Agenda Lecture Content: Discrete Probability Binomial and Combinatorial Identity Review UTS

Pokok Bahasan (Minggu 8-11) Minggu ke Topik 8 Peluang diskret Koefisien binomial Identitas kombinatorik 9 Algoritma rekursif Relasi rekurensi Penyelesaian relasi rekurensi Fungsi pembangkit 10 Graf dan terminologi graf Path dan cycle Euler cycle dan Hamiltonian cycle 11 Representasi graf Graf isomorfis Graf planar Permasalahan lintasan terpendek

Pokok Bahasan (Minggu 12-14) Minggu ke Topik 12 Terminologi dan karakterisasi pohon Spanning Trees Binary Trees 13 Tree traversals Decision trees Pohon isomorfis Game trees 14 Kombinatorial sirkuit Aljabar Boolean Ujian Akhir Semester (UAS)

Discrete Probability

Finite Probability If S is a finite sample space of equally likely outcomes, and E is an event, that is, a subset of S, then the probability of E is P(E) = |E| / |S|

Example 1 An urn contains four blue balls and five red balls. What is the probability that a ball chosen from the urn is blue? Answer: 4/9 What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7? Possible outcome: 36 Successful outcome: 6  (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) - Answer: 6/36

Example 2 In a lottery, players win a large prize when they pick four digits that match, in the correct order, four digits selected by a random mechanical process. What is the probability that a player wins the large prize? To choose 4 digits = 104 = 10.000 ways To choose all 4 digits correctly: 1 Answer: 1/10.000 = 0.0001

Example 3 What is the probability that a person picks the correct six numbers out of 40? The total number of ways to choose six numbers out of 40 is: C(40,6) = 40! / (34! * 6!) = 3.838.380 Answer: 1 / 3.838.380 ≈ 0,00000026

Example 4 What is the probability that 11, 4, 17, 39, 23 are drawn in that order from a bin containing 50 balls labeled with the number 1,2,…50, if : The ball selected is not returned to the bin before the next ball is selected Answer: 1 / (50 * 49 * 48 * 47 * 46) 1 / 254.251.200 The ball is selected is returned to the bin before the next ball is selected Answer: 1 / (50 * 50 * 50 * 50 * 50) 1 / 312.500.000

Example 4 How many different license plates are available if each plate contains a sequence of three letters followed by three digits? 26 choices for each of the three letters and 10 choices for each of the three digits. Total: 26 . 26 . 26 . 10 . 10 . 10 = 17,576,000 possible license plates

Assigning Probability Let S be the sample space of an experiment and P(s) is the probability to each outcome s, then: (i) 0 ≤ P(s) ≤ 1 for each s  S (ii) ∑sS P(s) = 1 P: probability distribution

Example What probabilities should we assign to the outcomes H (heads) and T (tails) when a fair coin is flipped? P(H) = P(T) = ½ What probabilities should be assigned to these outcomes when the coin is biased so that heads comes up twice as often as tails? P(H) = 2 * P(T) P(H) + P(T) = 1 2 * P(T) + P(T) = 1 P(T) = 1/3; P(H) = 2/3

Uniform Distribution Suppose that S is a set with n elements. The uniform distribution assigns the probability 1/n to each element of S. The probability of the event E is the sum of the probabilities of the outcomes in E. That is: P(E) = ∑sE P(s)

Example Suppose that a dice is biased so that 3 appears twice as often as each other number but that the other five outcomes are equally likely. What is the probability that an odd number appears when we roll this die? E = {1, 3, 5} p(1) = p(2) = p(4) = p(5) = p(6) = 1/7 p(3) = 2/7 p(E) = p(1) + p(3) + p(5) = 4/7

p(E) = 1 – p(E) Complement of Events Let E be an event in a sample space S, the probability of the event E (The complementary event of E) is given by: p(E) = 1 – p(E)

Example A sequence of 10 bits is randomly generated. What is the probability that at least one of these bits is 0? E : be the event that at least one of the 10 bits is 0. E : be the event that all the bits are 1s. p(E) = |E| / |S| = 1/210 = 1/1024 p(E) = 1 – p(E) = 1 – 1/1024 = 1023/1024

Combination of events E1 and E2 Let E1 and E2 be events in the sample space S. Then p(E1E2) = p(E1) + p(E2) – p(E1E2)

Example 1 What is the probability that a positive integer selected at random from the set of positive integers not exceeding 100 is divisible by either 2 or 5? E1 : the event that the integer selected is divisible by 2 E2 : be the event that it is divisible by 5. E1  E2 : the event that it is divisible by either 2 or 5 E1  E2 : the event that it is divisible by both 2 and 5 (divisible by 10)

Example 1 |E1| = {2, 4, 6, …, 100} = 50 |E2| = {5, 10, …, 100} = 20 |E1  E2| = {10, 20, …, 100} = 10 p(E1E2) = p(E1) + p(E2) – p(E1E2) = 50/100 + 20/100 – 10/100 = 3/5

Mutually Exclusive of events E1 and E2 Let E1 and E2 be events in the sample space S. Then p(E1E2) = p(E1) + p(E2)

Conditional Probability: Illustration Suppose that we flip a coin three times, and all 8 possibilities are equally likely. Suppose we know that the event F, that the first flip comes up tails, occurs. Given this information, what is the probability of the event E, that an odd number of tails appears? Because the first flip comes up tails, there are only four possible outcomes: TTT, TTH, THT, and THH, where H and T represent heads and tails, respectively. An odd number of tails appears only for the outcomes TTT and THH. p(E) = 2/4 = 1/2 , given F occurs.  the conditional probability of E given F

Conditional Probability Let E and F be events with p(F) > 0. the conditional probability of E given F, denoted by p(E | F), is defined as: p(E | F) = p(E ∩ F) / p(F)

Example 1 A bit string of length 4 is generated at random so that each of the 16 bit strings of length four is equally likely. What is the probability that it contains at least two consecutive 0s, given that its first bit is a 0? E: a bit string of length 4 contains at least two consecutive 0s. F: the first bit of a bit string of length 4 is a 0. p(E|F) =p(E  F)/p(F)

Example 1 (E  F) = {0000, 0001, 0010, 0011, 0100} p(E  F) = 5/16 p(F) = 8/16 p(E | F) = 5/8

Example 2 What is the conditional probability that a family with two children has two boys, given they have at least one boy? Possibilities: BB, BG, GB, GG E: a family with 2 children has two boys, {BB) F: a family with 2 children has at least one boy {BB, BG, GB} E  F = {BB} p(F) = 3/4 ; p(E  F) = ¼ p(E|F) = p(E  F) / p(F) = 1/3

Independent Events Two events are independent if the occurrence of one of the events gives no information about the probability that the other event occurs. The events E and F are independent if and only if p(E  F) = p(E) * p(F)

 E and F are independent Example Suppose E is the event that a randomly generated bit string of length four begins with a 1 and F is the event that a randomly generated bit string of length four contains an even number of 1s. Are E and F independent, if the 16 bit strings of length four are equally likely? E: begin with a one: 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111 F: contains even numbers of ones: 0000, 0011, 0101, 0110, 1001, 1010, 1100, 1111 p(E) = p(F) = 8/16 = ½ E  F = {1111, 1100, 1010, 1001} p(E  F) = 4/16 = ¼ = p(E) * p(F)  E and F are independent

 E and F are not independent Example E: a family with 2 children has two boys F: a family with 2 children has at least one boy Independent? Possibilities: BB, BG, GB, GG E: {BB) F: {BB, BG, GB} E  F = {BB} p(E) = ¼; p(F) = 3/4 ; p(E  F) = ¼ p(E) * p(F) = ¼ * ¾ = 3/16 p(E  F)  P(E) * p(F)  E and F are not independent

Binomial Coefficients & Combinatorial Identities

Expansion (a+b)n using Combination (a+b)3 = (a+b)(a+b)(a+b) = aaa+ aab+ aba+ abb+ baa+ bab+ bba+ bbb = a3+ a2b+ a2b+ ab2+ a2b+ ab2+ ab2+ b3 = a3+ 3a2b+ 3ab2+ b3 (a+b)n = C(n,0)anb0 + C(n,1)an-1b1 + C(n,2)an-2b2 + ··· + C(n,n-1)a1bn-1 + C(n,n)a0bn

Binomial Theorem  

Pascal’s Triangle & Combinatorial Identity The border consists of 1’s, and any interior value is the sum of the two numbers above it.  see next slide Theorem (Pascal’s Triangle) C(n+ 1,k) = C(n,k–1) + C(n,k) for 1 ≤ k ≤ n. Combinatorial identity An identity that results from some counting process Combinatorial argument The argument that leads to the formulation of a combinatorial identity The binomial coefficients satisfy many different identities

Pascal’s Traingle The binomial coefficients satisfy many different identities

Combinatorial Identities Ref: Rosen, Discrete mathematics and its applications