Introduction To Probability

Introduction to Probability Rohit Vishal Kumar March 2, 2007

1 Introduction The idea of probability, chance or randomness is quite old; where as its rigorous axiomisation in mathematical terms have occurred relatively recently. Many of the ideas of probability theory originated in the study of game of chance. In this century, the mathematical theory of probability has been applied to a wide variety of phenomena for example — It has been used in genetic theory to understand mutation and gene sequence, In Information management, probability has been used in designing and optimising various operating systems, to model the length of various queues, In communication theory, probability has been used to study noise in electrical devices and communication systems, In atmospheric research turbulence is modeled using probability, Actuarial sciences, used by insurance companies, rely heavily on the theory of probability to determine premium etc. In this article, we treat the basic idea of probability and statistics. In the first part we explain the theory of probability and subsequently we deal with problems of probability.

2 Common Terms Probability theory is concerned with situations in which the outcomes occur randomly. Generally, such situations are called experiments, and the set of all possible outcome is known as the sample space corresponding to the experiment. The sample space is generally denoted by S and an element of S is demoted by ω. Example A Driving to work, a commuter passes through a sequence of three intersection with traffic lights. At each light, she either stops s, or continues c. The sample space is the set of all possible outcomes: S = {ccc, ccs, csc, scc, css, scs, ssc, sss} where csc, for example, denotes the outcome that the commuter continues through the first light, stops at the second light and continues through the third light.  Example B The number of jobs in a print queue of a mainframe computer may be modeled at random. Theoretically the sample space would consist of all non negative integers up to infinity. In practice, there would be an upper limit N as to how large the print queue can be. The sample space can be defined as S = {0, 1, 2, 3, . . . , N }  We are often interested in a particular subset of S, which in the language of probability, are called events. In Example A, the event that the commuter stops at the first light is a subset of S and is given by A = {sss, ssc, scc, scs} Events of subsets, are usually denoted by uppercase roman letters. Similarly, in Example B, the event that the print queue has fewer than five jobs can be denoted by A = {0, 1, 2, 3, 4} The algebra of set theory is directly applicable to the events in probability theory. The union of two events A and B, denoted by A ∪ B is defined as the event such that either A occurs or B occurs or both 1

3 THEORY OF PROBABILITY

2

Commutative Law Complimentary Law

Involution Law Idempotency Law Associative Law Distributive Law De Morgan’s Law

A∪B A∩B A ∪ A′ A ∩ A′ A∪S A∩S A∪φ A∩φ (A′ )′ A∩A A∪A (A ∪ B) ∪ C (A ∩ B) ∩ C (A ∪ B) ∩ C (A ∩ B) ∪ C ′ (A ∪ B) ′ (A ∩ B)

= = = = = = = = = = = = = = = = =

B∪A B∩A S φ S A A φ A A A A ∪ (B ∪ C) A ∩ (B ∩ C) (A ∩ C) ∪ (B ∩ C) (A ∪ C) ∩ (B ∪ C) A′ ∩ B ′ A′ ∪ B ′

Table 1: Laws of Set Theory used in Probability occurs. The intersection of two events, denoted by A ∩ B is defined the event such that A and B both occur. The compliment of event A, denoted by A or Ac or A′ , is the event that A does not occur and thus consist of all the elements of S which are not in A. An empty set, denoted by φ, is a set which has no elements i.e. it is the event with no outcomes. If there are say two events A and B and if A ∩ B = φ then the events A and B are said to be disjoint events. The laws of set theory are extensively used in probability theory. Table 1 gives some of the laws of set theory which are used frequently in statistics. You are advised to check the validity of the laws using Venn Diagram

3 Theory of Probability 3.1 Definition’s of Probability Even though the meaning and understanding of probability was clear to the world for a long time; there was significant disagreement amongst the theoreticians as to how to define probability. The earliest definition of probability was the classical definition, which was later rejected because it lacked certain desirable properties. Subsequently, the mathematical definition, which defined probability in terms of limits was also used and later disputed. The current definition of probability is based on three axioms and is known as the axiomatic definition of probability. All the three definitions are provided here for the sake of completeness: 3.1.1

Classical Definition of Probability

Suppose an event results in n mutually exclusive, exhaustive and equally likely cases. Let m be the number of events which are favorable to the event A. Then the probability of event A, denoted as P (A) is defined as: P (A) = =

N o. of Cases F avorable to the Event A T otal number of cases m n

(1)

The classical definition introduced some concepts which are defined here. Consider an experiment which though repeated under essentially identical conditions does not give unique results but may result in any one of the several possible outcomes. Then the experiment is known as a Trial and the outcomes

3 THEORY OF PROBABILITY

3

are known as events or cases. The total number of possible outcome in any trial is known as the exhaustive event. Exhaustive events corresponds to the sample space of set theory. Events are said to be mutually exclusive if the occurrence of any one of them precludes (or prevents) the occurrence of all other events. Outcomes of a trial are said to be equally likely if taking into consideration all the relevant evidences, there is no reason to expect one event in preference to any other event. Favorable Events in a trial are the number of events, the occurrence of which leads to the occurrence of the defined event.

ab b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b ddd e e e ddd e e e ddd e e e ddd e e e dd e e fggggggggggggggggggggggggggggggggggggggggh U NDERSTANDING

THE

C LASSICAL P ROBABILITY D EFINITION

Let us consider that we toss a dice. Then, tossing of the dice is an Trial and getting 1 (or 2 or 3 or 4 or 5 or 6) is an event. Thus the above trial leads to the following events S = {1, 2, 3, 4, 5, 6} which is the sample space of the outcomes. If we assume that the dice is unbiased, we don’t know which event will occur. So we cannot favor any event over any other event. Hence all the events are said to be equally likely. Now if after tossing the dice, we get 1 (say), then in this toss any other number (2,3,4,5,6) cannot occur. Hence the outcomes of the event are mutually exclusive. Now suppose we define a new event A as the outcome of an odd number when an unbiased dice is tossed. Then the number of outcomes which are favorable to event A are A = {1, 3, 5}. Then the probability of event A is P (A) = 3/6 = 1/2.

It can be easily seen that 0 ≤ m ≤ n and therefore the value of P (A) lies between 0 and 1, both inclusive. It can also be seen that the classical definition is dependent on a finite number of cases (n 6= 0). Sometimes, the expression (1) is also referred to as that “the odds in favor of event A are m : (n − m) or the odds against event A are (n − m) : n. In case the sample space is infinite then the classical definition fails to define probability. The classical definition also fails if the trials of an event are not equally likely. For example, suppose a candidate appears for a test then we normally assume that the candidate is equally likely to fail or pass. However, if we already know that the candidate has more than 50% chance of passing, then the outcomes are not equally likely and hence we cannnot apply the classical definition. Given these limitations of the classical definition, we now look at other definitions of probability 3.1.2

Empirical Definition of Probability

If a trial is repeated a number of times under essentially homogeneous and identical conditions, then the limiting value of the ratio of number of times the event happens (m) to the number of trials (n), as the number of trials becomes indefinitely large is called the probability of happening of the event (under the assumption that the limit is finite and unique). Symbolically, if in n trials, an event A occurs m times, the probability of event A denoted by P (A) 3.1.3

=

lim

n→∞

m n

(2)

Axiomatic Definition of Probability

A probability measure on S is a function P which assigns a non-negative real number to every event A which satisfies the following axioms: 1. P (S) = 1 2. For each A ∈ S, P (A) is defined, is real and P (A) ≥ 0. 3. If A1 , A2 , A3 , . . . , An , . . . are mutually disjoint, then ! ∞ ∞ X [ P (Ai ) Ai = P i=1

i=1

4 PERMUTATION AND COMBINATION

4

The first two axioms are obviously desirable. Since S consists of all possible events, hence P (S) = 1. The second axiom simply states that probability of any event A is defined and non-negative. Let us first understand the third axiom in terms of two events A1 and A2 which are disjoint in nature i.e. they have no outcome in common; then P (A1 ∪ A2 ) = P (A1 ) + P (A2 ). Thus what the third axiom says is that if there are a large number of events, A1 , A2 , A3 , . . . , An , . . ., defined on the sample space S and they are all disjoint, then the probability of the union of these events is nothing but the sum of probabilities of the individual events. The following important properties derive from the axiomatic definition of probability: 1. The probability of an impossible event is zero i.e. P (φ) = 0 2. Probability of a complimentary event A′ is given by P (A′ ) = 1 − P (A) 3. If the event A is a subset of event B then P (A) ≤ P (B). 4. If A and B be any two events defined on the sample space S and are not disjoint, then P (A ∪ B) = P (A) + P (B) − P (A ∩ B). This is also known as the Addition Law of probability. You are advised to Prove all the Properties

4 Permutation and Combination A permutation is an ordered arrangement of objects. Suppose that from a set containing n objects we are to choose r objects and list them in order. The question is, in how many ways we can do this? The answer depends on whether we are allowed to duplicate objects from the list or not. If we are allowed to duplicate objects we are sampling with replacement and if we are not allowed to duplicate, then we are sampling without replacement. First suppose that we are allowed to duplicate i.e. we are doing sampling with replacement. So the first object can be chosen in any of the n ways. After we have chosen the first object, we can put the object back into the set and choose another object from the full set of n objects. So the second object can be chosen in another n ways. So there are nr ways of choosing an r objects from a set of n objects. Now suppose that the sampling is done without replacement. The first object can be chosen in n ways. The second object can be chosen in (n − 1) ways, the third object in (n − 2) and the rth object can be chosen in (n − r + 1) ways. Thus the total number of ways in which we can choose r objects from a set of n objects without replacement is n(n − 1)(n − 2) . . . (n − r + 1) ways. This is known as permutation and is usually denoted by n Pr or by (n )r .

ab b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b e ddd e e e ddd e e e ddd e e e ddd e e fggggggggggggggggggggggggggggggggggggggggh S OME P ROPOSITIONS

TO

R EMEMBER

P ROPOSITION A: For a set of size n and a sample of size r, there are nr different ordered n! samples with replacement and n Pr = (n−r)! = n(n − 1)(n − 2) . . . (n − r + 1) different ordered samples without replacement.  P ROPOSITION B: The number of unordered samples of r objects selected from n objects withn!  out replacement is n Cr = (r!)(n−r)! C OROLLARY A.1: The number of orderings of n elements is n(n − 1)(n − 2) . . . 1 = n!



Let us now consider a special case, in which we are not interested in ordered samples, but in the constituents of the sample regardless of the order in which they have been obtained. In particular, we ask the following question: If r objects are taken from a set of n objects without replacement and disregarding order of selection, then how many different samples are possible? Now we know that the number of ordered samples without replacement is n(n − 1)(n − 2) . . . (n − r + 1) and since a sample of size r can be ordered in r! ways, the number of unordered samples is n(n − 1)(n − 2) . . . (n − r + 1)/r!.
This is known as combination and is denoted by n Cr or by nr .

5 CONDITIONAL PROBABILITY

n

Cr

5

= =

n! r!(n − r)! n(n − 1)(n − 2) . . . (n − r + 1) r(r − 1)(r − 2) . . . 3.2.1

5 Conditional Probability 5.1 Introduction To introduce the aspect of conditional probability we take a help of an example. Digitalis therapy is often beneficial to patients who have suffered congestive heart failure — a type of cardiac disease. But giving Digitalis to patients has a serious side effect as the patient runs the risk of having Digitalis toxicity which can prove fatal. To improve the chance of correct diagnosis, the concentration of Digitalis in the blood can be measured. A study was conducted in 135 cardiac heart patients to find the concentration of Digitalis in the blood of the patients. The table below gives the results where the following notations are used: D+ congestive heart disease is present, D− the congestive heart disease is not present, T + there is high concentration of Digitalis in the blood and T − there is low concentration of Digitalis in the blood. T+ T− Total

D+ 25 18 43

D− 14 78 92

Total 39 96 135

Thus for example, 25 patients had high concentration of Digitalis in blood and the disease is present. Assuming that the findings of the the study holds for all the cardiac patients, the probability of having congestive heart disease is 43/135 = 0.318. But suppose now, that we have a patient and the patient shows a high concentration of toxicity in the blood. Then what would be the probability of the patient having the congestive heart disease? To answer this question we can restrict our attention to the first row of the table. We see that out of 39 cardiac patients who have high concentration of Digitalis in the blood, 25 suffer from congestive heart disease. Thus the probability of having congestive heart disease given that the patient has high concentration of Digitalis in blood is 25/ 39 = 0.640. Let us understand the results. The probability of having a congestive heart disease amongst cardiac patients P (D+) = 0.318, but when we get additional information of Digitalis concentration in blood the probability of having congestive heart disease becomes P (D + |T +) = 0.640 which is much higher than the P (D+). P (D+) is the unconditional probability of having congestive heart disease and P (D + |T +) is known as the conditional probability of having congestive heart disease given that we have the information T +. Conditional probability can also be looked upon as the probability of a particular event provided some additional information about the occurrence (or non occurrence) of the event is available.

5.2 Definition and Properties Let A and B be two events with P (B) 6= 0, the the conditional probability of A given B is defined as P (A|B) =

P (A ∩ B) P (B)

The following properties hold for the conditional probability: 1. P (S|A) = 1 where S is the sample space or sure event. 2. If A1 and A2 are two events such that A1 ∩ A2 = φ then P (A1 ∪ A2 |B) = P (A1 |B) + P (A2 |B). 3. P (A′ |B) = 1 − P (A|B). 4. Given the definition of conditional probability, we also have P (A|B).P (B) = P (A ∩ B). This is also know as the multiplication law of probability. You Are Advised To Prove All The Properties

5 CONDITIONAL PROBABILITY

6

5.3 Bayes Rule Let the events A1 , A2 , A3 , . . . , An be defined on the sample space S and let these events be exhaustive and mutually exclusive and let P (Ai ) > 0 ∀i . Then for any event B defined on the sample space we have n X P (Ai )P (B|Ai ) P (B) = i=1

Proof Since the events Ai ’s are exhaustive we have

∪ni=1 Ai = S Now we can write event B as B ∩ S (From Complimentary Law of Set Theory) and as such we have: B

=

B∩S

= =

B ∩ (∪ni=1 Ai ) ∪ni=1 (B ∩ Ai )

Since the events A1 , A2 , A3 , . . . , An are mutually exclusive, the events B ∩ A1 , B ∩ A2 , B ∩ A3 , . . . , B ∩ An are also mutually exclusive. And, therefore by the addition theorem of probability we have: P (B) = =

P (∪ni=1 (B ∩ Ai )) n X P (B ∩ Ai ) i=1

And from the multiplicative law of probability we know that P (B ∩ Ai ) = P (Ai )P (B|Ai ) hence P (B)

=

n X

P (Ai )P (B|Ai )

i=1

QED 

5.4 Independence of Events Two events, A and B are said to be stochastically independent, if the probability of occurrence of one of the events, does not depend on the occurrence or non occurrence of the other event. Thus the two events are said to be independent iff P (A|B) = P (A|B ′ ) = P (A). It can be shown that the two events are stochastically independent iff P (A ∩ B) = P (A)P (B). If the events A and B are independent, then the events (i) A′ and B, (ii) A and B ′ and (iii) A′ and ′ B are also independent.

ab b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b b e ddd e e e ddd e e e ddd e e e ddd e e e ddd e e fggggggggggggggggggggggggggggggggggggggggh You are advised to Prove the Above Independence’s

I NDEPENDENCE

OF

M ANY E VENTS

If there are more than two events say A1 , A2 , A3 , . . . An then for stochastic independence it is required that the following conditions are met: P (Ai ∪ Aj ) P (Ai ∪ Aj ∪ Ak )

= =

P (Ai )P (Aj ) P (Ai )P (Aj )P (Ak )

... ... ... P (Ai ∪ Aj ∪ Aj . . . ∪ An ) = P (Ai )P (Aj )P (Ak ) . . . P (An )

Note: For n events, (2n − n − 1) conditions are required to be met for stochastic independence

6 SOME CAUTIONS

7

6 Some Cautions The concept of probability is not easily understood by people and therefore at times is used to confuse the population at large. Take for example the following quote from the Los Angeles Times (August 24, 1987) which talks about AIDS: Several studies of people infected with the AIDS virus shows that a single act of unprotected sex with an has a surprising low risk of infecting partners — probably one in 100 to one in 1000. For an average, consider the risk to be 1 in 500. Statistically, 500 acts of unprotected sex with an infected partner or 100 acts with five partner leads to a 100% probability of infection. Have you spotted the flaw? There are many but we will consider only a few. First and foremost, the report says that 500 acts with one infected partner will lead to infection. So suppose a person has 1000 acts of sex with an infected partner, then what is his probability of getting infected? According to the report it is 2 (which, of course, is not possible theoretically). Let us assume that the probability of infection is 1/500, as reported in the news. Now suppose a person has 500 acts of sex with an infected partner. What is his probability of getting infected? Let us work it out. Assume that the sexual acts are independent of each other and each act has 1/500 probability of having the infection. Then the probability of non-infection in each act is 1 − (1/500) = 499/500. So in 500 acts of unprotected sex, the probability of non-infection is (499/500)500 = 0.3675. Therefore the probability of infection is 1 − 0.3675 = 0.6325 which is much less than 100% as claimed by the study. If you did not identify the flaws in the study, don’t despair. Research has shown that people are not too good at understanding probability. For example consider the following question “If Linda is a 31 year old woman who is outspoken on social issues such as disarmament and equal rights, which of the following statement is more likely to be true?” • Linda is a Bank Teller • Linda is a Bank Teller and active in feminist movement More than 80% of those questioned choose the second statement, despite the fact that the correct answer is the first statement. Even hardened professional’s have difficulty in answering probabilistic calculations. For example the following question was asked to 100 doctors: “In the absence of any special information, the probability that a woman has breast cancer is 1%. If the patient has breast cancer, the probability that the radiologist will correctly diagnose it is 80%. And if the patient has benign lession (no brest cancer) then the probability that the radiologist will incorrectly diagnose it as breast cancer is 10%” Then what is the probability that a patient with a positive mammogram actually has breast cancer? 95 out of the 100 physicians estimated the probability to be about 75% However, the correct probability, as given by Bayes rule is 7.5% (You can check this). So even experts make mistakes. However, in spite of it’s misuse and lack of interpretation, probability is the cornerstone of all sciences and also of various subjects of humanities and management. So it is imperative that you have a clear idea of probability and understand the basics of probability well. This document can be obtained from: Rohit Vishal Kumar Reader, Department of Marketing Xavier Institute of Social Service P.O. Box No: 7, Purulia Road Ranchi - 834001, Jharkhand India Phone: (91-651) 2200-873 Ext. 308 Email: [email protected] Final Print on: March 2, 2007 c 2007, Rohit Vishal Kumar