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

2. 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

3. 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)

4. Discrete Probability

5. 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|

6. 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

7. 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 = ways
To choose all 4 digits correctly: 1
Answer: 1/ =

8. 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!) =
Answer: 1 / ≈ 0,

9. 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 /
The ball is selected is returned to the bin before the next ball is selected
Answer: 1 / (50 * 50 * 50 * 50 * 50)
1 /

10. 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: = 17,576,000 possible license plates

11. 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

12. 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

13. 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)

14. 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

15. 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)

16. 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

17. 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)

18. 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)

19. 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 – 10/100
= 3/5

20. 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)

21. 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

22. 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)

23. 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)

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

25. 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

26. 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)

27.  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

28.  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

29. Binomial Coefficients & Combinatorial Identities

30. 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

31. Binomial Theorem

32. 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

33. Pascal’s Traingle
The binomial coefficients satisfy many different identities

34. Combinatorial Identities
Ref: Rosen, Discrete mathematics and its applications