ความน่าจะเป็นแบบดีสครีต (Discrete Probability)

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1 ความน่าจะเป็นแบบดีสครีต (Discrete Probability)
รศ.ดร. สาธิต อินทจักร์ ภาควิชาวิศวกรรมการวัดคุม คณะวิศวกรรมศาสตร์ สถาบันเทคโนโลยีพระจอมเกล้าเจ้าคุณทหาร ลาดกระบัง

2 Why Probability? Originally devised for gambling by Pascal and Laplace over 200 years ago. Current applications of probability include genetics study (e.g., to understand inheritance of traits) and computer science (e.g., to determine average-case complexity of algorithms).

3 Definitions Experiments: a procedure that yields one of a given set of possible outcomes. Sample space of an experiment: the set of possible outcomes. Event: a subset of the sample space whose outcome is of our interest. Probability of an event, p(E) = |E|/|S|

4 Examples Ex 1: What’s the probability of drawing a blue ball from an urn containing four blue balls and five red balls? [4/(4+5) = 4/9] Ex 2: What’s the probability that when two dice are rolled, the sum of the numbers on the two dice is 7? Number of possible outcomes, |S|, = 6*6 = 36 Events where the sum of the numbers is 7 include (1,6), (2,5), (3, 4), (4, 3), (5, 2), (6, 1)  |E| = 6 The probability of such event, p(E), = |E|/|S| = 6/36 = 1/6

5 More examples Ex 3: Find the probability that a hand of five cards in poker contains four cards of one kind? Sample space, S, is number of possible ways to choose 5 cards out of 52 cards  |S| = C(52, 5) Event, E, to select four cards of one kind, includes first select 1 kind out of 13 kinds (C(13, 1)), then select 4 cards of this kind from the four in the deck of this kind (C(4, 4)), and finally select 1 last card out of the 48 cards left (C(48, 1)) Probability, p(E) = |E|/|S| =

6 Complementary probability
Theorem 1: Let E be an event in a sample space S. The probability of the event , the complementary event of E, is given by Ex: A sequence of ten bits is randomly generated. What’s the probability that at least one of these bits is 0? E = event that at least one of ten bits is 0  = event that all bits are 1s. p(E) = 1 – p( ) = = 1 – 1/210 = 1 – 1/1024 = 1023/1024

7 Probability of union of two events
Theorem 2: Let E1 and E2 be events in the sample space S. Then (by the inclusion-exclusion principle), p(E1E2) = p(E1) + p(E2) – p(E1E2)

8 Properties of probability values
The probability of each outcome is a nonnegative real number no greater than 1. That is, 0  p(xi)  1 for i = 1, 2, …, n The sum of the probabilities of all possible outcomes should be 1. That is, Such a p is called a probability distribution Definition 2: The probability of the event E is the sum of the probabilities of the outcomes in E. That is,

9 Probability The probability p = p(E)  [0,1] of an event E is a real number representing our degree of certainty that E will occur. If p(E) = 1, then E is absolutely certain to occur, If p(E) = 0, then E is absolutely certain not to occur, If p(E) = ½, then we are completely uncertain about whether E will occur. What about other cases?

10 An Example Ex: What’s the probability that an odd number appears when we roll a die with equally likely outcomes? probability of the event an odd number appears, E, = {1, 3, 5}. Each event has probability p(1) = p(3) = p(5) = 1/6 Therefore, p(E) = p(1) + p(3) = p(5) = 3/6 = 1/2

11 Random Variables A random variable V is a variable whose value is unknown, or that depends on the situation. E.g., the number of students in class today the grades students receive in this class Whether it will rain tonight (Boolean variable) The proposition V=vi may be uncertain, and be assigned a probability.

12 Mutually Exclusive Events
Two events E1, E2 are called mutually exclusive if they are disjoint: E1E2 =  Note that two mutually exclusive events cannot both occur in the same instance of a given experiment. For mutually exclusive events, p(E1  E2) = p(E1) + p(E2).

13 Exhaustive Sets of Events
A set E = {E1, E2, …} of events in the sample space S is exhaustive if An exhaustive set of events that are all mutually exclusive with each other has the property that

14 Conditional Probability
Let E, F be events such that p(F)>0. Then, the conditional probability of E given F, written p(E|F), is defined as p(E|F) = p(EF)/p(F). This is the probability that E would turn out to be true, given just the information that F is true. If E and F are independent, p(E|F) = p(E).

15 An Example Ex 1: A bit string of length four is generated at random so that each of the 16 bit strings of length four is equally likely. What’s the probability that it contains at least two consecutive 0s given that its first bit is 0? Let E = event that a bit string of length four contains at least two consecutive 0s, and let F = event that the first bit of a bit string of length four is a 0. Then, p(E|F) = p(EF)/p(F) E  F = {0000, 0001, 0010, 0011, 0100}  p(EF) = 5/24 = 5/16. Since half of the bit string of length four must begin with 0 (the other half begins with 1), p(F) = 8/16 = ½. Therefore, p(E|F) = (5/16)/(1/2) = 5/8

16 Independent Events Two events E, F are independent if and only if p(EF) = p(E)·p(F). Relates to product rule for number of ways of doing two independent tasks Example: Flip a coin, and roll a die. p( quarter is heads  die is 1 ) = p(quarter is heads) × p(die is 1)

17 Bernoulli Trials Theorem 2: The probability of exactly k successes in n independent Bernoulli trials, with probability of success p and probability of failure q = 1 – p, is C(n, k) pk qn-k

18 An Example Ex: What’s the probability that exactly four heads come up when a fair coin is flipped seven times, assuming that the flips are independent. n = 7, k = 4, n – k = 7 – 4 = 3 p = probability of success (getting head) = ½ q = 1 – p = ½ Therefore, p(gets 4 heads out of 7 flips) = C(n, k) pk qn-k = C(7, 4) (1/2)4(1/2)3

19 Bayes’s Theorem Allows one to compute the probability that a hypothesis H is correct, given data D: Easy to prove from def’n of conditional prob. Extremely useful in artificial intelligence apps: Data mining, automated diagnosis, pattern recognition, statistical modeling, evaluating scientific hypotheses.

20 Expectation Values For a random variable X(s) on the sample space S is equal to E(X) = ∑sS p(s)X(s) The term “expected value” is widely used, but misleading since the expected value might be totally unexpected or impossible! Ex 1: Let X be the number that comes up when a die is rolled. What’s the expected value of X? Random variable X takes the values 1, 2, 3, 4, 5, or 6, each with probability 1/6. So E(X) = (1/6) 1 + (1/6) 2 + (1/6) 3 + (1/6) 4 + (1/6) 5 + (1/6) 6 = 21/6 = 7/2

21 Linearity of Expectation
Let X1, X2 be any two random variables derived from the same sample space, and if a and b are real numbers. Then: E(X1+X2) = E(X1) + E(X2) E(aX1 + b) = aE(X1) + b

22 Independent Random Variables
Definition 3: The random variables X and Y are independent if p(X = r1 and Y = r2) = p(X = r1)*p(Y = r2) Theorem 5: If X and Y are independent random variables, then E(XY) = E(X)E(Y).

23 Variance The variance V(X) = σ2(X) of a random variable X is the expected value of the square of the difference between the value of X and its expectation value E(X): The standard deviation or root-mean-square (RMS) difference of X, σ(X) :≡ V(X)1/2. Theorem 6: If X is a random variable on a sample space S, then V(X) = E(X2) – E(X)2

24 Example => Daily sales records for a shop selling electric appliances show that it will sell zero, one, two or three air-conditioners with the probabilities: Number of Sales Probability Calculate the expected value, variance and standard variation for daily sales.

25 Example => Expected value
= (0)(0.5) + (1)(0.3) + (2)(0.15) + (3)(0.05) = 0.75 Variance = ( )2(0.5) + (1 – 0.75)2(0.3) (2 – 0.75)2(0.15) + (3 – 0.75)2(0.05) = Standard deviation = =

26 Introducing binomial distribution
Consider the following random variables: a) X : no.of “6” obtained in 10 rolls of a fair die. b) X : no. of tails obtained in 100 tosses of a fair coin. c) X : no. of defective light bulb in a batch of 1000. d) X : no. of boys in a family of 5 children.

27 In each case, a basic experiment is repeated a number of times
In each case, a basic experiment is repeated a number of times. For example, the basic experiment in case (a) is rolling the die once.

28 The following are common characteristics of the random variables in cases (a) to (d):
1) The number of trials n of the basic experiment is fixed in advance. 2) Each trial has two possible outcomes which may be called “success” and “failure”. 3) The trials are independent. 4) The probability of success is fixed.

29 Binomial random variable
A random variable X defined to be the number of successes among n trials called a binomial random variable if the properties (1) to (4) are satisfied. Mathematically, we write X ~ Bin(n, ), where n = no. of trials, and  = prob. of success.

30 The p.d. of a binomial r.v. If X ~ Bin(n, ), then
p(x) = P(X = x) = nCxx(1-)n-x where x = 0, 1, 2, …, n .

31 Example (Binomial 1) A fair coin is tossed 8 times.
Find the probability of obtaining 5 heads. Let X be the number of heads obtained in 8 tosses. Then X ~ Bin(8, 1/2). P(5 heads) = = 7/32

32 Example (Binomial 2) There are 10 multiple-choice questions in a test and each question has 5 options. Suppose a student answers all 10 questions by randomly picking an option in each question. Find the probability that (a) he will answer six questions correctly, (b) he will get at least 3 correct answers.

33 Example (Binomial 2) Then X ~ Bin(10, 0.2).
Let X be the number of correct answers he will get. Then X ~ Bin(10, 0.2). (a) P(X = 6) = 10C6(0.2)6(1-0.2)10-6 = (b)P(at least 3 correct answers) = 1 – P(X = 0) – P(X = 1) – P(X = 2) = 1 – 10C0(0.8)10 – 10C1(0.2)(0.8)9 – C2(0.2)2(0.8)8 = 0.322

34 Example (Binomial 3) Binary digits 0 and 1 are transmitted along a data channel in which the presence of noise results in the fact that each digit may be wrongly received with a probabilty of Each message is transmitted in blocks of 2000 digits. What is the probabilty that at least one digit in a block is wrongly received? If a certain message has a length of 20 blocks, find the probability that 2 or more blocks are wrongly received.

35 Example (Binomial 3) (a) Let X be the number of digits wrongly received in a block of 2000 digits. Then X ~ Bin(2000, ) . P(X  1) = 1 – P(X = 0) = 1 – =

36 Example (Binomial 3) (b) Let Y be the number of block that are wrongly received among the 20 blocks. Then Y~ Bin(20, ). P(Y  2) = 1 – P(Y = 0) – P(Y = 1) = 1 – (1 – )20 – 20C1(0.0392)(1 – )19 = 0.184

37 Pr(ตัวแทนที่ประกอบด้วยกรรมการชาย 2 และ หญิง 1 คน) = 6/10
Event สมมุติมีคณะกรรมการอยู่ 5 คน เป็นผู้ชาย 3 และ ผู้หญิง 2 คน จะเลือกตัวแทน 3 คน จึงทำการจับฉลาก โดยให้มีผู้ชาย 2 และผู้หญิงหนึ่งคน Sampling with Replacement N3 = 53 = 125 ชุด M1 M1 M1, M1 M1 M2, M1 M1 M3, M1 M1 F1, M1 M1 F2 : : : : : F2 F2 F2, F2 F2 F1, F2 F2 M3, F2 F2 M2, F2 F2 M1 Sampling without Replacement M1 M2 M3, M1 M2 F1, M1 M2 F2, M1 M3 F1, M1 M3 F2 M1 F1 F2, M2 M3 F1, M2 M2 F2, M2 F1 F2, M3 F1 F2 Pr(ตัวแทนที่ประกอบด้วยกรรมการชาย 2 และ หญิง 1 คน) = 6/10

38 Hypergeometric Distribution
ในกลุ่มตัวอย่างขนาด n จะมีการแจกแจงแบบ Hypergeometric ถ้า X มีฟังก์ชันความน่าจะเป็นดังนี้ โดยที่ k เป็นส่วนที่ของข้อมูลที่เราให้ความสนใจ N เป็นจำนวนประชากรทั้งหมด n เป็นจำนวนประชากรของตัวอย่าง

39 ตัวอย่างที่ 1.4 กล่องใบหนึ่งบรรจุตัวต้านทาน 100 ตัว โดยมีตัวต้านทานที่เสื่อมคุณภาพอยู่ 5 ตัวปะปนอยู่ เพื่อตรวจสอบคุณภาพของตัวต้านทานทั้งกล่อง ผู้ซื้อสุ่มตัวอย่างตัวต้านทานมา 10 ตัว เพื่อนำไปตรวจสอบคุณภาพถ้าใน 10 ตัว ถ้าพบว่ามีตัวเสื่อมคุณภาพปะปนอยู่สองตัวผู้ซื้อจะไม่ยอมซื้อ จงหาความน่าจะเป็นที่จะมีตัวต้านทานที่เสื่อมคุณภาพปนอยู่ 2 ตัวในตัวอย่างสุ่ม จงคำนวณค่าความน่าจะเป็นในข้อแรก ด้วยวิธีการแจกแจงแบบทวินาม

40 ประมาณค่า Pr(x=2) ด้วย Binomial Distribution
ในการทดลอง n ครั้ง การแจกแจงทวินามจะมีลักษณะเหมือนกับการกระจายทวินาม n เทอมคือ โดยทั่วไป Hypergeometric Distribution สามารถประมาณด้วย Binomial Distribution ถ้าหาก n/N  0.1 นั่นคือ k/N = p เป็นโอกาสเลือกได้หน่วยที่สนใจมาเป็นตัวอย่าง (N-k)/N = q เป็นโอกาสที่เลือกได้หน่วยที่ไม่สนใจมาเป็นตัวอย่าง

41 Normal distribution =>
A random variable X is said to have a Normal distribution, with parameters  and 2 if it can take any real value and has p.d.f. In this case, we write X ~ N(, 2). It can be shown that  is the expected value of X and 2 is the variance.

42 P(a < X < b) = = This integral cannot be done algebraically and its value has to be found by numerical methods of integration. The value of P(a < X < b) can also be viewed as the area under the curve of f(x) from x = a to x = b.

43

44

45 Characteristics of normal distribution
1. Bell shape 2. Symmetric 3. Mean = Mode = Median 4. The 2 tails extend indefinitely

46 Standard normal distribution
A standard normal distribution is the normal distribution with  = 0 and 2 = 1, i.e. the N(0,1) distribution. A standard normal random variable is often denoted by Z. If Z ~ N(0,1), its c.d.f. is usually written as (z) = P(Z  z) =

47 Note: (1) (z) may be interpreted as the area to the left of z under the standard normal curve. (2) P(Z < -z) = P(Z > z), since the standard normal curve is symmetrical about the line Z = 0. (3) The area between Z= -1 and +1 is 68% Z= -2 and +2 is 95% Z= -3 and +3 is 99%

48 Example => Given Z ~ N(0,1), find the following probabilities using the standard normal table. (a) p( Z  1.25) (b) P( Z > 2.33) (c) P(0.5 < Z < 1.5) (d) P( Z < -1.25) (e) p (-1.5< Z <-0.5)

49 Example (a) P(Z  1.25) = 0.8944 P(Z > 2.33) = 1 - (2.33)
= 1 – = (c) P(0.5 < Z < 1.5) = (1.5) - (0.5) = – =0.2417 (d) P( Z < -1.25) = P(Z > 1.25) = 1 - (1.25) = 1 – = (e) P(-1.5 < Z < -0.5) =

50

51 Example (a) P(Z  1.25) =

52 Example P(Z > 2.33) = 1 - (2.33) = 1 – =

53 Example (c) P(0.5 < Z < 1.5) = (1.5) - (0.5)
= – =0.2417

54 Example Given Z ~ N(0,1), find the value of c if (a) P(Z  c) = 0.8888
(b) P( Z > c) = 0.37 (c) P(Z < c) = 0.025 (d) P(0  Z  c) =

55 Example (a) (c) = 0.8888 c = 1.22 (b) 1 - (c) = 0.37 (c) = 0.63

56 Recall that P(Z < -z) = P(Z > z)
Example (c) Obviously c is negative and the standard normal table cannot be used directly. Recall that P(Z < -z) = P(Z > z) 0.025 = P(Z < c) = P(Z > -c) = 1 –(-c) (-c) = 0.975 -c = 1.96 c = -1.96

57 Example (d) P(0  Z  c) = (c) – 0.5 = (c) = c = 2.43

58 Standardization of normal random variables
If X ~ N(, 2), then it can be shown that Z = ~ N(0,1) . In other words is the standard normal random variable. The process of transforming X to Z is called standardization of the random variable X.

59 Example Suppose X ~ N(10, 6.25). Find the following probabilities
(a) P(X < 13) (b) P(X > 5) (c) P(8 < X < 15)

60 Example Let Z = . (a) P(X < 13) = P(Z < ) = P(Z < 1.2)
=

61 Example (b) P(X > 5) = P(Z > ) = P(Z > -2) = P(Z <2)
=

62 Example (c) P(8 < X < 15) = P( ) = P(-0.8 < Z < 2)
= (2) - (-0.8) = (2) – (1 - (0.8)) = – (1 – ) =