##### Document Text Contents

Page 1

Handout 04 Probability Theory (1)

Muhammad Naeem, Assistant Professor Department of Mathematics

4. Elementary Probability

We live in a world in which we are unable to forecast the future with complete certainty. Our

need to cope with uncertainty leads us to the study and use of probability theory. In many instances, we,

as concerned citizen, will have some knowledge about the possible outcomes of a decision. By organizing

this information and considering it systematically, we will be able to recognize our assumptions,

communicate our reasoning to others, and make a sounder decision than we could by using a shot-in-the

dark approach.

Basic Terminology in Probability

In general, probability is the chance something will happen. Probabilities are expressed as

fractions ( 1/6, ½ , 8/9 ) or as decimals (0.167, 0.500, 0.889) between zero and one. Assigning a probability

zero means that something can never happen; a probability 1 indicates that something will always happen.

(1) A Random experiment

An experiment which produces different results even though it is repeated a large number of times under

essentially similar conditions is called a random experiment

(2) Sample Space and Events

The set S of all possible outcomes of a random experiment is called a Sample space. An element

in S, is called a sample point or a simple event.

(3) A Simple Event

A simple event is the most basic outcome of a random experiment.

(4) An Event

An event A is a collection of simple events. or, in other words, a subset of the Sample space S.

The empty set � is called the impossible event, and S, the sample space is called the certain or sure event.

(5) The Probability of an Event

The probability of event A is defined as

P(A) =

n(A)

n(S)

=

no. of favourable cases to A

Total possible cases

This is a classical approach.

We observe that

(i) 0 ≤ P(A) ≤ 1

(ii) P(�) = 0

(iii) P(S) = 1

(6) Probability Space

If the probability of each sample point in a sample space is known then it is called the probability

space, such that

(i) pi � 0

(ii) � pi = 1

(7) Complementary Events

The complementary of an event A is the event

�

A that occurs when A does not occur., i.e. the event

consisting of all simple events that are not in event A. We will denote the complement of A by Ac or

�

A.

Page 2

Handout 04 Probability Theory (2)

Muhammad Naeem, Assistant Professor Department of Mathematics

(8) Complementary Relationships

The sum of the probabilities of complementary events equals 1. That is

P(A)+P(A) = 1

Example (1)

A fair coin is tossed ten times and the up face is recorded after each toss. What is the probability

of event A: {observe at least one head}

Solution

Since we know the probability of the complement of A, we use the relation ship for the

complementary events:

P(A) = 1 – P(A) = 1 -

1

1024

=

1023

1024

= .999

That is, we are virtually certain of observing at least one head in 10 tosses of the coin.

Properties of Events

If A and B are two events then they may be

(i) Equally Likely

(ii) Exhaustive

(iii) Mutually Exclusive

(1) Equally Likely Events

Two events A and B are said to be equally likely, when one event is as likely to occur as the

other. In other words, each event should occur in equal number in repeated trials. For example when a fair

coin is tossed, the head is as likely to appear as the tail.

(2) Exhaustive Events

Events are said to be collectively exhaustive, when the union of mutually exclusive events is the

entire sample space S. A group of mutually exclusive and exhaustive events is called a partition of the

sample space. For instance, event A and Ac form a partition as they are mutually exclusive and their union

is the entire sample space.

(3) Mutually Exclusive Events

Two events A and B of a single experiment are said to be mutually exclusive or disjoint if and only if they

cannot both occur at the same time.

Example (2)

Give a collectively exhaustive list of the possible outcomes of tossing two dice. Also give the probability

for each of the following totals in the rolling of two dice: 1, 2, 5, 6, 7, 10 and 11.

Solution

An exhaustive list of all possible outcomes of tossing two dice: (dice 1, dice 2)

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

P(1) = 0/36 = 0, P(2) = 1/36, P(5) = 4/36, P(6) = 5/36, P(7) = 6/36, P(10) = 3/36, P(11) = 2/36

Page 6

Handout 04 Probability Theory (6)

Muhammad Naeem, Assistant Professor Department of Mathematics

P(B) = 0.40 P(Bc) = 0.60

P(A B) = 0.30

Now

Ac Bc = (A B)c denote the event that no car starts

Therefore 1- P( (A B)c ) denotes that at least one car starts

i.e. 1- P( (A B)c ) = 1 – P(Ac Bc) = 1 – {P(Ac) P(Bc) } = 1 – (0.20 0.60) = 1 – 0.12 = 0.88

alternatively we find the probability:

P(A Bc) + P(Ac B) + P(A B) = 0.80 0.60 + 0.20 0.40 + 0.80 0.40 = 0.88

Exercise 23.1

(1) What is the probability of obtaining at least one head in tossing six fair coins?

Sol. total possible cases = 26 = 64

Event A that at least one head is obtained

So Ac is the event of obtaining no head

As P(A) + P(Ac) = 1

P(A) = 1 – P(Ac)

Now Ac = {TTTTTT} that all six outcomes are tails

Therefore P(Ac) = 1/64

Hence P(A) = 1 – 1/64 = 63/64

(2) In rolling two fair dice. What is the probability of obtaining a sum greater than 10 or a sum

divisible by 6?

Sol. Two fair dice are rolled, n(S) = 36

Event A that sum is greater than 10 and event B is that sum divisible by 6

n(A) = 3, n(B) = 6 and n(A B) = 1

so P(A B) = P(A) + P(B) – P(A B)

= 3/36 + 6/36 – 1/36 = 8/36 = 2/9

(3) Three screws are drawn at random from a lot of 100 screws. 10 of which are defective. Find the

probability of the event that all 3 screws drawn are non defective, assuming that we draw (a) with

replacement (b) without replacement.

Sol. Defective = 10, good = 90 and total = 100

Experiment : 3 are drawn

(i) with replacement

required probability = 90/100 × 90/100 × 90/100 = (0.9)3 = 72.9%

(ii) without replacement

required probability = 90/100 × 89/99 × 88/98 = 72.65%

(4) Three boxes contain five chips each, numbered from 1 to 5 and one chip is drawn from each box.

Find the probability of the event E that the sum of the numbers on the drawn chips is greater than

4.

Sol. Box I = {1,2,3,4,5}, Box II = {1,2,3,4,5} and Box III = {1,2,3,4,5}

one number is drawn from each box, so total possible drawl = (5)3 = 125

event E that the sum of the numbers is greater than 4

so Ec = {(1,1,1), (1,1,2), (1,2,1), (2,1,1)}

therefore P(Ec) = 4/125

P(E) = 1 – 4/125 = 121/125

(5) A batch of 100 iron rods consists of 25 oversized rods, 25 undersized rods, and 50 rods of the

desired length. If two rods are drawn at random without replacement, what is the probability of

obtaining (a) two rods of the desired length (b) one of the desired length (c) none of the desired

length (d) two undersized rods.

Sol. oversized = 25, undersized = 25, desired length = 50, total = 100

Page 7

Handout 04 Probability Theory (7)

Muhammad Naeem, Assistant Professor Department of Mathematics

Experiment : two are drawn without replacement

(a) P(2 are of desired length) =

50

100

49

99

= 0.2474

This is the case of without replacement

(b) P(1 is of desired length) = 2

50

100

50

99

= 0.505

Actually these are the two either or cases (i) 1st of desired length and 2nd is not of desired

length (ii) 1st is not of desired length and 2nd of desired length

(c) P(no of desired length) = 1 -

50

100

49

99

= 1 - 0.2474 = 0.7526

(d) P(2 are undersized) =

50

100

49

99

= 0.2474

(7) If a certain kind of tire has a life exceeding 25000 miles with probability 0.95. What is the

probability that a set of these tires on a car will last longer than 25000 miles?

Sol. p = 0.95, i.e. the probability of a tire having life exceeding 25000 miles

P(a set of 4) = (0.95)( 0.95)( 0.95)( 0.95) = (0.95)4

(8) If a certain kind of tire has a life exceeding 25000 miles with probability 0.95. What is the

probability that at least one of the tires will not last for 25000 miles?

Sol. Probability that a tire has life exceeding 25000 miles = 0.95

Let A be the event that

P(no tire has a life exceeding 25000 miles) = (0.05)4

therefore P(at least one) = 1 (0.05)4

(11) If we inspect sheets of paper by drawing 3 sheets without replacement from every lot of 100

sheets, what is the probability of getting 3 clean sheets although 8% of the sheets contains

impurities?

Sol. p = 0.92, the probability of a sheet to contain purity

P(3 clean sheets) = (0.92)3

Independent Events

Events A and B are independent if the occurrence of B does not alter the probability that A has

occurred, i.e., events A and B are independent if

P(A/B) = P(A)

When events A and B are independent it will also be true that

P(B/A) = P(B)

Events that are not independent are said to be dependent.

Example (8)

Consider an experiment of tossing a fair coin twice and recording the up face on each toss. The

following events are defined:

A: {first toss is a head}

B: {second toss is a head}

Does knowing that event A has occurred affect the probability that B will occur?

Solution

Intuitively the answer should be no, since what occurs on the first toss should in no way affect

what occurs on the second toss. The sample space for this experiment is

S = {HH, TH, HT, TT} and A = { HH, HT}, B = { HH, TH}

Each of these simple events has a probability of ¼. Thus,

P(B) = P(HH) + P(TH) = ¼ + ¼ = ½

and P(A) = P(HH) + P(HT) = ¼ + ¼ = ½

Page 12

Handout 04 Probability Theory (12)

Muhammad Naeem, Assistant Professor Department of Mathematics

Find P(D�E), P(E/D). Check P(D/E) = P(D) and P(E/D) = P(E). Also check P(D�E)

For mutually exclusive, check P(D�E) = 0

Example (17)

The probability that Mr. Khan will get an offer on the first job he applied for is 0.5 and the

probability that he will get an offer on the second job he applied for is 0.6. He thinks that the probability

that he will get an offer on both jobs is 0.15.

(a) Define the events involved and use probability notation to show the probability

information.

(b) What is the probability that Mr Khan gets an offer on the second job given that he

receives an offer for the first job.

(c) What is the probability that Mr Khan gets an offer on at least one of the jobs he applied

for.

(d) What is the probability that Mr Khan gets an offer only of the job he applied for?

(e) Are the jobs offers independent? Explain.

Exercise

(1) A ball is drawn at random from a box containing 6 red, 4 white and 5 blue balls.

(i) Determine the probability that the ball drawn is (a) red, (b) white, (c) blue, (d) not red and

(e) red or white.

(ii) Find the probability that they are drawn in the order red, white and blue if each ball is (a)

replaced (ii) not replaced.

(2) A fair die is rolled twice. Find the probability of getting a 4, 5 or 6 on the first rolling or and a 1,

2, 3 or 4 on the second rolling. (solve with both methods directly and using multiplication law)

(3) Consider the experiment of tossing a fair die and define the following events

A = {observe an even number}

B = {observe a number less than or equal to 4}

Are events A and B are independent?

(4) A box contains 15 items, 4 of which are defective and 11 are good. Two items are selected. What

is the probability that the first is good and second is defective?

(5) An urn contains 10 white and 3 black balls. Another contains 3 white and 5 black balls. Two balls

are transferred from first urn and placed into second and then one ball is taken from the latter.

What is the probability that it is a white ball?

(6) Examine whether each of the following statements is correct. Explain.

a. P(A) = 0.40, P(A�B) = 0.25

b. P(A) = 0.40, P(A�B) = 0.60, so that A and B are independent.

c. P(A) = 0.20, P(B) = 0.60, P(A�B) = 0.12, the events A and B are independent.

d. If P(A) = 0.80, P(B) = 0.60, the events A and B cannot be mutually exclusive.

Assignment (1)

In a study involving manufacturing process the probability is 0.10 that a part is tested turns out to

be defective and the probability that a part was produced on machine A is 0.30. Given that a part was

produced on machine A there is a 0.15 probability that it is defective.

(f) What is the probability that a part tested is both defective and produced by machine A?

(g) If a part is found defective, what is the probability that it came from machine A?

(h) Is finding a defective part independent of its being produced on machine A?

(i) What is the probability of the part being defective or produced by machine A?

(j) Are the events ‘a defective part’ and ‘produced by machine A’ mutually exclusive

events? Explain.

Handout 04 Probability Theory (1)

Muhammad Naeem, Assistant Professor Department of Mathematics

4. Elementary Probability

We live in a world in which we are unable to forecast the future with complete certainty. Our

need to cope with uncertainty leads us to the study and use of probability theory. In many instances, we,

as concerned citizen, will have some knowledge about the possible outcomes of a decision. By organizing

this information and considering it systematically, we will be able to recognize our assumptions,

communicate our reasoning to others, and make a sounder decision than we could by using a shot-in-the

dark approach.

Basic Terminology in Probability

In general, probability is the chance something will happen. Probabilities are expressed as

fractions ( 1/6, ½ , 8/9 ) or as decimals (0.167, 0.500, 0.889) between zero and one. Assigning a probability

zero means that something can never happen; a probability 1 indicates that something will always happen.

(1) A Random experiment

An experiment which produces different results even though it is repeated a large number of times under

essentially similar conditions is called a random experiment

(2) Sample Space and Events

The set S of all possible outcomes of a random experiment is called a Sample space. An element

in S, is called a sample point or a simple event.

(3) A Simple Event

A simple event is the most basic outcome of a random experiment.

(4) An Event

An event A is a collection of simple events. or, in other words, a subset of the Sample space S.

The empty set � is called the impossible event, and S, the sample space is called the certain or sure event.

(5) The Probability of an Event

The probability of event A is defined as

P(A) =

n(A)

n(S)

=

no. of favourable cases to A

Total possible cases

This is a classical approach.

We observe that

(i) 0 ≤ P(A) ≤ 1

(ii) P(�) = 0

(iii) P(S) = 1

(6) Probability Space

If the probability of each sample point in a sample space is known then it is called the probability

space, such that

(i) pi � 0

(ii) � pi = 1

(7) Complementary Events

The complementary of an event A is the event

�

A that occurs when A does not occur., i.e. the event

consisting of all simple events that are not in event A. We will denote the complement of A by Ac or

�

A.

Page 2

Handout 04 Probability Theory (2)

Muhammad Naeem, Assistant Professor Department of Mathematics

(8) Complementary Relationships

The sum of the probabilities of complementary events equals 1. That is

P(A)+P(A) = 1

Example (1)

A fair coin is tossed ten times and the up face is recorded after each toss. What is the probability

of event A: {observe at least one head}

Solution

Since we know the probability of the complement of A, we use the relation ship for the

complementary events:

P(A) = 1 – P(A) = 1 -

1

1024

=

1023

1024

= .999

That is, we are virtually certain of observing at least one head in 10 tosses of the coin.

Properties of Events

If A and B are two events then they may be

(i) Equally Likely

(ii) Exhaustive

(iii) Mutually Exclusive

(1) Equally Likely Events

Two events A and B are said to be equally likely, when one event is as likely to occur as the

other. In other words, each event should occur in equal number in repeated trials. For example when a fair

coin is tossed, the head is as likely to appear as the tail.

(2) Exhaustive Events

Events are said to be collectively exhaustive, when the union of mutually exclusive events is the

entire sample space S. A group of mutually exclusive and exhaustive events is called a partition of the

sample space. For instance, event A and Ac form a partition as they are mutually exclusive and their union

is the entire sample space.

(3) Mutually Exclusive Events

Two events A and B of a single experiment are said to be mutually exclusive or disjoint if and only if they

cannot both occur at the same time.

Example (2)

Give a collectively exhaustive list of the possible outcomes of tossing two dice. Also give the probability

for each of the following totals in the rolling of two dice: 1, 2, 5, 6, 7, 10 and 11.

Solution

An exhaustive list of all possible outcomes of tossing two dice: (dice 1, dice 2)

S = { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6),

(2,1), (2,2), (2,3), (2,4), (2,5), (2,6),

(3,1), (3,2), (3,3), (3,4), (3,5), (3,6),

(4,1), (4,2), (4,3), (4,4), (4,5), (4,6),

(5,1), (5,2), (5,3), (5,4), (5,5), (5,6),

(6,1), (6,2), (6,3), (6,4), (6,5), (6,6) }

P(1) = 0/36 = 0, P(2) = 1/36, P(5) = 4/36, P(6) = 5/36, P(7) = 6/36, P(10) = 3/36, P(11) = 2/36

Page 6

Handout 04 Probability Theory (6)

Muhammad Naeem, Assistant Professor Department of Mathematics

P(B) = 0.40 P(Bc) = 0.60

P(A B) = 0.30

Now

Ac Bc = (A B)c denote the event that no car starts

Therefore 1- P( (A B)c ) denotes that at least one car starts

i.e. 1- P( (A B)c ) = 1 – P(Ac Bc) = 1 – {P(Ac) P(Bc) } = 1 – (0.20 0.60) = 1 – 0.12 = 0.88

alternatively we find the probability:

P(A Bc) + P(Ac B) + P(A B) = 0.80 0.60 + 0.20 0.40 + 0.80 0.40 = 0.88

Exercise 23.1

(1) What is the probability of obtaining at least one head in tossing six fair coins?

Sol. total possible cases = 26 = 64

Event A that at least one head is obtained

So Ac is the event of obtaining no head

As P(A) + P(Ac) = 1

P(A) = 1 – P(Ac)

Now Ac = {TTTTTT} that all six outcomes are tails

Therefore P(Ac) = 1/64

Hence P(A) = 1 – 1/64 = 63/64

(2) In rolling two fair dice. What is the probability of obtaining a sum greater than 10 or a sum

divisible by 6?

Sol. Two fair dice are rolled, n(S) = 36

Event A that sum is greater than 10 and event B is that sum divisible by 6

n(A) = 3, n(B) = 6 and n(A B) = 1

so P(A B) = P(A) + P(B) – P(A B)

= 3/36 + 6/36 – 1/36 = 8/36 = 2/9

(3) Three screws are drawn at random from a lot of 100 screws. 10 of which are defective. Find the

probability of the event that all 3 screws drawn are non defective, assuming that we draw (a) with

replacement (b) without replacement.

Sol. Defective = 10, good = 90 and total = 100

Experiment : 3 are drawn

(i) with replacement

required probability = 90/100 × 90/100 × 90/100 = (0.9)3 = 72.9%

(ii) without replacement

required probability = 90/100 × 89/99 × 88/98 = 72.65%

(4) Three boxes contain five chips each, numbered from 1 to 5 and one chip is drawn from each box.

Find the probability of the event E that the sum of the numbers on the drawn chips is greater than

4.

Sol. Box I = {1,2,3,4,5}, Box II = {1,2,3,4,5} and Box III = {1,2,3,4,5}

one number is drawn from each box, so total possible drawl = (5)3 = 125

event E that the sum of the numbers is greater than 4

so Ec = {(1,1,1), (1,1,2), (1,2,1), (2,1,1)}

therefore P(Ec) = 4/125

P(E) = 1 – 4/125 = 121/125

(5) A batch of 100 iron rods consists of 25 oversized rods, 25 undersized rods, and 50 rods of the

desired length. If two rods are drawn at random without replacement, what is the probability of

obtaining (a) two rods of the desired length (b) one of the desired length (c) none of the desired

length (d) two undersized rods.

Sol. oversized = 25, undersized = 25, desired length = 50, total = 100

Page 7

Handout 04 Probability Theory (7)

Muhammad Naeem, Assistant Professor Department of Mathematics

Experiment : two are drawn without replacement

(a) P(2 are of desired length) =

50

100

49

99

= 0.2474

This is the case of without replacement

(b) P(1 is of desired length) = 2

50

100

50

99

= 0.505

Actually these are the two either or cases (i) 1st of desired length and 2nd is not of desired

length (ii) 1st is not of desired length and 2nd of desired length

(c) P(no of desired length) = 1 -

50

100

49

99

= 1 - 0.2474 = 0.7526

(d) P(2 are undersized) =

50

100

49

99

= 0.2474

(7) If a certain kind of tire has a life exceeding 25000 miles with probability 0.95. What is the

probability that a set of these tires on a car will last longer than 25000 miles?

Sol. p = 0.95, i.e. the probability of a tire having life exceeding 25000 miles

P(a set of 4) = (0.95)( 0.95)( 0.95)( 0.95) = (0.95)4

(8) If a certain kind of tire has a life exceeding 25000 miles with probability 0.95. What is the

probability that at least one of the tires will not last for 25000 miles?

Sol. Probability that a tire has life exceeding 25000 miles = 0.95

Let A be the event that

P(no tire has a life exceeding 25000 miles) = (0.05)4

therefore P(at least one) = 1 (0.05)4

(11) If we inspect sheets of paper by drawing 3 sheets without replacement from every lot of 100

sheets, what is the probability of getting 3 clean sheets although 8% of the sheets contains

impurities?

Sol. p = 0.92, the probability of a sheet to contain purity

P(3 clean sheets) = (0.92)3

Independent Events

Events A and B are independent if the occurrence of B does not alter the probability that A has

occurred, i.e., events A and B are independent if

P(A/B) = P(A)

When events A and B are independent it will also be true that

P(B/A) = P(B)

Events that are not independent are said to be dependent.

Example (8)

Consider an experiment of tossing a fair coin twice and recording the up face on each toss. The

following events are defined:

A: {first toss is a head}

B: {second toss is a head}

Does knowing that event A has occurred affect the probability that B will occur?

Solution

Intuitively the answer should be no, since what occurs on the first toss should in no way affect

what occurs on the second toss. The sample space for this experiment is

S = {HH, TH, HT, TT} and A = { HH, HT}, B = { HH, TH}

Each of these simple events has a probability of ¼. Thus,

P(B) = P(HH) + P(TH) = ¼ + ¼ = ½

and P(A) = P(HH) + P(HT) = ¼ + ¼ = ½

Page 12

Handout 04 Probability Theory (12)

Muhammad Naeem, Assistant Professor Department of Mathematics

Find P(D�E), P(E/D). Check P(D/E) = P(D) and P(E/D) = P(E). Also check P(D�E)

For mutually exclusive, check P(D�E) = 0

Example (17)

The probability that Mr. Khan will get an offer on the first job he applied for is 0.5 and the

probability that he will get an offer on the second job he applied for is 0.6. He thinks that the probability

that he will get an offer on both jobs is 0.15.

(a) Define the events involved and use probability notation to show the probability

information.

(b) What is the probability that Mr Khan gets an offer on the second job given that he

receives an offer for the first job.

(c) What is the probability that Mr Khan gets an offer on at least one of the jobs he applied

for.

(d) What is the probability that Mr Khan gets an offer only of the job he applied for?

(e) Are the jobs offers independent? Explain.

Exercise

(1) A ball is drawn at random from a box containing 6 red, 4 white and 5 blue balls.

(i) Determine the probability that the ball drawn is (a) red, (b) white, (c) blue, (d) not red and

(e) red or white.

(ii) Find the probability that they are drawn in the order red, white and blue if each ball is (a)

replaced (ii) not replaced.

(2) A fair die is rolled twice. Find the probability of getting a 4, 5 or 6 on the first rolling or and a 1,

2, 3 or 4 on the second rolling. (solve with both methods directly and using multiplication law)

(3) Consider the experiment of tossing a fair die and define the following events

A = {observe an even number}

B = {observe a number less than or equal to 4}

Are events A and B are independent?

(4) A box contains 15 items, 4 of which are defective and 11 are good. Two items are selected. What

is the probability that the first is good and second is defective?

(5) An urn contains 10 white and 3 black balls. Another contains 3 white and 5 black balls. Two balls

are transferred from first urn and placed into second and then one ball is taken from the latter.

What is the probability that it is a white ball?

(6) Examine whether each of the following statements is correct. Explain.

a. P(A) = 0.40, P(A�B) = 0.25

b. P(A) = 0.40, P(A�B) = 0.60, so that A and B are independent.

c. P(A) = 0.20, P(B) = 0.60, P(A�B) = 0.12, the events A and B are independent.

d. If P(A) = 0.80, P(B) = 0.60, the events A and B cannot be mutually exclusive.

Assignment (1)

In a study involving manufacturing process the probability is 0.10 that a part is tested turns out to

be defective and the probability that a part was produced on machine A is 0.30. Given that a part was

produced on machine A there is a 0.15 probability that it is defective.

(f) What is the probability that a part tested is both defective and produced by machine A?

(g) If a part is found defective, what is the probability that it came from machine A?

(h) Is finding a defective part independent of its being produced on machine A?

(i) What is the probability of the part being defective or produced by machine A?

(j) Are the events ‘a defective part’ and ‘produced by machine A’ mutually exclusive

events? Explain.