STA 1311 Probability - PDF

PROBABILITY [Course Code: STA 1311; Unit: 3] Compiled by: Dr. Abdulazeez G. Ahmad; Dr. Ibrahim Babura Mal. Adamu M. Umar, Dr. Abel I. Shaba, Mr. David A. Oluyori Mr. Khalid S. Ahmad, Mr. Tijani Mustapha Mr. Hassan Muhammad Cont...

PROBABILITY [Course Code: STA 1311; Unit: 3] Compiled by: Dr. Abdulazeez G. Ahmad; Dr. Ibrahim Babura Mal. Adamu M. Umar, Dr. Abel I. Shaba, Mr. David A. Oluyori Mr. Khalid S. Ahmad, Mr. Tijani Mustapha Mr. Hassan Muhammad Contents 1 Permutation and Combination 3 1.1 Fundamental Principle of Counting................................. 3 1.2 Addition (Disjunctive) Rule:..................................... 3 1.3 Product (Sequential Counting) Rule:................................ 4 1.4 Permutation (Ordered Selection).................................. 6 1.4.1 Permutation of different objects with repetition...................... 7 1.4.2 Permutation of n objects, not all different......................... 7 1.4.3 Permutation with restrictions on some of the quantities................. 8 1.5 Combination (Unordered Selection)................................. 9 1.5.1 Combination of different objects with no repetition but with restriction........ 11 1.5.2 Combination of different objects of different types with no repetition.......... 12 1.5.3 Combination of different objects of different types, with no repetition but with restrictions 12 1.6 Try Outs............................................... 14 2 Concept and Principles of Probability 16 2.1 Definition of Terms.......................................... 16 2.2 Addition Law of Probability..................................... 21 2.3 Multiplication Law of Probability.................................. 22 2.4 Conditional Probability....................................... 24 2.5 Independent Events......................................... 25 2.6 Baye’s Theorem (Inverse Probability)............................... 26 2.7 Try Outs............................................... 28 3 Random Variable 30 3.1 Introduction.............................................. 30 3.2 Expected Value of Random Variable................................ 32 3.3 Exercises............................................... 33 4 Discrete Probability Distribution 35 4.1 Discrete Probability Distribution.................................. 35 4.2 Continuous Random Variables.................................... 36 4.3 Continuous Probability Distribution................................ 37 4.4 Cumulative Distribution Function................................. 37 1  4.5 Tryouts................................................ 39 5 Special Probability Distributions 40 5.1 Bernoulli Distribution........................................ 40 5.2 Binomial Distribution........................................ 40 5.2.1 Properties of Binomial Distribution............................ 41 5.3 Poisson Distribution......................................... 44 5.3.1 Recurrence Formular for Poisson Distribution....................... 44 5.4 Normal Distribution......................................... 47 5.4.1 Properties of the Normal Curve............................... 48 5.5 Poisson Approximation to Binomial................................ 55 5.6 Uniform Distribution......................................... 58 5.6.1 Uniform Distribution For Discrete Random Variable................... 58 5.6.2 Uniform Distribution for Continuous Random Variable.................. 58 5.6.3 Graph of a Uniform Distribution.............................. 58 5.6.4 Properties of Uniform Distribution............................. 59 5.6.5 Real Life Examples of Uniform Distribution........................ 59 5.7 Geometric Distribution........................................ 60 5.7.1 Characteristics of a Geometric Distribution........................ 60 5.7.2 Probability Distribution of a Geometric Distribution................... 60 5.7.3 Real life Examples of Geometric Distribution....................... 61 5.8 Hypergeometric Distribution.................................... 62 5.8.1 Conditions for the Application of Hypergeometric Distribution............. 62 5.8.2 Probability Distribution of Hypergeometric Distribution................. 62 5.9 Exponential Distribution....................................... 64 5.9.1 Probability Distribution of Exponential Distribution................... 65 5.9.2 Properties of Exponential Distribution........................... 65 5.9.3 Graph of an Exponential Distribution........................... 65 5.10 Multinomial Distribution...................................... 66 5.10.1 Assumptions of Multinomial Distribution......................... 67 5.10.2 Probability Distribution of Multinomial Distribution................... 67 6 Sampling Theory 69 6.1 Definition of Terms.......................................... 69 6.2 Types of Sampling Theory...................................... 70 6.3 Random Sampling.......................................... 70 6.3.1 Advantages of Random Sampling:............................. 70 6.3.2 Disadvantages of Random Sampling:............................ 71 6.4 Purposive Sampling:......................................... 71 6.4.1 Advantages of Purposive Sampling............................. 71 6.4.2 Disadvantages of Purposive Sampling........................... 71 6.5 Stratified Sampling:......................................... 72 6.5.1 Advantages of Stratified Sampling............................. 72 6.5.2 Disadvantages of Stratified Sampling............................ 72 2  6.6 Systematic Sampling:........................................ 72 6.6.1 Systematic Sampling Algorithms.............................. 73 6.6.2 Advantages of Systematic Sampling:............................ 73 6.6.3 Disadvantages of Systematic Sampling:.......................... 73 6.6.4 Sampling Distribution.................................... 73 7 General Practice Questions 75 Learning Outcome At the end of this course, students should be able to: Understand Basic concepts like Probability, Permutation and combination Understand random variable in the case of discrete and continuous events. Learn about various probability distribution and their real life application to life and business. Deploy the use of sampling tool when dealing with a particular population. Introduction to the Probability The statement ”It will rain today” can be both true or false depending on the weather and the season the speaker is currently in. It will be false to say there will be rain during harmattan (especially around Decem- ber) in the Jigawa for example. Similarly, it will be very true to say there will be rain in August because it is known to be the peak of the rainy season. That is just an example of the likelihood of an event which is called Probability. This course teaches the mathematics of possibilities which is an obvious facts in games, betting, final score in a match, horse race, Moto G.P., Formular 1 etc. Events of life as known to be very probabilistic in nature. This course will prepare students better in handling the twists and turns that comes in life and work as a results in linearity (success) and non-linearity (failure) of the occurrence or non-occurrence of events. Game theory is zero-sum in nature whereby your loss is someone’s gain and your gain is somebody’s loss. Your grades are dependent on outcome of your performance in the test and exams. These and many more are very significant to the study. Wishing you a happy reading! 2 Chapter 1 Permutation and Combination Combinatorics is the branch of mathematics that seek to answer questions without enumerating all possible cases. It depends on two elementary rules namely: the sum (addition/disjunctive) and product (multiplica- tion) rules. 1.1 Fundamental Principle of Counting The Fundamental Principle of Counting states that if one thing can be done in a different ways, and when it is done in any one of these ways, a second thing can be done in b different ways, the third thing can be done in c ways · · · then all the things in succession can be done in a × b × c × · · · different ways. Example 1.1.1 Determine the number of permutations of the letter in the word ”BANANA” Solution: Recall that, the number of permutations P on n things taken all at a time. [P (n, n)] of which n1 are alike, n2 others are alike, n3 others are alike etc n! P (n, n) = n1 !n2 !n3 ! · · · With n1 + n2 + n3 + · · · = n B = n1 = 1, A = n2 = 3, N = n3 = 2, the sum of which (i.e. n1 + n2 + n3 = 6). 6! 6 × 5 × 4 × 3! P (6, 6) = = = 60 (1!3!2!) 1!3!2! Example 1.1.2 5! − 4! = (5 × 4!) − 4! = 4!(5 − 1) = 4!(4) = 4 × 3 × 2 × 1 × (4) = 96. In line with the above Fundamental Principle of counting comes in two flavours namely: The Sum Rule (The principle of disjunction counting) and The Multiplication rule (The principle of sequential counting). The description of these with examples follows in the next two sections. 1.2 Addition (Disjunctive) Rule: If X is the union of disjoint non-empty subsets S1 , S2 , · · · , Sn−1 , Sn then |X| = |S1 |+|S2 |+· · ·+|Sn−1 |+|Sn |. We may emphasize that the subsets S1 , S2 , · · · , Sn−1 , Sn must have no elements in common but since 3 X = S1 ∪ S2 ∪ · · · ∪ Sn−1 ∪ Sn each element of X is in exactly one of the subsets Si. In other words S1 , S2 , · · · , Sn−1 , Sn is a partition of X. But if subsets S1 , S2 , · · · , Sn−1 , Sn were allowed to overlap then a more profound principle will be required - The Principle of Inclusion and Exclusion which is stated as follows: ”If E1 , E2 , · · · , En−1 , En−2 are mutually exclusive events and E1 can happen in e1 ways, E2 can happen in e2 ways, · · · En can happen in en ways, then E1 or E2 or · · · or En can happen in e1 + e2 + e3 + · · · + en−1 + en ways.” Inaddition, the sum rule can be formulated in terms of choices: If an object can be selected from a reservoir in e1 ways and an object can be selected from a separate reservoir in e2 ways, then the selection of one object from either one reservoir or the other can be made in e1 + e2 ways. Example 1.2.1 If there are 18 boys and 12 girls in a class, there are 18 + 12 = 30 ways of selecting 1 students (either a boy or a girl) as Class representative. Example 1.2.2 Suppose E is the event of selecting a prime number less than 10 and F is the events of selecting an even number less than 10. Then E can happen in 4 ways and F can happen in 4 ways. But because 2 is an even prime, EorF can happen in only 4+4-1=8-1=7 ways. 1.3 Product (Sequential Counting) Rule: If S1 , S2 , · · · , Sn−1 , Sn are nonempty sets, then the number of elements in the Cartesian product S1 × S2 × S3 × · · · × Sn is the product Yn |Si | = |S1 × S2 × S3 × · · · × Sn | (1.1) i=1 The latter can be best illustrated with a tree diagram where S1 = {a1 , a2 , a3 , a4 , a5 } and S2 = {b1 , b2 , b3 }. From the diagram above, we see that there are 5 branches in the first stage corresponding to the 5 elements of S1 and to each of these branches there are three branches in the second stage corresponding to the 3 elements of S2 giving a total of 15 branches altogether. More generally, if a1 , a2 , · · · , an are the n distinct Sn elements of S1 and b1 , b2 , · · · , bm are the m distinct elements of S2 , then S1 × S2 = i=1 (ai × S2 ). For if x is an arbitrary element of S1 × S2 then x = (a, b) where a ∈ S1 and b ∈ S2. This, a = ai for some i and Sn b = bj for some j. Thus, x = (ai , bj ) ∈ (ai × S2 ) and therefore x ∈ i=1 (ai × S2 ). 4 In terms of choices, the product rule can be formulated as follows: If an event can occur in m ways and the second event can occur in n ways, and if the number of ways the second event occurs does not depend upon how the first event occurs, then the two events can occur simultaneously in mn ways. More generally, if Ei (i = 1, 2 · · · k) are k events and if E1 can occur in n1 ways, E2 ways in n2 ways (no matter how Ei occurs), E3 can occur in n3 ways (no matter how E1 and E2 occur) · · · Ek can occur in nk ways (no matter how the previous k − 1 events occur), then the k events can occur simultaneously, in n1 , n2 , · · · nk−1 , nk ways. Example 1.3.1 A book shelf holds 6 different English books, 8 different French books and 10 different Ger- man books. Find: (i) how many ways can we select 3 books, 1 in each language. (ii) how many ways can we select 1 book in any of the languages. (iii) how many ways can we select an English and a French book (iv) how many ways can we select an English and a German book (iv) how many ways can we select a French book and a German book (v) how many ways can we select. (vi) How many ways can we select 2 books of 2 languages. Solution: (i) There are 6 × 8 × 10 = 480 ways of selecting 3 books, one in each language (ii) There are 6 + 8 + 10 = 24 ways of selecting 1 book in any of the languages. (iii) An English and French book can be selected in 6 × 8 = 48 ways. (iv) An English and German book can be selected in 6 × 10 = 60 ways. (v) A French and German book can be selected in 8 × 10 = 80 ways. (vi) There are 48 + 60 + 80=188 ways of selecting 2 books of 2 languages. Example 1.3.2 If each of the 8 questions in a multiple-choice examination has 3 answers (1 correct and 2 wrong), the number of ways of answering all questions is 38 = 6561. Example 1.3.3 Suppose that the license plates of a certain state require 3 English letters followed by 4 digits. (a) How many different plates can be manufactured if repetition of letters and digits are allowed? (b) How many plates are possible if only the letters can be repeated? (c) How many are possible if only the digits can be repeated? (d) How many are possible if no repetitions are allowed at all? Proof 1.3.4 (a) 263 · 104 since there are 26 possibilities for each of the 3 letters and 10 possibilities for each of 4 digits. (b) 263 · 10 · 9 · 8 · 7 (c) 26 · 25 · 24 · 104 (d) 26 · 25 · 10 · 9 · 8 · 7 5 1.4 Permutation (Ordered Selection) Permutation is an ordered arrangement of a set of objects. Suppose that we are given n distinct objects and we wish to arrange r of these objects in a line. Since there are n − ways of choosing the first object, (n − 1) − ways of choosing the second object, · · · , and finally (n − r + 1) − ways of choosing the rth object. It follows by the fundamental principle of counting that the number of arrangements or permutations is n! n(n − 1)(n − 2) · · · (n − r + 1) =n Pr = (n − r)! As a particular case, if the number of different arrangement of distinct objects, taken all at a time (i.e. one time) is n!. This is called the permutation of n objects and this can be expressed as n Pn = n!. As earlier explained, n Pn = n! is a situation where all the objects that are to be arranged are distinct. Now suppose it is required to find the number of arrangements that can be formed from a collection of n objects of which n1 are of one type, n2 are of a second type, · · · nk are of a k th type, with n1 + n2 + n3 + · · · + nk = n. Then the permutations of the objects is given by n! n1 ! + n2 ! + · · · + nk−1 ! + nk ! Example 1.4.1 Find all the arrangements of two letters chosen out of (a, b, c) Solution: (a) Bad Method: ab, ba, ac, ca, bc, cb = 6, 3! 3! (b) Good Method: 3 P2 = (3−2)! = 1! = 3 × 2 × 1 = 6 ways. The first method is bad because it would take a long time using it to calculate the possible arrangements when the number of objects are large. 23·22·21·20 Example 1.4.2 Express 4·3·2·1 in factorial form? Solution: The numerator 23 · 22 · 21 · 20 · 19 · 18 23! 23 · 22 · 21 · 20 = =. 19 · 18 19! Now let’s combine both 23 · 22 · 21 · 20 4·3·2·1 8! Example 1.4.3 Evaluate 6!2! Solution: 8! 8·7·6·5·4·3·2·1 8.7 = = = 4 × 7 = 28. 6!2! 6·5·4·3·2·1 2.1 Example 1.4.4 In how many ways can the first, second and 3rd positions be awarded in a race of 8 compe- titions without giving more than one to the same competitor. Solution: The total number of ways is given by: 8 8! 8! 8 × 7 × 6 × 5! P3 = = = = 336 ways. (8 − 3)! 5! 5! 6 1.4.1 Permutation of different objects with repetition Let n, r ∈ Z+ , the permutation of n different objects taking r at a time with repetition is given by nr. Example 1.4.5 In how many ways can the letters a, b, c, d be arranged taking two at a time assuming repetition is allowed. Solution: Taking the relation nr , where n = 4 and r = 2 ⇒ 42 = 16 ways. 1.4.2 Permutation of n objects, not all different Example 1.4.6 How many ways can the letters of the words: (i) ENGINEERING (ii) CONSONANTS be arranged? Solution: (i) In the word ”ENIGINEERING”, we have 3 E’s, 3 N’s, 2 G’s and 2 I’s. Therefore, 11! 11 × 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 T he number of permutation = = = 277, 200 ways 3!3!2!2! 6×6×2×2 (ii) In the word ”CONSONANTS”, we have 3 N’s, 2 O’s and 2 S’s. Thus, 10! T he number of possible arrangement = = 151, 200 ways 3!2!2! Example 1.4.7 Find the number of ways of arranging the letter EIGHT (a) If all are taken at a time (b) If 2 are taken at a time. Solution: Recall that there are 5 distinct words in the letter (a) The required number of arrangements = 5! = 5 × 4 × 3 × 2 × 1 = 120 (b) The required number of arrangements if two are taken at a time is 5 5! 5! 5×4×3×2×1 P2 = = = = 20 (5 − 2)! 3! 3×2×1 Example 1.4.8 Find the number of ways of arranging the letter ”ARRANGE” taking: (a) all at a time (b) four at a time? Solution: There are 7 words in the letter out of which two are repeated twice. (a) If all are taken at a time, then the required arrangements 7! 7·6·5·4·3·2·1 = = 1260 ways. 2!2! 2·1×2·1 (b) If four letters are taken at a time, then the number of arrangements 7! 7·6·5·4 = = 210 ways. (7 − 4)!2!2! 2·1×2·1 Example 1.4.9 (i)How many 6-letter words can be formed from the word ”NUMBER”. (ii) How many four letter words can be formed from the 6 - letter word ”NUMBER” 7 Solution: (i) There are P (6, 6) = 6! = 6 × 5 × 4 × 3 × 2 × 1 = 720. Thus, there are 720 ”6-letter” words that can be formed from the letters of the word. (ii) Here, n = 6, r = 4. Thus, there are 6! 6 × 5 × 4 × 3 × 2! P (6, 4) = = = 360 (6 − 4)! 2! Thus there are 360 ”four-letter” words that can be formed from the word ”NUMBER”. 1.4.3 Permutation with restrictions on some of the quantities Consider permutations of n objects taking r at a time where q of the n objects occupy fixed positions. If the q objects cannot be rearranged then the q space can be filled in only one way and we have n−q Pr−q ways of arranging the remaining n − q objects taking r − q at a time. If the q objects can be arranged in the occupied spaces, we have q!×n−q Pr−q ways of arranging the n objects taking r at a time with restriction on q. The central idea here is that ”Permutation of n different things, taking r at a time in which q particular thing is to be excluded in each arrangement.” Now, consider the permutation of n objects taking r at a time. Since the q is to be excluded from the n−objects, then we are left with n − q things to select r things from, the number of possible arrangements is given by: n−q Pr (1.2) Example 1.4.10 Suppose we have the letters A, B, C, D, E, F, G, H, I, J. Find the number of possible arrangements of three letters from the total letters if: (a) A must be the first letter (b) A and D must be the first two letters. Solution: (a) To arrange so that ”A” must be the first out of 3. A, [x1 , x2 ] where x1 and x2 are the other 2 numbers. This corresponds to the relation: q! ×n−q Pr−q where q = 1, n = 10 and r = 3. We have that 1! ×10−1 P3−1 =9 P2 = 72 ways. (b) To arrange the three letter such that A and D must be the first two letters. [A, D], x; where x is the third letter. Since A and D must be together, the remaining number of ways is n−q Pr−q where n = 10, q = 2, r = 3 but A and D are non-identical, so can also be rearranged in 2! ways. The the number of possible arrangement is given by: q! ×n−q Pr−q = 2! ×10−2 P3−2 = 2! ×8 P1 = 2 × 8 = 16 ways. Problem 1.4.11 How many numbers can be formed out of digits 0, 1, 2, 3, 4, 5. Assuming no digits appear more than once. [Ans: (i) 6 ways (ii) 25 ways (iii) 100 ways (iv) 300 ways (v) 600 ways (vi) 1091 ways.] 8 1.5 Combination (Unordered Selection) Combination simply means an unordered selection of the set of objects. Suppose we are interested in selecting (choosing) r objects from a set of n ≥ r objects without regard to order. The set of r objects being selected is called the Combination of r objects. We see from the latter that an unordered selection of r out of n elements of X with r elements is an r-combination of X. The number of r-combinations or r-subsets of a set of n distinct objects denoted by C(n, r) [”n choose r”]. For each r-subset of X there is a unique complementary (n − r) subsets, whence the important relation: C(n, r) = C(n, n − r) (1.3) To evaluate C(n, r) note that an r-permutations of an n-set X is necessary a permutation of some r− subset of X. Moreover, distinct r -subsets generate distinct r -permutations. Hence by the sum rule, P (n, r) = P (r, r) + P (r, r) + P (r, r) + · · · + P (r, r) (1.4) The number of terms on the right is the number of r-subsets of X; i.e. C(n,r). Thus P (n, r) = C(n, r)P (r, r) = C(n, r)r! · · · (1.5) The following results summarizes the above assertions: n P (n, r) = (1.6) (n − r) and P (n, r) n! C(n, r) = = (1.7) r! (n − r)!r! Thus the formular for combination is equation (7) which can also be written as: P (n, r) n! C(n, r) = = (1.8) r! (n − r)!r! Example 1.5.1 Evaluate C(10, 4)? Solution: Here n = 10 and r = 4. Using the formula for combination we have, 10! 10! 10 × 9 × 8 × 7 × 6! 10 × 9 × 8 × 7 5040 C(10, 4) = = = = = = 210 (10 − 4)!4! 6!4! 6! × 4! 4×3×2×1 24 Example 1.5.2 From a class consisting of 12 Computer science majors, 10 Mathematics majors and 9 Statistics majors. A committee of 4 Computer majors, 4 Mathematics majors and 3 Statistics majors is to be formed. How many number of ways can we form the committee? Solution: n(C) = 12, n(M ) = 10, n(S) = 9. We are to choose (r out of n), r(C) = 4, r(M ) = 4, r(S) = 3. There are 12! 12! C(12, 4) = = = 495 (12 − 4)!4! 8!4! 9  10! 10! C(10, 4) = = = 210 (10 − 4)!4! 6!4! 9! 9! C(9, 3) = = = 84 (9 − 3)!3! 6!3! Thus, there are 495 ways to choose 4 majors from computer science, 210 ways to choose 4 majors from Mathematics and 84 ways to choose 3 Statistics majors. Therefore, by product rule, the number of ways of forming the committee is 495 × 210 × 84 = 8, 731, 800. Example 1.5.3 How many baseball teams of 9 members can be chosen from away 12 boys, without regard to the position played by each member? Solution: Since there is no regard for position, this is a combination problem (but when order is important, it is permutation). Thus the number of combinations of 9 players taken out of 12 at a time is 12! 12! (12 × 11 × 10 × 9!) 12 × 11 × 10 1320 C(12, 9) = = = = = = 220 (12 − 9)!9! 3!9! 3! × 9! 3×2×1 6 Thus, there are 220 possible teams that can be formed. Example 1.5.4 There are 15 married couples in a party. Find the number of ways of choosing a woman and a man such that two are (i) married to each other (ii) not married to each other. Solution: (i) A woman can be chosen in 15 ways and once a woman is chosen, her husband is automatically chosen as well. So the number of ways of choosing a married couple is 15. (ii) A woman can be chosen in 15 ways. Among the 15 men, one is her husband. Out of the 14 other men, one can be chosen in 14 ways. Using product rule, the number of ways of choose a man and a woman that are not married to each other is (15)(14) = 210 ways. Example 1.5.5 Simplify the following: (a) 5 C0 (b) 5 C5 (c) 7 C5 Solution: (a) 5 5! 5! 5! 5! C0 = = = = =1 (5 − 0)!0! 5!0! 5! × 1 5! (b) 5 5! 5! 5! 5! C5 = = = = =1 (5 − 5)!5! 0!5! 1 × 5! 5! (c) 7 7! 7! 7! 7×6 C5 = = = = = 21 (7 − 5)!5! 5!2! 2! × 5! 2×1 Example 1.5.6 In how many ways could a band of 2 singers and 5 drummers be selected from 4 singers and 6 drummers? Solution: 2 singers could be selected from 4 singers in 4 C2 ways. 5 drummers could be selected from 6 drummers in 6 C5 ways. Therefore, 2 singers and 5 drummers could be selected from 4 singers and 6 drummers in 4 4! 6! C2 ∩ 6 C5 ways =4 C2 ×6 C5 = × = 6 ways. 2!2! 5!1! We discuss Combinatorics under three different subtopics: 10 1.5.1 Combination of different objects with no repetition but with restriction Let n, r ∈ Z+ , r ≤ r ≤ n. Consider the combination of n different objects taking r objects at a time: (n−q) (a) If q must be included, the number of selection r out of n objects is Cr−q. (n−q) (b) If q must be excluded, the number of possible selection is Cr. Example 1.5.7 A committee of 4 is to be selected from a group of 5 men and 3 women, in how many ways can this be done if the Chairman of the committee must be a man? Solution: Since the Chairman of the committee must be a man, that is, a man must be included in our selection and there are three women. Automatically the fourth person cannot be a woman that is, the fourth person is the man which is the Chairman of the committee. Hence the restriction is discarded. Number of Men = 1, Number of Women = 3. Total number of people in the group = 5 + 3 = 8. The number of people to be selected = 4. 8! 8 × 7 × 6 × 5 × 4! 8×7×6×5 T otal number of ways of selection =8 C4 = = = = 70 ways. (8 − 4)!4! 4!4! 4×3×2×1 Example 1.5.8 In how many ways can two letters from A, B, C be selected if: (i) there is no restriction (ii) the selections must contain A (iii) B must not be selected. Solution: (i) since we are selecting 2 letters from 3 3 3! 3! C2 = = = 3 ways. (3 − 2)!2! 1!2! Alternatively, since there is no restriction, the number of possible selection = {(A, B), (A, C), (B, C)} = 3 ways. (ii) If the selection must contain A, it means that A must be included in our selection. The number of possible selections = {(A, B), (A, C)} or this selection conform with the relation n−q Cr − q where n = the total number of objects, r = the number of objects to be selected, q = the number of objects that must be included. We have that n = 3, r = 2 and q = 1. Thus 3−1 2! 2×1 C2−1 =2 C1 = = = 2 ways. (2 − 1)!1! 1!1! (iii) B must not be selected implies that B must be excluded. The number of possible selection is {(A, C)} or using the relation n−q Cr ; the number of objects to be excluded do not affect the total number of objects we are selecting: 3−1 2! 2! 2! 2! C2 =2 C2 = = = = = 1 ways. (2 − 2)!2! 0!2! 1 × 2! 2! Problem 1.5.9 In how many ways can 4 fruits be selected out of 10 fruits of different sizes such that: 11  (a) the smallest must be included (b) the largest must be excluded. [Hint: The inclusion and exclusion of smallest and the largest means that out of 10 objects one must be included and excluded respectively] 1.5.2 Combination of different objects of different types with no repetition Let m, r, s ∈ Z+ , x ≤ m and s ≤ n. Suppose we can select r and s objects from m and s objects respectively. The number of ways of selecting r and s objects from m and n objects is given by m Cr ×n Cr. Example 1.5.10 How many ways can a committee of 4 students and 2 teachers be formed from 12 students and 7 teachers? Solution: The number of ways of selecting 4 out of 12 students = 12 C4. The number of selecting 2 out of 7 teachers = 7 C2. Therefore, the number of ways of selecting a Committee comprising of 4 students and 2 teachers is given by 12 12! 7! 12! 7! C4 ×7 C2 = × = × = 10395 ways. (12 − 4)!4! (7 − 2)!2! 8!4! 5!2! Example 1.5.11 A mixed hockey team containing 5 men and 6 women is to be chosen from 7 men and 9 women. In how many ways can this be done? Solution: The number of ways of choosing 5 out of 7 men = 7 C5 The number of ways of choosing 6 out of 9 women = 9 C6. The number of ways of forming a mixed hockey team containing 5 men and 6 women is 7 7! 9! C 5 ×9 C 6 = × = 1764 ways. 2!5! 3!6! 1.5.3 Combination of different objects of different types, with no repetition but with restrictions Let m, n, r, q, v, s ∈ Z+ , q ≤ r ≤ m and v ≤ s ≤ n. Suppose we can select r and s objects from m and n objects respectively. (i) If q must be included, the number of ways of selecting r objects from m is m−q Cr−q Also, if v must occur, the number of ways of selecting s objects from n is n−v Cs−v Therefore, if q and v must be included in r and s respectively, the number of ways of selecting r and s objects from m and n respectively is: m−q Cr−q ×n−v Cs−v 12  (ii) If q must be excluded, the number of ways of selecting r objects from m is m−q Cr. Also, if v must be excluded, the number of ways of selecting s objects from n is n−v Cs. Therefore, if q and v must be excluded from r and s respectively, the number of ways of selecting r and s objects from m and n respectively is: m−q Cr ×n−v Cr (iii) If q must be included from the selection of r from m and v must be excluded from the selection of s objects from n; the number of ways of selection is: m−q Cr−q ×n−v Cs and vice versa. Example 1.5.12 A committee of 5 members is to be formed from 3 females and 4 males. In how many ways can these be formed it: (a) Atleast one female is included as a member (b) Atleast one male is included as a member Solution: (a) Since we are selecting 5 members in all, atleast one female will include 1, 2 or 3 females such that we can have: (i) 1 female and 4 males (ii) 2 females and 3 males (iii) 3 females and 2 males. (i) Selection of 1 female and 4 males from 3 females and 4 males can be represented as 3 3! 4! C1 ×4 C4 = × = 3 ways. 2!1! 0!4! (ii) Selection of 2 females and 3 males from 3 females and 4 males is 3 3! 4! C2 ×4 C3 = × = 12 ways. 1!2! 1!3! (iii) Selection of 3 females and 2 males from 3 females and 4 males is 3 3! 4! C3 ×4 C2 = × = 6 ways. 0!3! 2!2! (b) Since we are selecting 5 members in all, atleast one male will include 1, 2, 3 or 4 males but selecting one male will not be possible because the number of females is not up to 4. So our selection can only include 2, 3 or 4 males. (i) Selection of 2 male and 3 males from 4 males and 3 females can be represented as 4 4! 3! C2 ×3 C3 = × = 6 ways. 2!2! 0!3! (ii) Selection of 3 males and 2 males from 4 males and 3 females is 4 4! 3! C3 ×3 C2 = × = 12 ways. 1!3! 1!2! 13 (iii) Selection of 4 males and 1 females from 4 males and 3 females is 4 4! 3! C4 ×3 C1 = × = 3 ways. 0!4! 2!1! ∴ Total number of ways of forming this committee including atleast 2 males is 6 + 12 + 3 = 21 ways. Problem 1.5.13 Out of 5 Mathematicians and 7 Chemists, a committee consisting of 2 Mathematician and 3 Chemists is to be formed. In how many ways can these be formed if: (a) Any Mathematician and any Chemist can be included (b) One particular Chemist must be on the committee (c) Two particular Mathematician cannot be in the committee [Hint: (a) 5 C2 ×7 C3 = 350 ways (b) 5 C2 ×6 C2 = 150 ways (c) 3 C2 ×7 C3 = 105ways] 1.6 Try Outs (1) Evaluate the followings (a) 6! + 11! (b) 2! + 3! (c) 5P3 (d) 8P3 (e) C(8 + 3) C(10,4) (f) C(8,2) C(5,3) (g) C(3,2) C(15,7) (h) C(10,5) (2) Determine the number of permutations of the letter in the word (i)”OSMOSIS” (ii) ”SOYYAYA” (3) How many 10-letter words can be formed from the word ”PRESIDENCY” (4) How many Five letter words can be formed from the 10-letter word ”PRESIDENCY” (5) How many 6-letter words can be formed from the word ”TINUBU” (6) How many 5- letter words can be formed from the 9-letter word ”BABANGIDA”. (7) A cross section of FUT Babura students consisting of 15 Computer science majors, 12 Mathematics majors and 10 Statistics majors. A committee of 7 Computer majors, 5 Mathematics majors and 4 Statistics majors is to be formed. How many number of ways can we form the committee? 14  (8) How many baseball teams of 7 members can be chosen from away 13 boys, without regard to the position played by each member? (9) There are 12 married couples in a party. Find the number of ways of choosing a woman and a man such that two are (i) married to each other (ii) not married to each other. (10) Five digits number are formed from digits 5, 6, 7, 8 and 9. How many of such numbers can be formed if repetition of digits is: (i) allowed (ii) not allowed. (11) Five females and eight male teachers applied for 4 vacancies in a school and all of them are equally qualified. Find the number of ways of employing the 4 teachers if: (a) atleast 2 of them are females (b) atmost 3 of them are males. (12) A number of four different digits is formed by using the digits 1, 2, 3, 4, 5, 6, 7 in all possible ways, each digit occurring once only. Find how many of such numbers can be formed. (13) In how many ways can 9 people be seated on a bench if only 3 places are available. (14) Assume that you and 15 of your friends have formed a company called Net-Media, an internet music and video provider. A committee consisting of a President, a Vice President and a three-member executive board will govern the company. In how many different ways can this committee be formed? (15) In how many ways can 5 children be arranged in a line such that: (a) two particular children of them are always together (b) two particular children of them are never together. (16) A boy has three library tickets and 8 books of his interest in the library of these 8, he does not want to borrow Mathematics part II, unless mathematics part I is also borrowed. In how many ways can he choose the three books to be borrowed? (17) n P3 Simplif y nC + n P0 2 15 Chapter 2 Concept and Principles of Probability Probability is a concept which numerically measures the degree of certainty (likelihood)/ uncertainty (un- likelihood) of the occurrence of an event. If an event A can happen in m ways and fail in n ways, all these ways being equally likely to occur. Then, N umber of f avourable cases m P b(Ahappens) = = (2.1) T otal number of mutually exclusive and equally likely cases m+n Also, N umber of unf avourable cases n P b(A will not happen) = = (2.2) T otal number of mutually exclusive and equally likely cases m+n Let p be the probability that A will happen and q be the probability that A will not happen then, m n m+n p+q = + = =1 m+n m+n m+n or p+q =1. 1 For example, in the toss of a coin, the probability of a Head is 2 and Tail is 12. 2.1 Definition of Terms Sample Space: The set of all possible outcomes of a single performance of an experiment is a Sample Space or an Exhaustive Event. Each outcome of the said experiment is called Sample Point. In the case of a football match, the sample space, S = {W in, Lose, Draw}, while the sample points are: Win, Loss, Draw. Also, in the toss of a coin, the Sample Space, S = {Head, T ail} where two outcomes constitute the exhaustive events (Head and Tail) because no other outcomes is possible. Random Experiments: Random experiments are events that produces different results even though they are performed under same 16 conditions. E.g. Tossing a coin or throwing a die is a random events. A Yoruba adage said: ”The rain that falls on sugar cane is the same that falls on bitter leaf”, one is sweet and the other is bitter but the same rain. Students are expose to the same lecture under the same class room but at the end of the semester or session, we have classes of students with different grades ranging from A − F in accordance to the produce (garner). Trials and Events: Performing a random experiment is called Trial and the outcome of an experiment is an Event. Tossing a coin is a trial and the turning up of Head or Tail is an event. Equally-likely Events: Two events are said to be ”equally likely’ is one of them cannot be expected in preference to the other. E.g. in the toss of a coin (die), the probability of any number outcome Either Head or Tail (any number ranging from 1 - 6) turning up is 1/2 (1/6). Independent Events: Two events are said to be independent if the occurrence of one does not affect the other. E.g. the event of getting H in the first coin and getting a T in the second coin is a simultaneous throw of two coins are independent. Mutually Exclusive Events: Two events are mutually exclusive if the occurrence of one excludes (stops) the occurrence of the other. For instance, on tossing a coin, either we get a Head or a Tail but not both. In the race of sperm cells to the egg (ovum) million cells starts but only one hits the egg and become fertilized. Thus the fact that you are born means million of healthy humans in form of sperm cells were lost because of you. You are the lucky one! Example 2.1.1 Indicate whether the following pair of events are mutually exclusive, independent or condi- tional (a) Winning and losing a football match (b) Plastering a house and watching a film (c) Planting maize and harvesting the maize (d) Cook yam and eat the yam (e) Causing trouble and getting arrested (f ) Live together and die together (g) Praising and rebuking a student (h) Being a Polygamist and a monogamist. 17 Compound Events: This is when two or more events occur in composition with each other, the simultaneous occurrence is called a Compound Event. E.g. when a die is thrown, getting a 5 or 6 is a Compound Event. Favourable Events: The events which ensure the required happening. For example, in the events leading to the 2023 general elections in Nigeria, the emergence of any of the top four presidential candidates namely: Asiwaju, Atiku, Peter Obi and Kwankwaso is a favourable events to individuals. Also, in the throw of a die, to have the even number (e.g. 2, 4, 6) is a favourable event over having an odd number (1, 3, 5). Expected Value: If p1 , p2 , · · · pn of the probabilities of the events x1 , x2 , · · · xn respectively. The expected value E(P ) = p1 , p2 , · · · , pn = Σnr=1 pr xr Fig 2.1: Outcomes when a Die is rolled (Top) Example 2.1.2 Find the probability of throwing (a) 5 (b) an even number in a biased 6 faced die. Solution: (a) There are 6 possible ways in which a die can fall but there is only one way the ”number 5” can fall. Therefore, N umber of f avourable way (5) P b(5) = T otal number of equally likely ways (b) Number of ways of having an even numbers (i.e. 2, 4, 6) = 3. The total number of ways of throwing a die is 6. Thus N umber of Even numbers 3 1 P b(Even N umbers) = = = T otal number of equally likely ways 6 2 Example 2.1.3 In the throw of two die. Find the probability of throwing (a) 9, (b) 12, (c) 8, (d) 10. Solution: 18  1 2 3 4 5 6 1 (1,1) (1,2) (1, 3) (1,4) (1,5) (1,6) 2 (2,1) (2,2) (2, 3) (2,4) (2,5) (2,6) 3 (3,1) (3,2) (3, 3) (3,4) (3,5) (3,6) 4 (4,1) (4,2) (4, 3) (4,4) (4,5) (4,6) 5 (5,1) (5,2) (5, 3) (5,4) (5,5) (5,6) 6 (6,1) (6,2) (6, 3) (6,4) (6,5) (6,6) From the table above, we see that the total number of possible outcome in the sample space, S = 6 × 6 = 36 Outcomes. (a) Number of ways of getting a 9 (from the table) are (3, 6), (6, 3), (4, 5) and (5, 4) is 4 ways. Thus 4 Pb(getting a 9) = 36 = 91. (b) Number of ways of getting a 12 (from the table) is (6,6) = 1 way. 1 Pb(getting a 12) = 12. (c) Number of ways of getting an 8 (from the table) are (2,6), (6,2), (3,5) , (5,3), (4, 4) = 5 ways. 5 Pb (getting an 8) = 36 (d) Number of ways of getting a 10 (from the table) are (4,6), (6,4), (5,5) = 3 ways. 3 1 Pb (getting a 10) = 36 = 12. Example 2.1.4 An unbiased coin is tossed three times. The probability of obtaining: (i) two head and tail (ii) three heads (iii) three tails (iv) two tails and one head Solution: The diagram below is best viewed when flipped by 90o Fig. 2.2: Outcomes of 3 Tosses of a Coin Sample Space = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T } = 8 3 1. Pb(2H, 1T) = {HHT, HT H, T HH} = 8 1 2. Pb(3H) = {HHH} = 8 1 3. Pb(3T) = {T T T } = 8 19  3 4. Pb(2T, 1H) = {HHT, T HT, T T H} = 8 Fig 2.3: A Standard 52 Deck of Cards Example 2.1.5 From a pack of 52 cards, One card is drawn at random. Find the probability of getting (a) a King (b) Queen of Spade (c) Jack of Club (d) King of Heart (e) Queen Solution: Using diagrams we explain the components of the 52 cards which can be further subdivided into suits and sub-suits in the following wise: Fig 2.4: Deck of Card Arrangement 4 1 (a) Pb(a King) = 52 = 13. Since there are 4 ways to select the king in the 4 suits. 1 1 (b) Pb(Queen of Spade) = 13 = 13. Since there is only one Queen in Spade. 1 1 (c) Pb(Jack of Club) = 13 = 13. Since there is only one way of selecting Jack in the suit, Club. 1 1 (d) Pb(King of Heart) = 13 = 13. Since there is only one way of selecting a King in the suit, Heart. 4 1 (e) Pb(a Queen) = 52 = 13. Since there are 4 ways to select the Queen in the 4 suits. Example 2.1.6 Hospital records indicated that maternity patients stayed in the hospital for the number of days shown in the distribution 20  No. of days stayed Frequency 3 15 4 32 5 56 6 19 7 5 Total 127 Find the following probabilities that a patient stayed: (a) exactly 5 days (b) less than 6 days (c) atmost 4 days (d) atleast 5 days. Solution: 56 (a) For exactly 5 days x = 5, thus P b(5) = 127 (b) Since less than 6 days is required this means the sum of the 3rd, 4th and 5th days. Thus 15 32 56 103 P b(less than 6 days) = + + = 127 127 127 127 (c) Atmost four days means 4 days and below (i.e. sume of the 3rd and 4th days). Therefore, 15 32 47 P b(atmost 4 days) = + = 127 127 127 (d) Atleast 5 days means 5 days and above that is the sum of the 5th, 6th and 7th days 56 19 5 56 + 19 + 5 80 P b(atleast 5 days) = + + = = 127 127 127 127 127 Example 2.1.7 In a class of 12 students, 5 are boys and the rest are girls. Find the probability that a student selected is a girl. Solution: No. of Boys, n(B) = 5, No. of Girls, n(G) = 12 − 5 = 7, Total No. of Students = 12. N umber of Girls 7 P b(Girl) = = T otal number 12 7 Problem 2.1.8 A bag contains 7 red and 8 black balls. Find the probability of drawing a red ball (Ans: 15 ) 2.2 Addition Law of Probability Theorem 2.2.1 If p1 , p2 , · · · , pn be seperated probabilities of mutually exclusive events, then the probabilities p that any of these events will happen is given by p = p1 + p2 + · · · + pn. Proof Let A, B, C, · · · be the events where the probabilities are respectively p1 , p2 , · · · , pn. Then P (A + B + C + · · · ) = pA + pB + pC + · · · Also, consider the case where two events A and B are not mutually exclusively. The probability of the event either A or B or both occurs is given as P (A ∪ B) = P (A) + P (B) − P (A ∩ B) The supporting diagram of addition probability is below: 21  Fig 2.5: Addition Law of Probability Example 2.2.2 An urn contains 10 black and 10 white balls. Find the probability of drawing 2 balls of the same colour? Solution: Let B = No. of black, n(B) = 10, W = No. of White, n(W ) = 10 and the total balls - 20. Thus 10 C2 P b(Drawing 2 black balls) = 20 C 2 10 10  10  C2 C2 C2 P b(Drawing 2 of same colour) = 20 + 20 2 20 = C2 C2 C2 10   C2 10 × 9 20 × 19 9 P b(Drawing 2 black balls) = 20 =2 ÷ = C2 2×1 2×1 19 Example 2.2.3 Three Machines I, II and III manufacture respectively 0.4, 0.5 and 0.1 of the total produc- tion. The percentage of defective items produced by Machine I, II and III is 2, 4 and 1 percent respectively. What is the total defective items produced by Machine I, II and III? Solution: The defective item produced by Machine I = 0.4×2 100 = 100 0.8 The defective item produced by Machine II = 0.5×4 2 100 = 100 = 50 1 0.1×1 0.1 The defective item produced by Machine III = 100 = 100 0.8 1 0.1 2.9 Total defective items produced by Machine I, II and III = 100 + 50 + 100 = 100 = 0.029 Therefore, the required probability is 0.029. 2.3 Multiplication Law of Probability If there are two independent events the respective probabilities of which are known, then the probability that both will happen is the probabilities of their happening respectively. P (AB) = P (A) × P (B) Example 2.3.1 An article manufactured by Zaf-Magem Company Limited consists of two parts A and B. in the process of manufacturing of part A, 9 out of 100 are likely to be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of B. Calculate the probability that the assembled article will not be defective. (Assuming that the events of finding the part A non-defective and that of B are independent). 22  9 9 Solution: Pb(Part A is defective) = 10 , Pb(Part A not defective) = 1 − 10 = 100−9 91 100 = 100 5 5 Pb(Part B is defective) = 10 , Pb(Part B not defective) = 1 − 10 = 100−5 95 100 = 100 Pb(that the assembled item will not be defective) = P b(A non − def ective) × P b(B non − def ective) 91 95 = 100 × 100 = 0.8645 Example 2.3.2 The probability that Machine A will be performing usual function in 5 years time is 14 , while the probability that Machine B will still be operating usefully at the end of the same period is 13. Find the probability that in the following cases that in 5 years time: (a) Both machines will be performing usual function (b) Neither of them will be operating (c) Only Machine B be operating (d) Atleast one of the machine will be operating Solution: Pb(A operates usefully) = p(A) = 14 , Pb(B operates usefully) = p(B) = 13 Pb(A is not operating) = q(A) = 1 − 14 = 34 , Pb(B is not operating) = q(B) = 1 − 31 = 2 3 1 1 1

(a) Pb(Both A and B operating) = p(A) × p(B) = 4 3 = 12 (b) Pb(Neither A nor B will be operating) = q(A) × q(B) = 43 2 2 1

3 = 4 = 2 (c) Pb(Only B will be operating) = p(B) × q(A) = 13 34 = 14.

1 (d) Pb(Atleast one of the machines will operating) = 1−Pb(Neither is operating) = 1 − 2 = 12. Example 2.3.3 An urn contains 9 balls, 2 of which are Red, 3 Blue and 4 Black. Three balls are drawn from the urn at random. What is the probability that: (a) 3 balls are of different colours (b) 3 balls are of the same colour. Solution: n(R) = 2, n(B) = 4, Total balls = 9. (a) The three balls will be of different colours if one is Red, one is Blue and the other Black. 2 C1 ×3 C1 ×4 C1 2×3×4 2 P b(3 balls of dif f erent colours) = 9C = = 3 84 7 (b) Pb (3 balls of the same colour either 3 blue balls or 3 black balls are drawn) = 3 4 C3 C3 3! 4! 9! 1+4 5 9C + 9C = + ÷ = = 3 3 (3 − 3)!3! (4 − 3)!3! (9 − 3)!3! 12 × 7 84 23 2.4 Conditional Probability Let A be any event in the sample space S and P (A) > 0. The probability that an event B occurs subject to the condition that A has already occurred is called the Conditional Probability of B given that A has already occurred. It is denoted by P (B/A). Similarly, the conditional probability of A given that B has occurred is P (A/B). Let A and B be two events of a sample space S and let p(B) 6= 0. The conditional probability of the event B given that A has occurred is denoted by P (A ∩ B) P (B/A) = P (A) where P (A) 6= 0 (i.e. P (B/A) is defined if and only if P (A) 6= 0 or P (A) > 0). and the conditional probability of the event A given that B has occurred us P (A ∩ B) P (A/B) = P (B) where P (B) > 0 (i.e. P (A/B) is defined if and only if P (B) 6= 0 or P (B) > 0). Theorem 2.4.1 If the event A and B defined on a sample space S of a random experiment are independent then P (A/B) = P (A) and P (B/A) = P (B). Solution: Given two independent events, A and B, P (A and B) = P (A) ∗ P (B) which implies that P (A ∩ B) P (A) ∗ P (B) P (A/B) = = = P (A) P (B) P (B) P (B ∩ A) P (B) ∗ P (A) P (B/A) = = = P (B) P (B) P (B) Example 2.4.2 Let A and B be events with P (A) = 31 , P (B) = 14 , P (A ∩ B) = 1 12. Find (i) P (A/B), (ii) P (B/A), (iii) P (B/Ac ), (iv) P (A ∩ B c ) Solution: (i) 1 P (A ∩ B) 12 1 P (A/B) = = 1 = P (B) 4 3 (ii) 1 P (A ∩ B) 12 1 3 1 P (B/A) = = 1 = × = P (A) 3 12 1 4 (iii) Now, P (B ∩ Ac ) P (B) − P (B ∩ A) ( 14 ) − 12 1 1 3 1 P (B/Ac ) = c = = 1 = × = P (A ) 1 − P (A) (1 − 3 ) 6 2 4 (iv) 1 1 3 1 P (A ∩ B c ) = P (A) − P (A ∩ B) = − = =. 3 12 12 4 24 2.5 Independent Events The events A and B are said to be Independent if P (AB) = P (A) · P (B). Thus in case of independent events the above multiplication theorem becomes P (A ∩ B) = P (A) × P (B) [∵ P (A/B) = P (A) and P (B/A) = P (B).] Example 2.5.1 Given that P (A) = 83 , P (B) = 5 8 and P (A ∪ B) = 34. Find (i) P (A/B) (ii) P (B/A) and (iii) Show whether A and B are independent? Solution: Given 3 3 5 3 5 3 1 P (A ∪ B) = P (A) + P (B) − P (A ∩ B) = = + − P (A ∩ B) =⇒ P (A ∩ B) = + − = 4 8 8 8 8 4 4 (i) Now, 1 P (A ∩ B) 4 1 8 2 P (A/B) = = 5 = × = P (B) 8 4 5 5 (ii) 1 P (A ∩ B) 4 1 8 2 P (B/A) = = 3 = × = P (A) 8 4 3 3 (iii) Again, 1 3 5 15 P (A ∩ B) = and P (A) × P (B) = × =. 4 8 8 64 We know that A and B are independent events, if P (A ∩ B) = P (A) · P (B). Thus, we notice that P (A ∩ B) 6= P (A) · P (B), so the events A and B are not independent. Example 2.5.2 The odds against a student X solving a STA 1311 problem are 8 to 6 and odds in favour of the student Y solving the same problem are 14 to 16. (i) What is the chance that the problem will be solved if they both try independent of each other? (ii) What is the probability that none of them is able to solve the problem? 6 6 Solution: Let A = the event that the student X solves the problem. Thus P (A) = 8+6 = 14. 14 6 14 Let B = the event that the student Y solves the problem. Thus P (B) = 14+16 = 14 = 30. (i) Probability that the problem will be solved = P(Atleast one of them solves the problem) = P (A ∪ B) = 6 P (A) + P (B) − P (A ∩ B) = P (A) + P (B) − P (A) · P (B) = 14 + 14 6 14 73 30 − 14 × 30 = 105. 8 (ii) Probability that none will solve the problem = P (Ac )×P (B c ) = (1−P (A))(1−P (B)) = 14 × 16 30 = 32 105 25 Problem 2.5.3 Two sets of candidates are competing for the positions on the Board of Directors of a Company. The probability that the first set and the second set will win are 0.6 and 0.4 respectively. If the first set wins, the probability of introducing a new product is 0.8 and the corresponding probability if the second set wins is 0.3. What is the probability that the new product will be introduced? [Hint: P(New Product) = P(1st set × New Product) + P(2nd Set × New Product) = P (A1 )P (B/A1 ) + P (A2 )P (B/A2 ) = 0.8 × 0.6 + 0.3 × 0.4 = 0.60] Problem 2.5.4 A candidate is selected for interviews for three positions. For the first position there was 3 candidates, for the second 4 and for the third 2. What is the probability that the candidate is selected for atleast one position? [Hint: P (1st P osition) = 31 , P (2nd P osition) = 14 , P (3rd P osition) = 12. Then P (atleast getting one position) = 1 − P (not getting all the three positions) = 43 ] Problem 2.5.5 An article manufactured by a company consists of two parts A and B. In the process of manufacturing part A, 9 out of 100 are likely to be defective. Similarly, 5 out of 100 are likely to be defective in the manufacture of part B. Calculate the probability that the assembled part will not be defective. 9 5 [Hint: Given, P (X) = 100 and P (B) = 100. Then P(the assembled part will not be defective) = P (X ∩ Y )= 91 95 P (X) × P (Y ) = 100 × 100 = 0.8645] 2.6 Baye’s Theorem (Inverse Probability) If B1 , B2 , · · · , Bn are mutually exclusive events with P (Bi ) 6= 0, (i = 1, 2, · · · n) of a random experiment then for any arbitrary event A of the sample space of the above experiment with P (A) > 0, we have P (A ∩ Bi ) P (Bi )P (A/Bi ) P (Bi /A) = = Pn P (A) i=1 P (Bi )P (A/Bi ) For n = 3 P (B2 )P (A/B2 ) P (B2 /A) = P (B1 )P (A/B1 ) + P (B2 )P (A/B2 ) + P (B3 )P (A/B3 ) Note: P (B) is the probability of occurrence of B. If we are told that the event A has already occurred. On knowing about the event A, P (B) is changed to P (B/A). Thus with the help of Baye’s theorem we can calculate P (B/A). Example 2.6.1 An urn I contains 3 white and 4 red balls and an urn II contains 5 white and 6 red balls. One ball is drawn at random from one of the urns and is found to be white. Find the probability that it was drawn from urn I. Solution: Let U1 : the ball drawn from Urn I, U2 : the ball drawn from Urn II, W : the ball is white. We have P (U1 )P (W/U1 ) P (U1 /W ) = (2.3) P (U1 )P (W/U1 ) + P (U2 )P (W/U2 ) 26  1 Since two urns are equally likely to be selected, P (U1 ) = P (U2 ) = 2 3 P (W/U1 ) = P (a white ball is drawn f rom U rn I) = 7 5 P (W/U2 ) = P (a white ball is drawn f rom U rn II) = 11 1 3 2 × 7 33 P (U1 /W ) = 1 3 1 5
=. 2 × 7 + 2 × 11 68 Example 2.6.2 Three urns contains 6 red, 4 black; 4 red, 6 black; 5 red, 5 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is red, find the probability that it is drawn from the first urn. Solution: Let U1 : the ball is drawn is U1 , U2 : the ball is drawn U1 , U2 , U3. R : the ball is Red. We ahve to find P (U1 /R). Since the three urns are equally likelt to be selected. 1 P (U1 ) = P (U2 ) = P (U3 ) = 3. Also, 6 P (R/U1 ) = P (a is drawn f rom U rn I) = 10 4 P (R/U2 ) = P (a is drawn f rom U rn II) = 10 5 P (R/U1 ) = P (a is drawn f rom U rn III) = 10 Baye’s theorem, P (U1 )P (R/U1 ) P (U1 /R) = P (U1 )P (R/U1 ) + P (U2 )P (R/U2 ) + P (U3 )P (R/U3 ) 1 6 3 × 10 2 P (U1 /R) = 1 6 1 4 1 5 = 3 × 10 + 3 × 10 + 3 × 10 5 Example 2.6.3 In a bolt factory, Machine A, B and C manufacture respectively 25%, 35% and 40% of the total. If their output 5, 4 and 2 per cent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by Machine B? Solution: We denote thus, A: bolt manufactured by Machine A, B: bolt manufactured by Machine B and C: bolt manufactured by Machine C. 25 35 40 P (A) = = 0.25; P (B) = = 0.35; P (C) = = 0.40 100 100 100 The probability of drawing a defective bolt manufactured by Machine A is P (D/A) = 0.05 The probability of drawing a defective bolt manufactured by Machine B is P (D/B) = 0.04 The probability of drawing a defective bolt manufactured by Machine C is P (D/C) = 0.02. By Baye’s Theorem, 27  P (B)P (D/B) P (B/D) = P (A)P (D/A) + P (B)P (D/B) + P (C)P (D/C) 0.35 × 0.04 P (U1 /R) = = 0.41 (0.25 × 0.05) + (0.35 × 0.04) + (0.40 × 0.02) 2.7 Try Outs (1) A bag contains of 10 white and 15 black balls. Two balls are drawn in quick succession. What is the probability that the first is white and the second is black [Ans: 41 ] (2) A committee is to be formed by choosing 2 boys and 4 girls out o a group of 5 boys and 6 girls. What is the probability that a particular boy named A and a particular girl named B are selected in the 4 committee. [Ans: 15 ]. (3) A husband and wife appear in an interview for two vacancies in the same post. The probability of husband selection is 17 and that of wife’s selection is 15. What is the probability that: (a) both will be selected. (b) both will be selected (b) only one of them will be selected. (c) None of them will be selected. 1 [Ans: (i) 35 ; (ii) 27 ; (iii) 24 35 ]. (4) A Mathematical problem is given to three students Aminu, Babawo and Hanatu whose chances of solving it are 32 , 34 , 14 respectively. (i) What is the probability that it will not be solved? (ii) What is the probability that it can be solved? 1 (iii) What is the probability that only two of them were able to solve it? [Ans: (i) 35 ; (ii) 72 ; (iii) 24 35 ] (5) In a bolt factory, Machine A, B and C manufacture respectively 15%, 20% and 45% of the total. If their output 7, 5 and 4 per cent are defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it was manufactured by Machine C? (6) Three urns contains 7 red, 3 black; 6 red, 4 black; 4 red, 6 black balls respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black, find the probability that it is drawn from the third urn. (7) In a bolt factory, Machines A, B and C manufacture 25%, 35% and 40% of the total output respectively. Of their outputs, 5%, 4% and 2% are defective bolts. A bolt is chosen at random and found to be 25 28 16 defective. What is the probability that the bolt came from Machine A, B or C? [Ans: 69 , 69 , 69 ] (8) In a bolt factory machine A, B, C manufacture 20%, 30% and 50% of the total of the output and 6%, 3% and 2% are defective. A bolt is drawn at random and found to be defective. Find the probability that it is manufactured from Machine (i) A (ii) B (iii) C [Ans: 12 9 10 31 , 31 , 31 ] 28  (9) Of the three men the chances that a politician, business man or an academician will be appointed as a Vice Chancellor of FUT Babura are 0.5, 0.3 and 0.2 respectively. The probability at that research is promoted by those persons if they are appointed as VC are 0.3, 0.7 and 0.8 respectively. (i) Determine the probability that research is promoted. (ii) The probability that if research is promoted, what is the probability that the VC is an academician. 4 [Ans: (i) 0.52; (ii) 13 ] (10) The procurement department of the Federal University of Technology, Babura purchased a particular product from two suppliers A and B. The quality of the product supplied is varied and the Procurement Department of FUTB has a policy of checking the product supplied and returning those items classified as faulty to the supplier. Faulty Not Faulty Total Items Supplier A 15 60 75 Supplier B 5 45 50 20 105 125 Table 1: Items supplied last month Thus a total of 20 out of 125 items were found to be defective upon delivery. This month another batch of items has been received from both suppliers. Unfortunately, the relevant invoices have been lost by the warehouse department and there is no way of knowing for certain which item were sent by which suppliet. Upon checking the quantity of goods supplied the Procurement department has found faulty items. Using Baye’s theorem. Determine: (a) Find the probability that the item came from supplier A? "   # A P (A) · P ( F A) 0.12 Ans : P = = = 0.75 F P (A) · P ( F F A ) + P (B) · P ( B ) 0.16 (b) The probability that the supplied item came from supplier B? "   # F B P (B) · P ( B ) 0.04 Ans : P = = = 0.25 F P (A) · P ( F F A ) + P (B) · P ( B ) 0.16 29 Chapter 3 Random Variable 3.1 Introduction In the study of probability, any process of observation is referred to as an experiment. The results of an observation are called the outcomes of the experiment. An experiment is called a random experiment if its outcome cannot be predicted. Typical examples of a random experiment are the roll of a die, the toss of a coin, drawing a card from a deck, or selecting a message signal for transmission from several messages. Consider a random experiment with sample space S. A random variable X(ξ) is a single-valued real function that assigns a real number called the value of X(ξ) to each sample point ξ of S. Usually, we use a simple letter X for X(ξ) and r. v. denotes the random variable. The sample space S is termed the domain of the r.v. and the collection of all numbers [values of X(ξ)] is called the range of the r. v. X. Thus the range of X is a certain subset of the set of all real numbers. Note: The main purpose of random variable (r. v.) X is so that we can define certain probability functions that make it both convenient to compute the probabilities of various events. Also, two or more different sample points might give the same value of X(ξ), but two different numbers in the range cannot be assigned to the same sample point. 30  Fig 3.2: Toss of Coins Example 3.1.1 In the experiment of tossing a fair coin three times, the sample space S, consists of eight equally likely sample points S = {HHH, HHT, HT H, HT T, T HH, T HT, T T H, T T T }. If X is the random variable giving the number of heads obtained. Find (a) P (X = 2); (b) P (X < 2). Solution: (a) Let A ⊂ S, be the event defined by X = 2. Then, we have A = (X = 2) = {ξ : X(ξ) = 2} = {HHT, HT H, T HH} Since the sample points are equally likely, we have 3 P (X = 2) = P (A) =. 8 (b) Let B ⊂ S, be the event defined by X < 2. Then B = (X < 2) = {ξ : X(ξ) < 2} = {HT T, T HT, T T H, T T T } and 4 1 P (X < 2) = P (B) = = 8 2 Example 3.1.2 Consider the experiment of throwing a fair die. Let X be the r. v. which assigns 1 if the number that appears is even and 0 if the number that appears is odd. (a) What is the range of X? (b) Find P (X = 1) and P (X = 0)? Solution: The sample space S on which X is defined consists of 6 points which are equally likely: S = {1, 2, 3, 4, 5, 6} 31  (a) The range of X is Rx = {0, 1} (b) (X = 1) = {2, 4, 6}. Thus, P (X = 1) = 36 = 12. Similarly, (X = 0) = {1, 3, 5} and P (X = 0) = 36 = 12. Example 3.1.3 A can hit a target three times in 5 shots, B two times in 5 shots and C three times in 4 shots. All of them fire one shot each simultaneously at the target. What is the probability that: (i) 2 shots hit (ii) Atleast 2 shots hits. Solution: Pb(A hits the target) = p(A) = 53 , q(A) = Pb (A misses the target) = 1 − 3 5 = 2 5 Pb(B hits the target) = p(B) = 52 , q(B) = Pb (B misses the target) = 1 − 52 = 53 Pb(C hits the target) = p(C) = 34 , q(C) = Pb (C misses the target) = 1 − 43 = 14 (i) Pb(2 shots hits the target) = p(A)p(B)q(C) + p(A)p(C)q(B) + p(B)p(C)q(A)       3 2 1 3 3 3 2 3 2 6 + 27 + 12 45 9 = × × + × × + × × = = = 5 5 4 5 4 5 5 4 5 100 100 20 (ii) Pb(Atleast 2 shots hits the target) = Pb(Only 2 shots hits the target) + Pb(All the 3 shots hits the 9 target) = 20 + p(A)p(B)p(C)   9 3 2 3 9 18 45 + 18 63 = + × × = + = = 20 5 5 4 20 100 100 100 3.2 Expected Value of Random Variable If p1 , p2 , · · · , pn of the probabilities of the events x1 , x2 , · · · , xn−1 , xn respectively. The expected value n X E(X) = p1 x1 , p1 x1 , · · · , pn−1 xn−1 , pn xn = p r xr r=1 Example 3.2.1 For a biased die, the probabilities for the different faces to turn up are given as follows: Face (x) 1 2 3 4 5 6 Probability 0.1 0.32 0.21 0.15 0.05 0.17 Find the expectation for (i) 2, (ii) 3 and 5, (iii) 2 or 4, (iv) Atleast 4, (v) Atmost 3. Solution: Recall the formula for expectation n X E(X) = p1 x1 , p1 x1 , · · · , pn−1 xn−1 , pn xn = pr xr r=1 For n = 6, (i) E(X = 2) = p2 x2 = 0.32 × 2 = 0.64 (ii) E(X = 3) × E(X = 5) = p3 x3 × p5 x5 = (3 × 0.21) × (5 × 0.05) = 0.63 × 0.25 = 0.1575 (iii) E(X = 2) + E(X = 4) = p2 x2 + p4 x4 = 2 × 0.32 + 4 × 0.15 = 0.64 + 0.60 = 1.24 32 (iv) E(Atleast 4) = E(X = 4) + E(X = 5) + E(X = 6) = 4×0.15+5×0.05+6×0.17 = 0.60+0.25+1.02 = 1.87 (v) E(Atmost 3) = E(X=3) + E(X = 2) + E(X = 1) = 3×0.21+2×0.32+1×0.1 = 0.63+0.64+0.1 = 1.28 Example 3.2.2 Suppose that a game is to be played with a single die assumed fair. In this game a player wins $20 if a 2 turns up; $40 if a 4 turns up; loses $30 if a 6 turns up; while the player neither wins nor loses if any other face turns up. Find the (i) expected sum of money to be won (ii) variance (iii) standard deviation. Solution: (i) Let X be the r.v. giving the amount of money won on any toss. The possible amounts won when the die turns up 1, 2, · · · ,6 are x1 , x2 , · · · , x6 respectively, while the probabilities of these are f (x1 ), f (x2 ), · · · , f (x6 ). The probability function for X is given by: x 0 + 20 0 + 40 0 - 30 1 1 1 1 1 1 f (x) 6 6 6 6 6 6 Therefore, the expected value or expectation is             1 1 1 1 1 1 E(X) = µ = 0 + 20 +0 + 40 +0 − 30 = 5. 6 6 6 6 6 6 It follows that the player can expect to win $5. In a fair game, therefore, the player should expect to pay $5 in order to play the game. = E(X − µ) = (0 − 5)2 16 + (20 − 5)2 16 + (0 − 5)2 16 + (40 − 5)2 16 + (0 − 5)2 16 + 2




(ii) Variance, σX (−30 − 5)2 16 = 2750
6 = 458.333 (iii) Standard Deviation, σX is the square root of the variance, q √ σX2 = 458.333 = 21.40872096 3.3 Exercises (1) Consider the experiment of tossing a fair coin three times. Let X be the r. v. that counts the number of heads in each sample point. Find the following probabilities: (a) P (X ≤ 1) (b) P (X > 1) (c) P (0 < X < 3) [Ans: (a) 21 , (b) 21 , (c) 34 ] (2) Consider the experiment of throwing two fair die. Let X be the random variable indicating the sum of the numbers that appear. (a) What is the range of X? 1 (b) Find (i) P(X = 3), (ii) P (X ≤ 4), (iii) P (3 ≤ X < 7) [Ans: (i) 18 ; (ii) 16 ; (iii) 12 ] 33 (3) Let X and Y be the random independent events of rolling a fair die. Compute the expected value of X + Y and the variance of X + Y. [Ans: Expected Value = 7; Variance = 5.8333]. (4) Suppose that a game is to be played with a single die assumed fair. In this game a player wins $ 10 if 1 turns up; $ 30 if 3 turns up; loses $ 20 if a 6 turns up; while the player neither wins nor loses if any other face turns up. Find the: (i) expected sum of money to be won (ii) variance (iii) standard deviation. 34 Chapter 4 Discrete Probability Distribution Suppose that to each point on a sample space we assign a number. We then have a function defined on the sample space. This function is called a Random Variable (or Stochastic Variable) or more precisely Random Function. It is usually denoted with a capital letter X or Y. In general a r.v. has some specified physical and geometrical significance. A r.v. that takes a finite or countably infinite number of values is called a Discrete Random Variable while the one that takes on an infinite or non-countably infinite number of values is called a Continuous Random Variable. 4.1 Discrete Probability Distribution Let X be the corresponding probabilities of the values x1 , x2 , · · · , xn of a variate X be p1 , p2 , · · · , pn respec- tively. if these are such that (i) 0 ≤ pi ≤ 1, i = 1, 2, · · · , n Pn (i) p1 + p2 + · · · + pn = i=1 pi = 1, Thus we say the variate X satisfy the Probability Distribution. In tabular form, it is expressed as X x1 x2 x3 ··· xn P (X) p1 p2 p3 ··· p1 n X M ean, µ = pi xi i=1 n X V ariance = pi x2i − µ2 i=1 We can also rephrase the above as follows: Let X be a discrete r.v. and suppose that the possible values that it can assume are given x1 , x2 , · · · , arranged in some order. Suppose also that these values are assumed with probabilities given by P (X = xk ) = f (xk ) k = 1, 2, · · · (∗) 35 Thus, the probability distribution is given by P (X = x) = f (x) · · · (∗∗). In general, f (x) is a probability distribution function if (i) f (x) ≥ 0 P (ii) x f (x) = 1 The latter condition (ii) states that when the sum is taken over all possible values x. Example 4.1.1 Suppose that a coin is tossed twice. Let X represent the number of heads that can come up, with each sample point we can associate a number for X as follows: Sam