PERMUTATIONS, COMBINATIONS AND DISCRETE PROBABILITY

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1 Friends, we continue the discussion with fundamentals of discrete probability in the second session of third chapter of our course in Discrete Mathematics. The conditional probability and Baye s theorem are intro duced in this session. As an application of the theory, information and mutual information also are introduced. Chapter 3.2 PERMUTATIONS, COMBINATIONS AND DISCRETE PROBABILITY Module INTRODUCTION TO DISCRETE PROBABILITY Anxiety to know future is just common to all of us. Impatience is another common thing. What will be the immediate consequence of these two? The answer is very simple. One would try to guess the possible outputs much before the completion of any process. Is there any scientific support for your guess? Of course the answer is YES. The entire probability theory comes for our help. There are many processes like tossing a coin, throwing a die, etc. In which there are only a finite number of possible outputs. If it is unable to predict the exact output before the completion of the process, then the process is called a random experiment. Each output of the random experiment is termed an outcome. The set of all possible outcomes is called the sample space. Any subset of the sample space is called an event. A collection of events is said to be exhaustive if their union is the sample space. Two events are said to be mutually exclusive (or disjoint) if their intersection is an empty set. For example, in the random experiment of throwing a die, the sample space is the set {, 2, 3, 4, 5, 6 }. The event of getting an odd number is the set {, 3, 5 }. The events of an odd number, a prime number, perfect squares and perfect number are exhaustive. These events are respectively {, 3, 5 }, { 2, 3, 5 }, {, 4 } and { 6 }. Their union is the sample space. The events of a prime number and perfect squares are mutually exclusive as there are no outcomes common in them. Events are actually sets. So all the set operations are applicable to them. Definition.. (Probability) The probability of an event is defined as Probability = number of outcomes in the event number of outcomes in the sample space

2 For any event, the complement also is an event. Probability of complement = probability of the event. Example.2. In the random experiment of throwing a fair die, there are six outcomes in the sample space. In the event of prime numbers, (which is the set {2, 3, 5}), there are three outcomes. Thus the probability of getting a prime number when a fair die is thrown is = 6 3 = 2 Example.3. In the random experiment of tossing three fair coins, the sample space is the set {(H,H,H), (H,H,T), (H,T,H), (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)} consisting of eight outcomes where H stands for a Head and T for a Tail. The probability of getting exactly one head is 3 8, since the set {(H,T,T), (T,H,T), (T,T,H)} is the corresponding event. Example.4. Consider the random experiment of forming three digit num bers using the digits 0,, 3, 5 and 6. There are = 00 outcomes. The event of getting an odd 60 number has = 60 outcomes and it s probability is 00 =0.6 Example.5. Consider the random experiment of tossing two fair dice. There are 36 outcomes in the sample space. The event of getting the sum of numbers on the dice as 8 or consists of 7 outcomes which are (2, 6), (3, 5), (4, 4), (5, 3), (6, 2), (5, 6) and (6, 5). 7 So it s probability is 36 Example.6. Digital data received from a remote site might fill up 0 to 32 buffers. Let the sample space be S = {0,, 2,, 32}, where the sample i denotes that i of the buffers are full. It is given that P (i) = 56! (33 i). Let A denote the event that at most 6 buffers are full and B denote the event that an odd number of buffers are full. Then A = {0,, 2,, 6} B = {, 3, 5,, 3} A B = {, 3, 5,, 5} and

3 6 p( A)= 56! i= (33 i)= ! = p(b)= 56! odd i= (33 i)= ! = p( A B)= 56! odd i= (33 i)= ! =0.357 The principle of inclusion and exclusion can be used to determine the number of outcomes in the union of events. The probability of the union could be obtained as P(A B) = P(A) + P(B) - P(A B). This is known as the addition theorem on probability. Example.7. Consider the random selection of a number from the first 30 natural numbers. Take A as the event of getting a multiple of 2 and B as the event of getting a multiple of 5. Then we can write the following events and the corresponding probabilities. A = {2,4,6,8,0,2,4,6,8,20,22,24,26,28,30} B = {5,0,5,20,25,30} and A B = {0,20,30}. Also P( A)= 5 30 P(B)= 6 30 P( A B)= 3 30 Now P( A)+P(B) P (A B)= =8 30 Here the event A B ={2,4,5,6,8,0,2,4,5,6,8,20,22,24,25,26,28,30} and its probability 8 is P(A B) = 30 Thus the addition theorem is verified.

4 Module 2 CONDITIONAL PROBABILITY The occurence of certain events can make changes in the probability of other events. For instance, in a cricket match, the captain who had won the toss may opt batting first. But if there was rain in the previous night or a snow fall in the morning, instead, he may opt fielding. So the event that the captain opts batting first depends upon the event that there was rain/snow fall. These types of situations are studied in conditional probability. Definition 2.. (Conditional probability) : The event A under the condition that another event B has already been happened is called the conditional event. It is denoted by A/B (read as A given B). The probability of such an event is called the conditional probability and is denoted by It is defined as P (A/B) = P( A B) P( A /B)= P (B) Example 2.2. Consider the following data. Discipline Boys Girls T otal Arts Science T otal Let A be the event that a student selected is a boy and B the event that a student selected studies a science subject. Then P( A)= 0 70 P( A)= P( A B)= Here, the probability that a science student called is a boy will be the conditional probability

5 P (A/B) = = Definition 2.3. (Independent Events) : If P (A/B) = P (A), then we say that A is independent of B. For independent events A and B, we have P (A B) = P (A) P (B), which is used as the condition to check the independence of the events. It is easy to prove the following for two independent events A and B.. A c and B c are independent events. 2. A and B c are independent events. 3. A c and B are independent events. Sometimes, finding the probability of the complementary event will be simpler than that of the event. For this we use the equality the probability of an event = probability of the complementary event. Example 2.4. Consider the case that power to a room comes from three different lines which are named as A, B and C. Let these lines be live in 80%, 90 % and 95% of the cases. It is required to find the probability that there will be light in the room on certain night. Here if A, B and C represent the events that there will be power in the lines and A c, B c and C c respectively represent the complementary events, then P(A c ) = 0.2, P(B c ) = 0. and P(C c ) = The required probability is that of the event A B C. It is the probability of the complement of A c B c C c. Now P(A c B c C c ) = P(A c ) P(B c ) P(C c ) = = So the required probability is 0.00 = or 99.9 %.

6 Example 2.5. Consider A as the event of drawing a red card from a well shuffled pack of cards and B as the event of drawing a number card. It is P( A)= = 36, P(B) 2 52 = 9 3, P ( A B)= 8 52 = 9 26 Now P( A) X P (B)= 2 X 9 3 = 9 26 Here P(A B) = P(A) P(B). So A and B are independent events. Module 3 BAYE S THEOREM Let E, E2,,En be mutually exclusive and exhaustive events and let A be any event. Then the Baye s theorem states that P(E i / A)= P(E i/ A). P( E i ) n = P(E i / A ). P(E i ) = P( A E i) n i= P ( A E i ) The conditional probability and Baye s theorem have a variety of applications in many reallife situations. As an example, consider the following situation. Let there be three surgeons named A, B and C in a hospital who are available respectively on 2, 3 and 2 days a week. Any patient admitted in the hospital will be attended by one of them. The success rates, (the probability that a patient is recovered after treatment), of these surgeons are respectively 0.8, 0.9 and A patient was admitted in critical state for surgery and died after the treatment. We can find the probability that the patient was attended by the surgeon C. Let us denote the events that the surgeons attend a patient respectively as A, B and C. Let the event that a patient dies after treatment be denoted as D.

7 Then from the given facts it follows that P(A) = 2 7, P(B) = 3 7 and P(C) = 2 7. Also, P(D/A) = 0.8 = 0.2, P(D/B) = 0.9 = 0. and P(D/C) = 0.95 = It is required to find P(C/D). Here the events E, E2 and E3 mentioned in Baye s theorem are respectively A, B and C. Also, D is the event mentioned as A in the theorem. So, P(D /C ) X P(C) P(C / D)= P( D/ A )X P( A)+P (D/ B) X P(B)+P(D /C )X P(C ) 0.05 X X X X 2 7 = 8 Module - 4 INFORMATION AND MUTUAL INFORMATION Usually messages/statements convey information. If a statement asserts the occurrence of an event which is very much likely to happen, we say that the statement contains only a small amount of information. On the other hand, if a statement asserts the occurrence of an event which is very rare, then we say that the statement contains a large amount of information. In other words, if an event is highly specific, its probability will be small and the amount of information will be larger. If the event is vaguely told, then its probability will be large and so the amount of information will be small. For example, the amount of information conveyed by the statement it is going to rain soon when told in the months of July and March are widely different. In July, rain is very

8 much likely to occur and so the amount of information is small whereas in March, rain is not likely to occur and so the amount of information is large. Definition 4.. If p is the probability of the occurrence of an event, then the information contained in a statement asserting the occurrence of the event is log p = log p. Here the logarithm is taken with base as 2. Note 4.2. Consider any event with probability p, which is always a number less than or equal to. So, log p will be non-positive or log p will be a non negative number. Moreover, when p is small, the amount of information will be large and when p is large, the amount of information will be small. Suppose that we receive from the computer as output a binary digit which is either 0 or with equal probability of occurrence. If it is told that the output is surely, then the amount of information received is log =. 2 For the output 0 also the same will be the amount of information. We may take the unit for the amount of information as a bit. Now if we receive 32 binary digits from a computer as output and if all the 2 32 possibilities are equally likely, then the information we receive is log = 32 bits In the following example, it is shown how an information is processed from the given data. Example 4.3. Consider the problem of examining the likelihood that there will be a -hour examination when the teacher is scheduled to go out of the town. Let S = {x, x2, x3, x4 } be the sample space, where x : teacher is out of town and examination is given x2 : teacher is out of town and examination is not given x3 : teacher is in town and examination is given x4 : teacher is in town and examination is not given Let p(x ) =, 2 p(x2 ) = 6, p(x3 ) = 3 6 and p(x4 ) = 4. Let A denote the event that an examination is given, and B the event that the teacher is

9 out of town. Note that p(a ) = = 6 and P(A / B ) = = 8 9 The information needed to determine that an examination will be given is log p(a) = log 6 = log + log 6 = = 0.54 bits and the information provided by the fact that the teacher is out of town on the fact that an examination will be given is I(A, B) = log + log = log + log 6 + log 8 log 9 = = 0.37 bits. Let C denote the event that the teacher is in town. since P(A/C) = = we have I(A, C) = log 6 + log 3 7

10 = log + log 6 + log 3 log 7 = = 0.69 bits. The fact that the teacher is in town makes it less likely that an examination will be given. Consequently, the mutual information provided by the presence of the teacher on the occurrence of an examination is a negative quantity. Exercises The following questions are provided for you to practise.. What is the number of different outcomes when four fair dies are rolled? 2. A student randomly marks the answers True or False in an examination consisting of 0 questions. What is the probability that the answers are alternatively different? 3. When four boys and two girls are seated around a table, what is the probability that the two girls will not sit together? 4. When three digit numbers are formed using the digits, 2, 4, 5, 6 and 7, what is the probability that a number randomly selected is a multiple of 4? 5. Let p be the probability that a man aged 60 will die within one year. If there are 0 such men, what is the probability that at least one will die within one year? References The following books are recommended as reference.. Elements of Discrete Mathematics, C.L.Liu., Tata McGraw-Hill Publishing Company Ltds. 2nd edition 3 2. Miller & Freund s Probability and statistics for Engineers, 5th edition, Richard a Johnson, Prentice Hall India Private Limited, 200. Hope you have followed this session. We will continue our discussion with Relations and Functions in the next chapter. Till then Good Bye.

Mathematics. ( : Focus on free Education) (Chapter 16) (Probability) (Class XI) Exercise 16.2

Mathematics. ( : Focus on free Education) (Chapter 16) (Probability) (Class XI) Exercise 16.2 ( : Focus on free Education) Exercise 16.2 Question 1: A die is rolled. Let E be the event die shows 4 and F be the event die shows even number. Are E and F mutually exclusive? Answer 1: When a die is