2.6 Tools for Counting sample points

Transcription

1 2.6 Tools for Counting sample points When the number of simple events in S is too large, manual enumeration of every sample point in S is tedious or even impossible. (Example) If S contains N equiprobable sample points and an event A contains exactly n a sample points, P (A) = n a /N. How about too large n a or N? (Theorem 2.1)(mn rule) With m elements a 1, a 2,..., a m and n elements b 1, b 2,..., b n it is possible to form mn = m n pairs containing one element from each group. (Proof) [Note] (Generalization of mn rule) If R groups are such that each group may contains n r elements where r = 1,..., R, it is possible to form n 1 n 2 n R pairs from R groups. 19

2 (Example 2.5) An experiment involves tossing a pair of dice and observing the numbers on the upper faces. Find the number of sample points in S, the sample space for the experiments. (Example) Lottery From the numbers 1,2,...,44, a person may pick any six for her ticket. The winning number is then decided by randomly selecting six numbers from the forty-four. How many possible lottery tickets? 1 Ordered, without replacement : 2 Ordered, with replacement : 3 Unordered, without replacement : 20

3 Two important factors in counting rules are Order and Replacement Number of possible arrangements of size r from n objects : With Without Replacement Replacement Ordered n r Permutation Unordered Combination (Definition 2.7)(permutation) An ordered arrangement of r distinct objects is called a permutation. The number of ways of ordering n distinct objects taken r at a time will be designated by the symbol P n r. (Theorem 2.2) P n r = n(n 1)(n 2) (n r + 1) = n! (n r)!. Note that if n = r, P n r = n! = n(n 1) (2)(1) and 0! = 1. 21

4 (Example) From 10 persons, choose a president, a vice president, a secretary and a treasurer for a club. How many possible results? (Definition 2.8)(combination) The number of combinations of n objects taken r at a time is the number of subsets of size r chosen (without replacement) from n objets. This number will be denoted by Cr n or ( ) n r. (Theorem 2.4) ( n r ) = C n r = P n r r! = n! r!(n r)! = n! (n r)!r! = Cn n r (Proof) (Example) In a class, there are 20 male students. Choose 2 male students. How many possible choices? If one chooses 5 students from 20 students? 22

5 The terms ( ) n r are generally referred to as binomial coefficients, because they occur in the expansion of the binomial expansion (x + y) n = n ( n i=0 i ) x n i y i. (Example 2.12) Let A denote the event that exactly one of the two best applicants appears in a selection of two out of five. Find the number of sample points in A and P (A). (Example) There are 10 balls, of which 3 are basket balls, and 7 are footballs. If we want to put them in a row, how many possible arrangement?(there are two methods, using Cr n and Cn r n.) 23

6 We can use the following theorem to determine the number of subsets of various sizes that can be formed by partitioning a set of n distinct objects into k nonoverlapping groups. (Theorem 2.3) The number of ways of partitioning n distinct objects into k distinct groups containing n 1, n 2,..., n k objects, respectively, where each object appears in exactly one and k i=1 n i = n, is (Proof) N = ( n n 1 n 2 n k ) = n! n 1!n 2! n k!. The terms ( ) n n 1 n are often called multinomial coefficients, because they occur in the ex- 2 n k pansion of the multinomial term y 1 +y y k raised to the nth power: (y 1 + y y k ) n = ( n ) n y 1 n 1 n 2 n 1 yn 2 2 yn k k k where n n k = n. 24

7 (Example 2.10) A labor disputes has arisen concerning the distribution of 20 laborers to four different construction jobs. The first job required 6 laborers; the second, third, and fourth utilized 4,5, and 5 laborers, respectively. Determine the number of ways the 20 laborers can be divided into groups of the appropriate sizes to fill all of the jobs. (Example) There are 10 balls, of which 3 basket balls, 2 footballs, 3 are volleyballs and 2 soccer balls put all balls in a row. How many possible arrangement? 25

8 [Summary for counting rules] Number of possible arrangements of size r from n objects : With Replacement Without Replacement Ordered n r P n r Unordered C n r 26

9 2.7 Conditional probability (Example) The probability of a 1 in the toss of one balanced die is 1/6. One has new information that an odd number has fallen. Then probability of a 1 is 1/3. The probability of an event will sometimes depend on whether we know that other events have occurred. Unconditional probability of an event: Conditional probability of an event: the probability of the event given the fact that one or more events have already occurred. (Example) The unconditional probability of a 1 in the toss of one balanced die is 1/6. If we know that an even number has fallen, the conditional probability of the occurrence of a 1 is. 27

10 (Def 2.9) The conditional probability of an event A, given that an event B has occurred, is equal to P (A B) = P (A B) P (B) provided P (B) > 0. P (A B) is read probability of A given B. Note that the outcome of event B allows us to update the probability of A. (Example 2.14) Suppose that a balanced die is tossed once. 1) Use (Def 2.9) to find the probability of a 1, given that an odd number was obtained. 2) Use (Def 2.9) to find the probability of a 4, given that an odd number was obtained. 28

11 (Example) In a class, there 44 Junior and Senior students, 30 are men and 40 Junior of whom 12 are women. If a student is randomly selected from this class, (a) What is the probability that the selected student is a Junior male student? (b) If the selected student is female, what is the probability that she is a Senior? (Example) A child mixes ten good and three dead batteries. To find the dead batteries, his father tests them one by one and without replacement. What is the probability that his father find all three dead batteries at the fifth test? 29

12 If probability of the occurrence of an event A is unaffected by the occurrence or nonoccurrence of event B, we would say that events A and B are independent. (Def 2.10) Two events A and B are said to be independent, if any one of the following holds: P (A B) = P (A), P (B A) = P (B), P (A B) = P (A)P (B). Otherwise, the events are said to be dependent. (Example 2.15) Consider the following events in the toss of a single die: A : Observe an odd number. B : Observe an even number. C : Observe 1 or 2. a) Are A and B independent events? b) Are A and C independent events? 31

13 (Example) A red fair dice and a white fair dice are rolled: A : 4 on the red die. B : Sum of dice is odd. C : Sum of dice is 10. a) Are A and B independent events? b) Are A and C independent events? (Theorem) If A and B are independent, then the following pairs of events are also independent. (Proof) (a) (a)a and B, (b)ā and B, (c)ā and B. 32

14 Extension of (Def 2.10) Three events A, B and C are said to be mutually independent, if and only if the following two conditions hold: (a) They are pairwise independent : P (A B) = P (A)P (B), P (A C) = P (A)P (C), P (B C) = P (B)P (C), (b) P (A B C) = P (A)P (B)P (C). Otherwise, the events are said to be dependent. (Example) An urn contains 3 red, 2 white and 4 yellow balls. An ordered sample 3 is drawn from the urn with replacement. Find the probability of the sequence RWY(i.e., R in 1st, W in 2nd and Y is 3rd). A : R in 1st. B : W in 2nd. C : Y is 3rd. Are A, B and C mutually independent? 34

15 (Example) Consider tossing a fair coin three times and consider the following three events: A : Number of heads is even. B : The first two flips are the same C : The second two flips are heads. Are A, B and C mutually independent? (Example) An experiment consists of tossing different balanced dice, white and black. The sample space S of the outcomes consists of all ordered pairs (i, j)(i = 1,..., 6, j = 1,..., 6): S = (1, 1), (1, 2),..., (1, 6),..., (6, 1), (6, 2),..., (6, 6). Define the following events: E 1 = First die is 1, 2, or 3, E 2 = First die is 3, 4, or 5, E 3 = Sum of the faces is 9. Are E 1, E 2 and E 3 mutually independent? 35

16 Generally, many experiments consists of a sequence of n trials that are mutually independent. If the outcomes of the trials do not have anything to do with one another, then events, such that each is associated with a different trial, should be independent in probability sense. That is if A i is associated with the ith trial, i = 1,..., n, then P (A 1 A 2 A n ) = P (A 1 ) P (A 2 ) P (A n ). (Example) A fair six-sided die is rolled 6 independent times. Define an event A i as follows: A i = side i is observed on the i-th roll, i = 1,..., 6. What is the probability that none of A i occurs? 36

17 2.8 Two laws of probability Theorem 2.5 and 2.6 are useful for calculating the probabilities of unions and intersections of events. They play an important role in the event-composition approach(section 2.9). (Theorem 2.5) The probability of the intersection of two events A and B is P (A B) = P (A)P (B A) = P (B)P (A B) If A and B are independent, then P (A B) = P (A)P (B). (Proof) From (Def 2.9). Its extension: The probability of the intersection of any number of, say, k events, can be obtained in the following way: P (A 1 A 2 A k ) =P (A 1 )P (A 2 A 1 )P (A 3 A 1 A 2 ) P (A k A 1 A 2 A k 1 ). 37

18 (Theorem 2.6) The probability of the union of two events A and B is P (A B) = P (A) + P (B) P (A B) If A and B are mutually exclusive, P (A B) = 0 and P (A B) = P (A) + P (B). (Proof). Its extension for three events: (Theorem 2.7) If A is an event, then (Proof) P (A) = 1 P (Ā) Sometimes it is easier to calculate P (Ā) than to calculate P (A). In such cases, it is easier to find P (A) by the relationship P (A) = 1 P (Ā) than to find P (A) directly. 38

19 (Exercise 2.94) (Exercise 2.95) (Exercise 2.104) 39

20 2.9 Calculating the Probability of an event The event-composition method : calculate the probability of an event(defined on a discrete sample space), A, expresses A, as a composition involving unions and/or intersections of other events. The laws of probability are then applied to find P(A). A summary of the steps follows: 1. Define the experiment. 2. Visualize the nature of the sample points. Identify a few to clarify your thinking. 3. Write an equation expressing the event of interest, say, A, as a composition of two or more events, using unions, intersections, and/or complements. Make certain that event A and the event implied by the composition represent the same set of sample points. 4. Apply the additive and multiplicative laws of probability to the compositions obtained in step 3 to find P (A). 40

21 (Examples 2.20) It is known that a patient with a disease will respond to treatment with probability equal to 0.9. If three patients with disease are treated and respond independently, find the probability that at least one will respond. (Examples 2.22) (Exercise 2.111) An advertising agency notices that approximately 1 in 50 potential buyers of a product sees a given magazine ad, and 1 in 5 sees a corresponding ad on television. One in 100 see both. One in 3 actually purchases the product after seeing the ad, 1 in 10 without seeing it. What is the probability that a randomly selected potential customer will purchase the product? 41

22 2.10 Bayes Rule The event-composition approach is sometimes facilitated by viewing the sample space S as a union of mutually exclusive subsets and using the following law of total probability. (Def 2.11) For some positive integer k, let the sets B 1, B 2,..., B k be such that 1. S = B 1 B 2 B k. 2. B i B j = φ for i j. Note that B i, i = 1,..., k are mutually exclusive and exhaustive. Then the collection of sets {B 1, B 2,..., B k } is said to be a partition of S. If A is any subset of S and {B 1, B 2,..., B k } is a partition of S, A can be decomposed as follows: A = 42

23 (Theorem 2.8) Assume that {B 1, B 2,..., B k } is a partition of S such that P (B i ) > 0 for i = 1, 2,..., k. Then for any event A (Proof) P (A) = k i=1 P (A B i )P (B i ) (Theorem 2.9)[Bayes s Rule] Assume that {B 1, B 2,..., B k } is a partition of S such that P (B i ) > 0 for i = 1, 2,..., k. Then (Proof) P (B j A) = P (A B j)p (B j ) ki=1 P (A B i )P (B i ) 43

24 (Examples 2.23) (Exercise 2.124) A population of voters contains 40% Republicans and 60% Democrats. It is reported that 30% of the Republicans and 70% of the Democrats favor an election issue. A person chosen at random from this population is found to favor the issue in question. Find the conditional probability that this person is a Democrat. 44

25 (Example) Box B 1 contains 2 red and 4 white balls. Box B 2 contains 1 red and 2 white balls. Box B 3 contains 5 red and 4 white balls. The experiment consists of selecting a box and then drawing a ball from that box. The probability for selecting the boxes are not the same but given by P (B 1 ) = 1/3, P (B 2 ) = 1/6 and P (B 3 ) = 1/2, where B 1, B 2 and B 3 are the events that B 1, B 2 and B 3 are chosen, respectively. If the selected ball is red, find the conditional probability that it was drawn from box B 1. 45

26 (Example) A package, say P 1, of 24 balls, contains 8 green, 8 white, and 8 purple balls. A Package, say P 2, of 24 balls contains 6 are green, 6 white and 12 purple balls. One of the two packages is selected at random. (a) If 3 balls from this package were selected, all 3 are purple. Compute the conditional probability that package P 2 was selected. (b) If 3 balls from this package were selected, they are 1 green, 1 white, and 1 purple ball. Compute the conditional probability that package P 2 was selected. 46