Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
Pearson eText for Probability & Statistics for Engineers and Scientists with R -- Instant Access (Pearson+)
1st Edition
ISBN: 9780137548552
Author: Michael Akritas
Publisher: PEARSON+
Question
Book Icon
Chapter 2.2, Problem 8E
To determine

Prove that (AB)(BA)=(AB)(BA) from Exercise 4.

Prove that (AB)c=AcBc from Exercise 6.

Prove that (AB)C=(AC)(BC) from Exercise 7.

Expert Solution & Answer
Check Mark

Explanation of Solution

Calculation:

Define the events A and B. The events A and B are considered as equal when AB and BA. The event A would be subset of B for an element x, if xA then xB is true.

Pair of events (AB)(BA)=(AB)(BA):

Let x be an element.

x(AB)(BA)x(AB)orx(BA)(xAbutxB)or(xBbutxA)xAorxBx(AB)andx(AB)x(AB)(AB)

The event (AB)(BA)(AB)(BA) because x(AB)(BA) then x(AB)(AB) is also true.

x(AB)(AB)x(AB)andx(AB)xAorxB(xAbutxB)or(xBbutxA)x(AB)orx(BA)x(AB)(BA)

The event (AB)(BA)(AB)(BA) because x(AB)(AB) then x(AB)(BA) is also true.

Hence, it is proved that (AB)(BA)=(AB)(BA).

Pair of events (AB)c=AcBc:

x(AB)cxABx(AB)orx(BA)orx(AB)c[x(AB)orx(AB)c]or[x(BA)orx(AB)c]xBcorxAcx(AcBc)

The event (AB)cAcBc because x(AB)c then x(AcBc) is also true.

x(AcBc)xBcorxAc[x(AB)orx(AB)c]or[x(BA)orx(AB)c]x(AB)orx(BA)orx(AB)cxABx(AB)c

The event AcBc(AB)c because x(AcBc) then x(AB)c is also true.

Hence, it is proved that (AB)c=AcBc.

Pair of events (AB)C=(AC)(BC):

x(AB)Cx(AB)orxC[xAandxB]orxC[xAorxC]and[xBorxC]xACandxBCx(AC)(BC)

The event (AB)C(AC)(BC) because x(AB)C then x(AC)(BC) is also true.

x(AC)(BC)xACandxBC[xAorxC]and[xBorxC][xAandxB]orxCx(AB)orxCx(AB)C

The event (AC)(BC)(AB)C because x(AC)(BC) then x(AB)C is also true.

Hence, it is proved that (AB)C=(AC)(BC).

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
13. If X has the distribution function F(x) = 0 1 12 for x < -1 for -1x < 1 for 1x <3 2 3 for 3≤x≤5 4 1 for x≥5 find (a) P(X ≤3); (b) P(X = 3); (c) P(X < 3); (d) P(X≥1); (e) P(-0.4
Please solve the following Statistics and Probability Problem (show all work) :    The probability that a patient recovers from a rare blood disease is 0.4 and 10 people are known to havecontracted this disease. Let X denote the random variable which denotes the number of patient who survivefrom the disease.1. Plot the probability mass function (pmf) of X.2. Plot the cumulative distribution function (cdf) of X.3. What is the probability that at least 8 survive, i.e., P {X ≥ 8}?4. What is the probability that 3 to 8 survive, i.e., P {3 ≤ X ≤ 8}?
Please solve the following Probability and Statistics problem (show all work and double check solution is correct):  Suppose that a die is rolled twice. What are the possible values that the following random variables can take1. the maximum value to appear in the two rolls;2. the value of the first roll minus the value of the second roll?3. Calculate the probabilities associated with the above two random variables?
Knowledge Booster
Background pattern image
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
A First Course in Probability (10th Edition)
Probability
ISBN:9780134753119
Author:Sheldon Ross
Publisher:PEARSON
Text book image
A First Course in Probability
Probability
ISBN:9780321794772
Author:Sheldon Ross
Publisher:PEARSON