chapter 2.2, Problem 13E

A computer consulting firm presently has bids out on three projects. Let A_i = {awarded project i}, for i = 1, 2, 3, and suppose that P(A₁) = .22, P(A₂) = .25, P(A₃) = .28, P(A₁ ∩ A₂) = .11, P(A₁ ∩ A₃) = .05, P(A₂ ∩ A₃) = .07, P(A₁ ∩ A₂ ∩ A₃) = .01. Express in words each of the following events, and compute the probability of each event:

1. a. A₁ ∪ A₂
2. b. A₁′∩ A₂′ [Hint: (A₁ ∪ A₂)′ = A′₁ ∩ A′₂]
3. c. A₁ ∪ A₂ ∪ A₃
4. d. A′₁ ∩ A′₂ ∩ A′₃
5. e. A′₁ ∩ A′₂ ∩ A₃
6. f. (A′₁ ∩ A′₂ ) ∪ A₃

To determine

Compute $P\left(A_1\cup A_2\right)$.

Answer to Problem 13E

The probability of an event $\left(A_1\cup A_2\right)$ is 0.36.

Explanation of Solution

Given info:

The data represents the projects of the consulting firm.

Here, A₁  be awarded project 1,

A₂  be awarded project 2,

A₃  be awarded project .

Let $P\left(A_1\right)=0.22$, $P\left(A_2\right)=0.25$, $P\left(A_3\right)=0.28$, $P\left(A_1\cap A_2\right)=0.11$, $P\left(A_1\cap A_3\right)=0.05$, $P\left(A_2\cap A_3\right)=0.07$ and $P\left(A_1\cap A_2\cap A_3\right)=0.01$.

Calculation:

Addition rule:

For any two events A and B,

$P\left(A\cup B\right)=P\left(A\right)+P\left(B\right)-P\left(A\cap B\right)$

Complement:

The complement of the event A contains the set of all element that are contained in sample space S and not contained in event A. it is denoted as $A\text{'}$

Union:

The union of two events A₁  or A₂ contains set of all elements which are either in A₁,  A₂, or both the events.  It is denoted by $A_1\cup A_2$.

Intersection:

The intersection of two event A₁ and A₂ contains set all of element which are in both A₁ and A₂. It is denoted by $A_1\cap A_2$.

The probability of A₁ or A₂ can be obtained as

$\begin{array}{c}P\left(A_1\cup A_2\right)=P\left(A_1\right)+P\left(A_2\right)-P\left(A_1\cap A_2\right)\\ =0.22+0.25-0.11\\ =0.36\end{array}$

Thus, the probability of an event $\left(A_1\cup A_2\right)$ is 0.36.

To determine

Compute $P\left(A{\text{'}}_1\cap A{\text{'}}_2\right)$.

Answer to Problem 13E

The probability of an event $\left(A{\text{'}}_1\cap A{\text{'}}_2\right)$ is 0.64.

Explanation of Solution

Calculation:

The probability of $A{\text{'}}_1$ and $A{\text{'}}_2$ can be obtained as

$\begin{array}{c}P\left(\text{not in }{A}_1\text{ and not in }{A}_2\right)=P\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\\ P\left(A{\text{'}}_1\cap A{\text{'}}_2\right)=P\left(\left(A_1\cup A_2\right)\text{'}\right)\end{array}$

The event $\left(A{\text{'}}_1\cap A{\text{'}}_2\right)$ represents that is neither in A_1­ nor in A₂. It means that company awarded neither project 1 nor project 2.

$\begin{array}{c}P\left(\left(A_1\cup A_2\right)\text{'}\right)=1-P\left(A_1\cup A_2\right)\\ =1-0.36\\ =0.64\end{array}$

Thus, the probability of an event $\left(A{\text{'}}_1\cap A{\text{'}}_2\right)$ is 0.64.

To determine

Compute $P\left(A_1\cup A_2\cup A_3\right)$.

Answer to Problem 13E

The probability of an event $\left(A_1\cup A_2\cup A_3\right)$ is 0.53.

Explanation of Solution

Calculation:

Addition rule:

For any three events A, B and C ,

$P\left(A\cup B\cup C\right)=P\left(A\right)+P\left(B\right)+P\left(C\right)-P\left(A\cap B\right)-P\left(A\cap C\right)-P\left(B\cap C\right)+P\left(A\cap B\cap C\right)$

The event $\left(A_1\cup A_2\cup A_3\right)$ represents that is either in A_1­ or in A₂ or in A₃. It means that company at least awarded the project 1 or project 2 or project 3.

The probability of $A_1$ or $A_2$ or $A_3$ can be obtained as,

$\begin{array}{c}P\left(A_1\cup A_2\cup A_3\right)=P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right)-P\left(A_1\cap A_2\right)-P\left(A_1\cap A_3\right)-P\left(A_2\cap A_3\right)+P\left(A_1\cap A_2\cap A_3\right)\\ =0.22+0.25+0.28-0.11-0.05-0.07+0.01\\ =0.52+0.01\\ =0.53\end{array}$

Thus, the probability of an event $\left(A_1\cup A_2\cup A_3\right)$ is 0.53.

To determine

Compute $P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)$.

Answer to Problem 13E

The probability of an event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)$ is 0.47.

Explanation of Solution

Calculation:

The event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)$ represents that is neither A_1­ nor A₂ nor A₃. It means that company awarded the none of the projects.

The probability of $A{\text{'}}_1$ and $A{\text{'}}_2$ and $A{\text{'}}_3$ can be obtained as

$\begin{array}{c}P\left(\text{not in }{A}_1{\text{ and not in A}}_2\text{and not in }{A}_3\right)=P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)\\ P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)=P\left(\left(A_1\cup A_2\cup A_3\right)\text{'}\right)\end{array}$

$\begin{array}{c}P\left(\left(A_1\cup A_2\cup A_3\right)\text{'}\right)=1-P\left(A_1\cup A_2\cup A_3\right)\\ =1-0.53\\ =0.47\end{array}$

Thus, the probability of an event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)$ is 0.47.

To determine

Compute $P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)$.

Answer to Problem 13E

The probability of an event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)$ is 0.17.

Explanation of Solution

Calculation:

The event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)$ represents that is  either A₃ and neither A_1­ nor A₂. It means that company awarded project 3 and not awarded project 1 and project 2.

The event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)$ is represented using Venn diagram:

$\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)$

From the Venn diagram, the probability of $A{\text{'}}_1$ and $A{\text{'}}_2$ and $A_3$ can be obtained as

$\begin{array}{c}P\left(\text{not in }{A}_1{\text{ and not in A}}_2\text{and }{A}_3\right)=P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)\\ P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)=P\left(A_3\right)-P\left(A_1\cap A_3\right)-P\left(A_2\cap A_3\right)+P\left(A_1\cap A_2\cap A_3\right)\end{array}$

$\begin{array}{c}P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)=P\left(A_3\right)-P\left(A_1\cap A_3\right)-P\left(A_2\cap A_3\right)+P\left(A_1\cap A_2\cap A_3\right)\\ =0.28-0.05-0.07+0.01\\ =0.17\end{array}$

Thus, the probability of an event $\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A_3\right)$ is 0.17.

To determine

Compute $P\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)$.

Answer to Problem 13E

The probability of an event $\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)$ is 0.75.

Explanation of Solution

Calculation:

The event $\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)$ represents that is either and neither of A₁ ,A₂, A₃. It means that company awarded neither of projects.

The event $\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)$ is represented using Venn diagram:

$\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)$

The probability of $A{\text{'}}_1$ and $A{\text{'}}_2$ or $A_3$ can be obtained as

$\begin{array}{c}P\left(\text{not in }\left(A_1\text{ and }{A}_2\right)\text{or }{A}_3\right)=P\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)\\ P\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)=P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)+P\left(A_3\right)\end{array}$

$\begin{array}{c}P\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)=P\left(A{\text{'}}_1\cap A{\text{'}}_2\cap A{\text{'}}_3\right)+P\left(A_3\right)\\ =0.47+0.28\\ =0.75\end{array}$

Thus, the probability of an event $\left(\left(A{\text{'}}_1\cap A{\text{'}}_2\right)\cup A_3\right)$ is 0.75.