# Chapter 8: Probability – Biostatistics

## Probability

#### Objectives

After completing this chapter, you can understand the following:

• The definition, meaning and significance of probability.
• The concept of addition theorem and multiplication theorem on probability and its applications.
• Baye’s theorem and its implications towards biological decision making situations.
##### 8.1 INTRODUCTION

The concept of probability was introduced in the late seventeenth century. This concept was introduced in problems relating to the coin-tossing game and playing cards. But the probability concept is now introduced in almost all areas of study such as economics, statistics, industry, engineering, business and biology. It refers the study of events which are going to happen or not.

Before defining the term probability, let us see some of the basic terms which are going to be used in the definition of probability.

#### Experiment

It refers an activity or measurement that results in an outcome.

Example: Tossing a single coin for 50 times.

#### Sample Space

It refers the collection of all possible events of an experiment and denoted by S.

Example: In a coin-tossing experiment, the sample space should contain the possible outcomes of a head/tail. S = [H, T ]

#### Event

It refers one or more of the possible outcomes of an experiment, a subset of a sample space.

Example: 1

In throwing a dice, S = [1,2,3,4,5,6] contains the face 1 is an event.

#### Equally Likely Events

In a sample space containing with at least two events, the chance of the occurrence of each of the event is same.

Example: In a coin-tossing experiment, having a head or tail in a trial is equal to ½ each.

#### Mutually Exclusive Events

Events are said to be mutually exclusive, if the outcome is only one element at a time. There is no chance that two or more events to happen at a time. Alternatively, it is called incompatible events.

Example: In a coin-tossing experiment, we can have either head or tail as an outcome. Clearly the occurrence of head prevents the occurrence of tail which implies that the two events are said to be mutually exclusive.

#### Outcome

The result of a random experiment is called an outcome.

Example: In coin-tossing, the two outcomes are head and tail.

##### 8.3 MEANING OF PROBABILITY

The term probability can be defined in two approaches. They are

1. The classical approach
2. The relative frequency approach

#### The Classical Approach

This approach describes the term probability as the proportion of times in event can be theoretically expected to happen.

Example: 2

Find the probability of having the face –1 in throwing a dice.

Selection of the face –1: It is one of the outcome of six possible outcomes [equally likely events], i.e., 1/6.

#### The Relative Frequency Approach

In this approach, probability is the proportion of times an event is observed to happen in a large number of trials.

#### Notation

The probability of an event A is denoted by P[A]. The value of P[A] should be in the range 0 P[A]1.

If the event A’ be the negation of the event A, then its probability can be defined as P[A’]. Clearly the range of P[A’] is 0 P[A’]1.

This implies that P[A] + P[A’] = 1. Also P[A] = 1 – P[A’] and P[A’] = 1 – P[A].

Note:

1. If P[A] = 1, then the event A is said to be a sure event.
2. If P[A] = 0, then the event A is said to be a null event.

Example: 3

If a coin is tossed, what is the chance of a head?

The sample space can be defined as, S = [H,T]; n[S] = 2.

Let A be the event that refers head, then A = [H]: n[A] = 1.

The probability of having head, P[A] = n[A]/n[S].

Here, n[A]: number of elements in the set A, and n[S]: number of elements in the set S. Then, P[A] = ½= 0.5.

Example: 4

Three fair coins are tossed once. Find the probability of [1] at least one tail, [2] exactly one head, [3] exactly two tails, [4] exactly three heads and [5] at least two tails.

The sample space can be defined as,

S = [HHH, HHT, HTH, HTT, THH, THT, TTH, TTT]; n[S] = 8.
1. Let A refers the event of at least one tail, then

A = [ΗΗΤ,ΗΤΗ,ΗΤΤ,ΤΗΗ,ΤΗΤ,ΤΤΗ,ΤΤΤ]; n[A] = 7;

Then, P[A] = n[A]/n[S] = 7/8.

2. Let B refers the event of exactly one head,

Β = [ΗΤΤ,ΤΗΤ,ΤΤΗ]; n[B] = 3;

Then, P[B] = n[B]/n[S] = 3/8.

3. Let C refers the event of exactly two tails,

C= [ΗΤΤ,ΤΗΤ,ΤΤΗ]; n[C] = 3;

Then, P[C] = n[C]/n[S] = 3/8.

4. Let D refers the event of exactly three heads,

D = [HHH]; n[D] = 1;

Then, P[D] = n[D]/n[S] = 1/8.

5. Let Ε refers the event of at least two tails,

Ε =[HTT,TTH,THT,TTT]; n[E] = 4;

Then, P[E] = n[E]/n[S] = 4/8 = ½

Example: 5

If a dice is tossed, what is the probability that the number appearing on top is [1] odd number, [2] less than 3 and [3] an even number less than 5.

The sample space can be defined as S = [1,2,3,4,5,6]; n[S] = 6.

1. Let A be the event of having odd numbers, A = [1,3,5];

n[A] = 3;

Then, P[A] = n[A]/n[S] = 3/6 = 1/2 = 0.5.

2. Let Β be the event of having the number less than 3, Β = [1,2]; n[B] = 2;

Then, P[B] = n[B]/n[S] = 2/6 = 1/3 = 0.333.

3. Let C be the event of having an even number less than 5, C = [2,4]; n[C] = 2;

Then, P[C] = n[C]/n[S] = 2/6 = 1/3.

Example: 6

What is the probability of setting 2 red balls in a draw of 2 balls from a box containing 4 white and 3 red balls?

Given,

Box contains:

 3 Red balls 4 White balls

Number of red balls = 3

Number of white balls = 4

Total number of balls = 7

Number of balls to be selected = 2.

Total number of ways of selecting

2 red balls out of 7 balls = 7C2 = [7 * 6]/[1 * 2] = 21.

Number of favourable chances of selecting 2 red

balls out of 3 red balls = 3C2 = [3 * 2]/[1 * 2] = 3.

P[selecting 2 red balls in 2 draws] = 3/21 = 1/7 = 0.143.

Example: 7

What is the chance that a leap year selected at random will contain 53 Mondays?

Number of weeks in a year = 52

Number of days = 52 * 7 = 364

Number of days in a leap year = 366

Difference in days between the leap

year and the normal year = 366 – 364 = 2

Clearly, we have 2 excess days.

The sample space of the 2 excess days can be given as

S = [[Sun, Mon], [Mon, Tue], [Tue, Wed], [Wed, Thr], [Thr, Fri], [Fri, Sat], [Sat, Sun]]

n[S ] = 7.

To get 53 Mondays, we have to look for the excess of one more Monday [53 – 52 = 1].

Let A be event of the occurrence of 53rd Monday.

Then, A = [[Sun, Mon], [Mon, Tue]]; n[A] = 2.

P[having 53 Mondays in a leap year] = n[A]/n[S ] = 2/7 = 0.286.

Example: 8

From a pack of 52 cards, one card is drawn at random. Find the chance of drawing a heart and a chance of not drawing a heart.

Total number of cards in a pack = 52

Number of cards to be selected = 1

Total chance of selecting one card out of 52 cards = 52C1 = 52

Number of cards having heart symbol = 13

Number of hearts to be selected = 1.

Total number of favourable chances = 13C1 = 13

Let A be the event of selection of one heart, then P[A] = 13/52 = 1/4 = 0.25.

We know that P[A] + P[A’] = 1.

P[A’] = 1 – P[A] = 1 – [1/4] = 0.75.

The chance of not drawing a heart is 0.75.

Example: 9

In a certain college, 55% of the students are women. Suppose we take a sample of two students. Use a probability tree to find the probability

1. that both the chosen students are women.
2. that at least one of the two students is a woman.

Let A and B are the events stand for the selected student being a male and female, respectively.

Given, P[B] = 0.55; then, P[A] = 1 – P[B] = 0.45; n = 100.

This implies that

No. of female No. of male Total
55
45
100

Total number of women students = 55

Number of women students selected = 2

Number of favourable cases = 55C2

Total number students = 100

Number of students selected = 2

Total number of cases = 100C2

Example: 10

In the United States, 44% of the population has type Ο blood, 42% are type A, 10% are type Β and 4% are type AB. Consider choosing someone at random and determining the person’s blood type. [The probability of a given blood type will correspond to the population percentage.]

The following table gives the information out of each 100 persons of the United States.

Let Ei [i = 1, 2, 3, and 4] stand for the event of selected persons to have the blood type Ο, Α, Β and AB respectively.

Example: 11

In a study of the relationship between health risk and income, a large group of people living in Massachusetts were asked a series of questions. Some of the results are shown in the following table.

1. What is the probability that someone in this study smokes?
2. What is the probability that someone in this study do not smoke and earning medium salary?

Let A be the event referring the selected person smokes and Β be the event referring the selected person do not smoke and earning medium salary.

Number of smoking persons =1,213;

Total number of persons selected for this study = 6,549;

Number of persons do not smoke and earning medium salary = 1,622;

Example: 12

In a certain population of the European starling, there are 5,000 nests with young. The distribution of brood size [number of young in a nest] is given in the accompanying table.

Brood size Frequency [no. of broods]
1
90
2
230
3
610
4
1,400
5
1,760
6
750
7
130
8
26
9
3
10
1
Total
5,000

Find [a] P[X = 3] [b] P[x > = 7] and [c] P[4 < = x < = 6]

• P[X = 3] = 610/5,000 = 0.122,
• P[x > = 7] = P[7] + P[8] + P[9] + P[10] = [130 + 26 + 3 + l]/5,000 = 0.032 and
• P[4 < = x < = 6] = P[4] + P[5] + P[6] = [1,400 + 1,760 + 750]/5,000 = 0.782.

#### 8.3.1 Addition Rules for Probability

There are situations where we wish to evaluate the probability that two or more of several events will occur in an experiment. The evaluation of such probabilities seeks the help of addition rules.

Events are not mutually exclusive.

When events are not mutually exclusive, two or more of them can happen at the same time. For this case, let us derive the condition based on two events.

#### 8.3.2 Addition Theorem on Probability

Result: 1

If A and Β be any two events, then the probability that at least one of the two events A and Β occurs can be denoted by P[AB] and the same can be defined as

P[AB] = P[A] + P[B] – P[AB].

Let S be the sample space, and A and Β be the two events of S.

Then, by definition,

We know that,

Dividing by n[S] on both sides of [2], we have

Note:

[3] can be generalized for any number of events.

Result: 2

Let us extend the result of Result 1 for any three events A, Β and C. Find P[ABC].

Let BC = D, then we have P[ABC] = P[AD]

The Results 1 and 2 can be deduced further based on certain conditions on the events.

Condition: 1

A, B and C are three mutually exclusive events. When the events are mutually exclusive, then only one event can occur at a time.

There is no chance for the occurrence of two or three events together.

The same thing can be expressed as follows:

1. P[AB] = 0
2. P[BC ] = 0
3. P[CA] = 0
4. P[ABC ] = 0

Hence, the Results 1 and 2 can be reduced as follows:

Condition: 2

The events A, B and C are three independent events.

When the events are independent, then we have,

1. P[AB] = P[A] * P[B]
2. P[CB] = P[C] * P[B]
3. P[AC] = P[A] * P[C]
4. P[ABC] = P[A] * P[B] * P[C]

Hence, the Results 1 and 2 can be reduced as follows:

Example: 13

In a study of the relationship between health risk and income, a large group of people living in Massachusetts were asked a series of questions. Some of the results are shown in the following table.

What is the probability that someone in this study either is smoking or has low income [or both]?

Let A be the event referring the selected person smokes and Β be the event referring the selected person having low income.

To find P[A + B].

By definition, P[AB] = P[A] + P[B] – P[AB].

Then, P[AB] = 0.1852 + 0.3787 – 0.0968 = 0.4671.

Example: 14

A fair dice is thrown. What is the chance that either an even number or a number greater than 3 will turn up?

The sample space S can be defined as S = [1,2,3,4,5,6]; n[S] = 6.

Let A be the event of having an even number, then A = [2,4,6]; n[A] = 3.

Let B be the event of having a number which is more than 3, then B = [4,5,6]; n[B] = 3.

To find P[AB]

The probability of either an even number or a number greater than 3 will turn up is 0.667.

Example: 15

The probability that a contractor will not get a plumbing contract is 1/3 and the probability that he will get an electric contract is 4/9. If the probability of setting at least one contract is 4/5, what is the probability that he will get both the contracts?

Let A and Β stand for the event of getting the plumbing and electrical contract, respectively.

Given, P[A’] = 1/3; P[B] = 4/9; and P[AB] = 4/5;

To find P[AB].

P[A’] = 1/3; it implies that P[A] = 1 – P[A’]; P[A] = 1 – 1/3 = 2/3.

By definition, P[AB] = P[A] + P[B] – P[AB]

The probability that he will set both the contract is 0.311.

#### 8.3.3 Multiplication Rule on Probability When Events Are Independent

Events are independent, when the occurrence of one event has no effect on the probability that another will occur, their joint probability is the product of their individual probabilities; then,
P[AB] = P[A] * P[B].

Note:

If two events A and Β be independent, then the following events are also independent.

1. A’ and B,
2. B’ and A and
3. A’ and B’

We have,

1. P[A’B] = P[A’] * P[B]
2. P[AB’] = P[A] * P[B’]
3. P[A’B’] = P[A’] * P[B’]

Example: 16

A candidate is selected for an interview for three posts. For the first post, there are 3 candidates, for the second there are 4 and for the third there are 2. What are the chances of his getting at least one post?

Let A, B and C stand for the events of setting selected for post 1, post 2 and post 3, respectively.

Number of candidates for the first post = 3; P[A] = 1/3 = 0.333.

Number of candidates for the second post = 4; P[B] = 1/4 = 0.125.

Number of candidates for the third post = 2; P[C] = 1/2 = 0.5.
To find P[ABC]. Here, the events A, Β and C are independent.
Let ABC = D, then we have, P[D] + P[D’] = l.

P[D] = 1 – P[D’] = 1 – P{[ABC ]}.

Using Demorgon’s property, [ABC]’ = A’B’C’

The chance of setting at least one post is 0.75.

#### 8.3.4 Compound Probability or Conditional Probability

When events A and Β are not independent, the occurrence of A will influence the probability that Β will take place. The multiplication rule when A and Β are independent can be given as:

P[AB] = P[A] * P[B/A] or P[B/A] = {P[AB]}/{P[A]}; where P[A] > 0.

Here P[B/A) is the conditional probability referring that the chance of Β has to occur after the occurrence of A. [The event A occurs first, then followed by the second event B takes place.]

In the same way, we can define the conditional probability of event A, given that Β has occurred.

P[AB] = P[B] * P[A/B] or P[A/B] = {P[AB]}/{P[B]}; where P[B] > 0.

Example: 17

A manager has drafted a scheme for the benefit of employees. To get an idea of the support for the scheme, he randomly polls literate workers [L] and illiterate workers [I]. He polls 30 of each group with the following results:

Opinion for Scheme L I

Strongly support [SS]

9
10

Mildly support [MS]

11
3

Undecided [U]

2
2

Mildly oppose [MO]

4
8

Strongly oppose [SO]

4
7
30
30
1. What is the probability that a literate worker selected randomly from the polled group mildly supports the scheme?
2. What is the probability that a worker [literate or illiterate] selected randomly from the polled group strongly or mildly supports the scheme?

[IGNOU, 2003]

Given:

1. Let the event L refers the selection of literate worker, then

P[L] = 30/60 = 0.5.

To find P[MS/L].

By definition, P[MS/L] = P[MSL]/P[L];

P[MSL] = 11/60;

Then, P[MS/L] = [11/60]/[0.5] = [11/30] = 0.367.

P[MS/L] = 0.367.

2. To find P[SS or MS].

Both the events are mutually exclusive.

Example: 18

The personnel department of a company has records that show the following analysis of its 200 engineers:

Age UG degree only PG degree only
<30
90
10
30–40
20
30
>40
40
10

If one engineer is selected at random from the company, find

1. The probability that he has only UG degree.
2. The probability that he has PG degree, given that he is over 40.
3. The probability that he is under 30, given that he has only a UG degree.

Given,

Let A, B, C and D be the events of selected personnel to have UG degree only, PG degree, with age more than 40 and age under 30, respectively.

To find [1] P[A], [2] P[B/C] and [3] P[D/A].

1. From the table, we have P[A] = 150/200 = 0.75;

P[A] = 0.75.

2. By definition, P[B/C ] = P[CB]/P[C ]

From the table, we have

3. By definition, P[D/A] = P[DA]/P[A]

From the table, we have

Hence,

The probability that he has only UG degree is 0.75.

The probability that he has PG degree given that age is over 40 is 0.2.

The probability that he is under 30, given that he has only a UG degree is 0.6.

Example: 19

A bag contains 8 red and 5 white balls. Two successive draws are made. Find the probability that the first draw will give 3 white balls and the second 3 red balls.

[1] With replacement and [2] Without replacement

Number of red balls = 8

Number of white balls = 5

Total number of balls = 13

1. With replacement:

First draw: 3 white balls

Total chances = 13C3

Number of favourable chances = 5C3

P[having 3 white balls in the first draw] = 5C3/13C3

= [10/286] = 0.035.

The three white balls selected in the first are replaced before the second draw.

Second draw: 3 red balls

Total chances = 13C3

Number of favourable chances = 8C3

P[second draw/first draw] = 8C3/13C3 = 56/286 = 0.196.

P[required] = 0.035 * 0.196 = 0.00686 = 0.0069.

2. Without replacement:

First draw: 3 white balls

P[first draw] = 0.035.

The three white balls selected in the first are not replaced before the second draw.

Second draw: Given that the balls are not replaced.

Total number of balls after the first draw = 13 – 3 = 10

Total chances = 10C3

Number of favourable chance = 8C3

P[second draw/first draw] = 8C3/10C3 = 56/120 = 0.467.

P[required] = 0.035 * 0.467 = 0.0163.

Example: 20

Suppose that a disease is inherited via a sex-linked mode of inheritance, so that a male offspring has a 50% chance of inheriting the disease, but a female offspring has no chance of inheriting the disease. Further suppose that 51.3% of births are male. What is the probability that a randomly chosen child will be affected by the disease?

Let M and F are the events of the person being a male and female, respectively.

Let D be the event of inheriting the disease.

Given, P[M] = 0.513; P[F] = 0.487; P[D/M] = 0.5 and P[D/F] = 0.

The event D can be happen if

1. The inheritance of the disease by a male [MD]
2. The inheritance of the disease by a female [FD]

D = [MD] ∪ [FD]. Both the events are mutually exclusive.

To find P[D) = P{[MD] ∪ [FD)} = P[MD] + P[FD]

By definition, P[MD] = P[D/M] * P[M] = 0.5 * 0.513 = 0.2565 and

P[FD] = P[D/F] * P[F] = 0 * 0.487 = 0.

Then, P[D] = 0.2565 + 0 = 0.2565.

Hence, the chance of the randomly selected child will be affected by the disease is 0.2565.

Example: 21

If a woman takes an early pregnancy test, she will either test positive, meaning that the test says she is pregnant, or test negative, meaning that the test says she is not pregnant. Suppose that if a woman is really pregnant, there is 98% chance that she will test positive. Also, suppose that if a woman really is not pregnant, there is a 99% chance that she will test negative.

• Suppose that 1,000 women take early pregnancy test and that the 100 of them are really pregnant. What is the probability that a randomly chosen woman from this group will test positive?
• Suppose that 1,000 women take early pregnancy tests and that 50 of them are really pregnant. What is the probability that a randomly chosen woman from this group will test negative?

Let Ρ and NP are the events of the selected female being pregnant and non-pregnant, respectively.

Let TP and TNP are the events of the test result positive and negative, respectively.

Example: 22

There are three men aged 60, 65 and 70 years. The probability to live 5 years more is 0.8 for a 60-year-old, 0.6 for a 65-year-old, and 0.3 for a 70-year-old person. Find the probability that at least two of the three persons will remain 5 years hence.

Let A, Β and C is the events of 60-years-old, 65-years-old and 70-years-old, person to live for 5 years hence, respectively, and all are independent.

Let D be the event of at least two of the three persons will remain 5 years hence.

Given,

1. P[A] = 0.8 = > P[A’] = 1 – P[A] = 0.2.
2. P[B] = 0.6 = > P[B’] = 1 – P[B] = 0.4.
3. P[C] = 0.3 = > P[C’]= 1 – P[C] = 0.7.

To find P[D].

The event corresponds to D are as follows:

1. ABC
2. AB′∩C
3. A′∩BC and
4. ABC

Hence, the probability that at least two of the three persons will remain alive after 5 years is 0.612.

Example: 23

In a study of the relationship between health risk and income, a large group of people living in Massachusetts were asked a series of questions. Some of the results are shown in the following table.

What is the conditional probability that someone in this study smokes, given that the person has high income?

Let A be the event referring the selected person smokes and Β be the event referring the selected person earning high salary.

To find P[A/B].

By definition, P[A/B] = P[AB]/P[B].

Then, P[A/B] = 0.0377/0.3787 = 0.1.

Example: 24

The following data table is taken from the study reported in the previous problem. Here ‘stressed’ means that the person reported that most days are extremely stressful or quite stressful; ‘not stressed’ means that the person reported that most days are a bit stressful, not very stressful, or not at all stressful.

Is being a stressed person independent of having medium salary?

Let A be the event referring the selected person having stress and Β be the event referring the selected person earning high salary.

By definition,

Since P[AB] ≠ P[A] * Ρ[Β]; it implies that A and Β are not independent.

##### 8.4 ΒAYE’S THEOREM

The extension concept of conditional probability is Baye’s theorem, which was introduced by Thomas Baye’s during 1700s. In this application of conditional probability, the stress is given as sequential events, especially information received from a second event is used to modify the probability that a first event has occurred.

Statement:

If Α1, Α2,…, An are mutually exclusive events with P[Ai] > 0; [i =1,2,…, n], for any event B which is a subset of [A1A2 ∪ … ∪ An] such that P[B] > 0, then

The statement can be explained through a diagram.

Obviously the events A1B, A2B,…, AnB are existing and all are mutually exclusive.

Then, B = [A1B] ∪ [A2B] ∪ … ∪ [AnB].

Then, the probability of B,

By definition,

Then, we have,

Using [2] in [1],

By definition,

Using [2] and [3] in [4],

Hence, the theorem is proved.

Example: 25

The chances that a doctor will diagnose a disease correctly are 60%. The chances that a patient will die by his treatment after correct diagnosis are 40% and the chances of death by wrong diagnosis are 70%. A patient of doctor, who had disease, was died. What is the chance that his disease was diagnosed correctly?

Events

B1: The doctor diagnosing the disease correctly.

B2: The doctor not diagnosing the disease correctly.

Prior probability

This is an initial probability based on the prior level of information on the basis,

P[B1] = 0.6, since the doctor diagnosing the disease correctly is 60%.

P[B2] = 0.4, [1 – 0.6 = 0.4].

Event

D: The patient who had a disease dies.

Posterior probability

This is the revised probability that has the benefit of additional information. It is a conditional probability and can be expressed P[D/Bi]

P[D/B1] = 0.4; P[D/B2] = 0.7;

Tabulate the prior and posterior probabilities:

To find P[B1/D].

The probability that the patient dies even the disease was diagnosed by the doctor correctly is 0.46154.

Example: 26

The residents of a locality are examined for cancer. The examination results are classified as positive, if malignancy is suspected, and as negative, if there is no indication of malignancy. If a person has cancer, the probability of a suspected malignancy is 0.95 and the probability of cancer where none existed is 0.12. If 8% of the residents have cancer, what is the probability of a person not having cancer if the examination is positive.

Events

B1: The resident having cancer.

B2: The resident not having cancer.

Prior probability

This is an initial probability based on the prior level of information on the basis,

Event

E: The examination for cancer is positive.

Posterior probability

This is the revised probability that has the benefit of additional information. It is a conditional probability and can be expressed P[E/Bi].

P[E/B1] = 0.95; P[E/B2] = 0.12.

Tabulate the prior and posterior probabilities:

To find P[B2/E].

The probability that the patient not having cancer but the examination becomes positive is 0.59227.

Example: 27

An absent-minded nurse is supposed to give the patient a pill each day. The probability that the nurse forgets to give the pill is 0.4. If the patient receives the pill, the probability that he will die is 0.25. If he does not get the pill, the probability that he will die is 0.8. The patient died. What is the probability that the nurse forgot to give the pill to the patient?

Events

B1: The nurse forgets to give the pill to the patient.

B2: The nurse gives the pill to the patient.

Prior probability

This is an initial probability based on the prior level of information on the basis,

Event

D: The patient dies.

Posterior probability

This is the revised probability that has the benefit of additional information. It is a conditional probability and can be expressed P[D/Bi].

P[D/B1] = 0.8; P[D/B2] = 0.25.

Tabulate the prior and posterior probabilities:

To find P[B1/D].

The probability that the patient dies even the disease was diagnosed by the doctor correctly is 0.46154.

Example: 28

A person has two coins; one is unbalanced and lands heads 60% of the time, the other is fair and lands heads 50% of the time. He selects one of the coins and flips it. The result is head.

• What is the prior probability that the fair coin was selected?
• Given additional information in the form of the single flip that came up as head, what is the revised probability that the coin is the fair one?

Event:

B1: The selected coin was unbalanced.

B2: The selected coin was fair.

A: To get head in a flip.

Tabulate the prior and posterior probabilities.

Hence,

• The prior probability of selection of a fair coin is 0.5.
• The probability to set the head in a single flip using a fair coin is 0.455.

Example: 29

There are two identical boxes containing 4 white and 3 red balls, and 3 white and 7 red balls, respectively. A box is chosen at random and a ball is drawn from it. If the ball is white, then what is the probability that it is from the first box?

Events:

B1 : Selection of the box 1.

B2 : Selection of the box 2.

A : Selection of white ball.

Box 1

 4 White 3 Red

Box 2

 3 White 7 Red

 Total balls = 7 Total balls = 10 Selection of one ball = 7C1 = 7 Selection of one ball = 10C1 = 10 Number of white balls = 4 Number of white balls = 3 Favourable chances of selection one white ball = 4C1 = 4 Favourable chances of selection one white ball = 3C1 = 3 P[A/B1] = 4/7 = 0.571. P[A/B2] = 3/10 = 0.3.

Tabulate the prior and posterior probabilities

To find P[B1/A], by definition,

The probability of selection of a white ball from box 1 is 0.656.

##### EXERCISES
1. Suppose that a student who is about to take a multiple-choice test has learned 40% of the material covered by the exam. Thus, there is 40% chance that she will know the answer to a question. However, even if she does not know the answer to a question, she still has 20% chance of getting the right answer by guessing. If we choose a question at random from the exam, what is the probability that she will get it right?
2. If two dice are thrown, what is the probability that the sum of numbers that appeared on them is [a] greater than 8? [b] neither 7 nor 11?
3. The probability that a student A solves a biology-related problem is 2/5 and the probability that a student Β solves it is 2/3. What is the probability that the problem is not solved, when they are working independently?
4. A box contains 6 red, 4 white and 5 black balls. A person draws 4 balls from the box at random. Find the probability that among the balls drawn, there is at least one ball of each colour.
5. Among 1,000 applicants for admission to MSc [Bio-tech] program in a university, 600 were biology with mathematics graduates and 400 were biology without mathematics graduates. 30% of biology with mathematics graduates applicants and 5% of biology without mathematics graduates obtained admission. If an applicant selected at random is found to have been given admission, what is the probability that he/she is a biology with mathematics graduate?
6. Consider a population of consumers consisting of two types. The upper class of customers comprises 35% of the population and each member has probability 0.8 of purchasing brand A of a product. Each of the rest of the population has probability 0.3 of purchasing brand A. A consumer chosen at random is a buyer of brand A. What is the probability that the buyer belongs to middle and upper class of consumers?
7. Explain the concept of conditional probability and Baye’s theorem.
8. There are two identical boxes containing 4 white and 3 red balls, and 3 white and 7 red balls, respectively. A box is chosen at random and a ball is drawn from it. If the ball is white, what is the probability that it is from the second box?
9. Urn I and II contain 4 white, 3 red and 3 blue balls, and 5 white, 4 red and 3 blue balls, respectively. If one ball is drawn at random from each urn, what is the probability that both the balls are of same colour?
10. P1, P2, P3 and P4 are four mutually exclusive and exhaustive events. If the odds against the events P2, P3, and P4 are 7:2, 7:5 and 13:5, find the odds in favour of the event A1. [Hint: use addition theorem.]
11. Two boxes contain 4 white, 3 red and 5 blue balls, and 6 white, 4 red and 5 blue balls, respectively. If one ball is drawn at random from each box, what is the probability that both the balls are of same colour?
12. A problem in biology is given to three students S1, S2 and S3 whose chances of solving it are 0.6, 0.5 and 0.4, respectively. If they try it individually, what is the chance that the biology problem will be solved?
13. Four balls are drawn at random from a bag containing 5 red and 7 blue balls. Compute the probability of getting [a] 4 red balls, [b] 2 red and 2 blue balls, [c] w blue balls and 1 red ball.
14. Tech Search Inc. specializes in placing technical managers. It classifies clients in terms of skills and years of experience. The skills are ‘research and development [R&D]’ and ‘design’. No candidate possesses both the skills. Experience categories are ‘2 years or less’, ‘between 2 and 10 years’ and ‘10 years or more’. At present, there are 100 executives on file with skills and experience summarized in the following table:

Suppose that you select at random one executive’s file. Determine each of the following probabilities:

• P[R&D]
• P[Design]
• P[R&D and 10 years or more experience]
• P[10 years or more experience R&D given an R&D executives is selected]
15. Two balls are drawn at random from a bag containing 6 white and 4 black balls. Find the chance that one is white and the other is black.
16. A problem in Bio-stat is given to three experts, A, B, and C whose chances of solving are 1/3, 1/4, and 1/5, respectively. What is the probability that the problem will be solved?
17. The probability that a manager’s job applicant has a PG degree in Biotechnology is 0.3, and he has had some work experience as a office chief is 0.7 and that he has both is 0.2. Out of 400 applicants, what number would have either a PG degree or some professional work experience or both?
18. Given that P[A] = 3/8; P[B] = 5/8 and P[AB] = 3/4; find P[A/B] and P[B/A]. Are A and Β independent?
19. In a group of equal number of men and women 10% men and 45% women are unemployed. What is the probability that a person selected at random is employed?
20. Mr. Sree Balaji is called for interview for 3 separate posts. At the first interview, there are 5 candidates; at the second, 4 candidates and at the third, 6 candidates. If selection of each candidate is equally likely, find the probability that Mr. Sree Balaji will be selected for [a] at least one post and [b] at least two posts.
21. It is possible to have a sample space in which P[A] = 0.7 and P[B] = 0.6 and P[AB] = 0.35. Given the information, would events A and B be mutually exclusive? Would they be independent?
22. A magician has two coins: one is unbalanced and lands heads 60% of the time, the other is fair and lands heads 50% of the time. A member of the audience randomly selects one of the coins and flips it. The result is head.
• What is the probability that the fair coin was selected?
• Given additional information in the form of the single flop that came up heads, what is the revised probability that the coin is fair one?
23. Suppose that 5 men out of 100 and 25 women out of 10,000 are colour blind. A colour-blind person is chosen at random. What is the probability of his being male? [assume that males and females are equal in number].
24. A doctor is to visit the patient, and from past experience it is known that the probabilities that he will come by train, bus or scooter are 0.3, 0.2 and 0.1, respectively; the probability that he will use some other means of transport being, therefore, 0.4. If he comes by train, the probability that he will be late is 0.25; if by bus 0.33 and if by scooter 1/12; if he uses some other means of transport it can be assumed that he will not be late. When he arrives he is late. What is the probability that [a] he comes by train and [b] he is not late.
25. An absent-minded nurse is supposed to give the patient a pill each day. The probability that the nurse forgets to give the pill is 0.4. If the patient receives the pill, the probability that he will die is 0.25. If he does not get the pill, the probability that he will die is 0.8. The patient died. What is the probability that the nurse give pill to the patient?
26. The residents of a locality are examined for cancer. The examination results are classified as positive, if malignancy is suspected, and as negative, if there is no indication of malignancy. If a person has cancer, the probability of a suspected malignancy is 0.95 and the probability of cancer where none existed is 0.12. If 8% of the residents have cancer, what is the probability of a person having cancer if the examination is positive.
27. The chances that a doctor will diagnose a disease correctly are 60%. The chances that, a patient will die by his treatment after correct diagnosis is 40% and the chance of death by wrong diagnosis is 70%. A patient of doctor, who had disease, was died. What is the chance that his disease was not diagnosed correctly?
1. The concept of probability was introduced in the late________________.
2. ________________refers the study of events which are going to happen or not.
3. Define the term sample space.
4. Define the term experiment.
5. Define the term event.
6. What do you mean by the events are mutually exclusive?
7. Probability can be studied based on the________________.
8. The formula for the classical approach is________________.
9. The formula for the relative frequency approach is________________.
10. The probability of an event A is denoted by________________.
11. The value of P[A] should be in the range________________.
12. If P[A] = 1, then the event A is said to be a________________.
13. If P[A] = 0, then the event A is said to be a________________.
14. State the addition theorem on probability.
15. If A and Β be any two events then the probability that at least one of the two events A and Β occurs can be denoted by P[AB] and the same can be defined as________________.
16. “ If A, B and C be any three events then the probability that at least one of the three events A, Β and C occurs can be denoted by P[ABC] and the same can be defined as P[ABC] = P[A] + P[B] + P[C] - P[A] * P[B] - P[C] * P[B] - P[A] * P[C] + P[A] * P[B] * P[C]” - Comment on this statement.
17. There is no chance for the occurrence of two or three events together if all the________________.
• events are mutually exclusive
• events are independent
• none
18. If P[ABC] = P[A] * P[B] * P[C]; then the events A, Β and C are________________.
• mutually exclusive
• independent
• none
19. State Baye’s theorem.
20. The conditional probability of P[A/B] can be defined as________________.
1. seventeenth century
2. Probability
3. Refer Section 8.2
4. Refer Section 8.2
5. Refer Section 8.2
6. Refer Section 8.2
7. classical approach and relative frequency approach
8. P[A]
9. 0<Ρ[A]<1.
10. sure event
11. null event
12. Refer Result-1 of Section 8.3.1
13. P[AB] = P[A] + P[B] - P[AB].
14. True
15. events are mutually exclusive
16. independent
17. Refer Section 8.4
18. P[A/B] = P[AB]/P[B].