I. WHAT IS PROBABILITY?

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1 C HAPTER 3 PROBABILITY Random Experiments I. WHAT IS PROBABILITY? The weatherman on 0 o clock news program states that there is a 20% chance that it will snow tomorrow, a 65% chance that it will rain and a 5% chance that the weather will be clear. What does this mean? What are the chances that it will not rain tomorrow? What are the chances that there will be some form of precipitation? Assuming that it will either snow, rain or be clear, which of the three will occur? The table below shows the results of a recent poll of 000 Americans on the President s economic policies: Strongly approve 22% Mildly approve 5% Mildly disapprove 7% Strongly disapprove 44% No opinion 2% If a person is selected at random from the set of people who were polled, what is the likelihood that the person approved of the President s economic policies? What is the likelihood that the person did not disapprove? An ordinary die is rolled. What is the likelihood that the number that turns up is a prime number? What are the chances that the number that turns up is not greater than 4? A two-headed coin is flipped. What is the likelihood that the coin will come up heads? Tails? Processes in which there are an observable set of possible outcomes are called experiments. There are two basic types of experiments. In the first three examples above, we do not know in advance what the outcome will be. We do not know for certainty what the weather will be tomorrow. If we pick a person at random from the poll group, we would 43

2 have no way of knowing in advance what the person s preference will turn out to be. In contrast, if we flip a two-headed coin we do know for sure what the result will be, heads! An experiment in which any one of number of possible outcomes may result is called a random experiment or probability experiment. In contrast, a process in which the outcome is known in advance (e.g., flipping a two-headed coin) is called a deterministic experiment. is the branch of mathematics that is concerned with modeling random experiments. That is, probability attempts to provide mathematical formulations (mathematical models) of random experiments. Since we will be concerned only with random experiments in the work that follows, the word experiment will mean random experiment. Sample Spaces Suppose we perform an experiment. The set S of possible outcomes is called the sample space for the experiment. Examples.: List, or describe, the sample space for each experiment.. Roll an ordinary die and record the number of dots on the upper face. 2. Draw a card from a standard 52-card deck and record its suit. 3. Draw a card from a standard deck. 4. Toss a fair coin. If the result is heads, stop. If the result is tails, toss the coin a second time. 5. A dartboard is in the shape of a square of side length 2 inches. Place a coordinate system on the board with the origin at the center of the square. Throw a dart at the board and record its position as an ordered pair of real numbers (a,b) Solutions:. Since any one of the six faces can land up, we set,2,3,4,5,6 S. 2. There are four suits: hearts (H), spades (S), diamonds (D), and clubs (C). Thus we set S H, S, D, C. 3. There are 52 possible outcomes: ace of hearts (AH), king of hearts (KH), two of S AH, KH,...,2H, AS, KS,..., 2C. Alternatively, hearts (2H), and so on. Thus, S c c is a card from a standard 52 - card deck. 44

3 4. A tree diagram is a convenient way to picture the outcomes of this experiment. H H T T Thus, S H, TH, TT. 5. Since the square has side length 2 inches and the origin is at the center, the coordinates of an arbitrary point (a,b) on the board must satisfy 6 a 6, 6 b 6. Therefore, S ( a, b) a and b are real numbers, and 6 a 6, 6 b 6. Finite and infinite sample spaces: Look at the experiments in Examples.. There is a significant difference between the sample spaces for experiments 4 and experiment 5, namely, the sample spaces in 4 have only finitely many elements, while the sample space in experiment has infinitely many elements. We will be concerned only with the finite case in this treatment of probability. Choosing a sample space: A sample space for an experiment is the set of possible outcomes. However, there may be different appropriate sample spaces for exactly the same experiment. The choice of the sample space depends on what you want to observe, on what questions you want to answer. Here are some simple examples. Examples.2:. a. Flip a fair coin twice. b. Flip a fair coin twice and record the total number of heads that come up. c. Flip a fair coin twice and record whether the results match or not. 2. a. Roll an ordinary die. 45

5 Events Suppose we perform an experiment and determine a sample space S with possible outcomes e e, e,..., e, 2 3 n. An event is a subset of S (including the empty set and the whole set S). A simple event is a one-element subset of S; the simple events represent each of the possible outcomes of the experiment. For example, if we roll an ordinary die, then S, 2, 3, 4, 5, 6, and e 2 2 represents the event that the number 2 turns up, and so on. e is the (simple) event that the number turns up, If an event E contains more than one element, then it is a compound event. It is important to note that, in considering random experiments, the sample space S plays the role of the universal set introduced in the Chapter Sets. Examples.3:. An experiment consists of recording the possible boy-girl compositions of a three-child family. a. How many simple events are there? b. Use a tree diagram to determine a sample space for the experiment. c. Let E be the event at least 2 girls. What is E? d. Let F be the event four boys. What is F? 2. An experiment consists of rolling a pair of dice, one red, the other green, and recording the sum of the numbers on the top faces. a. Give a sample space for the experiment. How many outcomes (simple events) are there? b. Let E be the event the sum is a prime number. What is E? c. Let F be the event the sum is greater than. What is F? 3. Give the sample space for each of the following experiments coin tossing experiments and find the event E = exactly one head turns up. a. Toss a fair coin twice. 47

6 b. Toss a penny and a nickel. c. Toss two identical fair coins. Solutions:. a. Since the first child is either a boy or a girl, the second is either a boy or a girl, and the third is either a boy or a girl, the number of possible outcomes is by the Multiplication Principle. B b. B G B G B G G B G B G B G Thus, S BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG. c. E BGG, GBG, GGB, GGG d. F 2. a. There are eleven possible outcomes. b. S 2, 3, 4, 5, 6, 7, 8, 9,0,,2 c. E 2, 3, 5, 7, d. F S. 3. a. & b. Tossing one coin twice and tossing two distinguishable coins (e.g., a penny 48

7 and a nickel) are equivalent; we can distinguish what each does. S = {HH, HT, TH, TT} and E = {HT, TH} c. If we toss two identical coins, then we can only record what we might see: S = {HH, HT, TT} and E = {HT}, a simple event. The outcomes HT and TH in the sample spaces above become one outcome here because we cannot tell which coin came up heads and which coin came up tails; the coins are indistinguishable. Another way to represent the sample space is in terms of the number of heads (or the number of tails) that turn up: S = {2,, 0}. Exercises 3.:. Roll a pair of dice and note the numbers that turn up. Give the sample space if: a. One of the dice is red and the other is green. b. The dice are identical. 2. An experiment consists of tossing a fair coin and then rolling a fair die. a. How many possible outcomes are there? b. Give the sample space. c. Find the event E = either the coin is heads or the die is odd. 3. A couple decides to have children until they have at least one of each sex, up to a maximum of three children. a. Draw a tree diagram for this experiment. b. Give the sample space. c. Find the event E = at least two girls. 4. A box contains two \$ bills, one \$5 bill and one \$0 bill. An experiment consists of drawing bills out of the box, without replacement, until the \$0 bill is drawn. a. Draw a tree diagram for this experiment. b. Give the sample space. c. Find the event E = the sum of the bills drawn is \$7. d. Find the event F = the sum of the bills drawn is less than \$5. 5. An urn contains three balls numbered, 2, and 3. 49

8 a. The balls are drawn out in succession, without replacement, and the number on the ball is recorded. Draw a tree diagram for this experiment and find all possible 3-digit numbers. b. Suppose three balls are drawn as follows: a ball is drawn, the number recorded, and then replaced before the next ball is drawn. How many elements are in the sample space? 6. Suppose a ball numbered 0 is added to the urn in Problem 5. The balls are drawn out in succession, without replacement, and the number on the ball is recorded. a. How many elements are in the sample space? b. How many 4-digit numbers are there? (Remember that a 4-digit number cannot begin with 0.) c. How many 4-digit numbers are odd? d. How many 4-digit numbers are less than 3000? 7. An experiment consists of rolling two fair dice. However, one of the dice has exactly one dot on two opposite faces, exactly two dots on two opposite faces, and exactly three dots on two opposite faces; the other one is a standard die. a. Give the sample space for this experiment. b. Let E = the sum of the numbers that turn up is a prime number. How many elements are in E? c. Let F = the sum of the numbers is 6. How many elements are in F? 8. An experiment consists of dealing a 5-card hand from a standard deck of 52 cards. a. How many different hands are there? That is, how many outcomes are in the sample space? b. How many hands have 4 aces? c. How many hands have all the cards in the same suit? d. How many hands have four cards of one kind (i.e., 4 kings, or 4 queens, and so on). e. How many hands have three cards of one kind and two of another kind? 50

9 II. PROBABILITY OF AN EVENT As indicated above, probability is a quantification, or a mathematical model, of a random experiment. This quantification is a measure of the likelihood that a given event will occur when the experiment is performed. A probability function is a function P defined on the subsets of a sample space S. To define this function we proceed as follows: of a Simple Event: Let S e e,...,, 2 e n be the sample space of an experiment. To each simple event i P ( e i assign a real number ) two conditions are satisfied: e we, called the probability of the event e i, such that the following (i) The probability of each simple event is a number between 0 and, inclusive. That is, 0 e i ), i,2,..., n (ii) The sum of the probabilities of the simple events space is. That is, e, e2,..., e in the sample n P ( e ) e2)... e ) Any assignment that satisfies conditions (i) and (ii) is an acceptable probability assignment. Examples 2.:. e, e, e, e e S 2 3 4, 5 is the sample space of an experiment. Which of the following are acceptable probability assignments? If the assignment is not acceptable, explain why. (a) P e ) 0.3, e ) 0.2, e ) 0.5, e ) 0., e ) ( (b) P e ) 0.3, e ) 0.2, e ) 0.5, e ) 0.3, e ) ( (c) P e ) 0.3, e ) 0.2, e ) 0.5, e ) 0., e ) ( Roll a die. Find the sample space and determine two different acceptable probability assignments. n 5

10 Solutions:. (a) Condition (i) is satisfied; each value is a number between 0 and. Checking condition (ii): The sum is not ; the assignment is not acceptable. (b) Pe ( 3) 0.5 ; condition (i) is not satisfied; the assignment is not acceptable. (c) Condition (i) is satisfied; each value is a number between 0 and. Checking condition (ii): This is an acceptable probability assignment. 2. The sample space : S {,2,3,4,5,6}. There are infinitely many solutions to this problem. We ll give 3: Assignment #: ), 2), 3), 4), 5), 6). This is the assignment we would make if we know that the die is fair; each number is as likely to turn up as any other number. For another assignment, we simply have to insure that conditions (i) and (ii) are satisfied; there are infinitely many choices. Here s one: Assignment #2: ), 2), 3), 4), 5), 6). This assignment would indicate that the die is loaded in such a way that an even number is twice as likely to come up as an odd number. Assignment #3: ) = 2) = 4) = 5) = 6) = 0, 3) =. This is the assignment we would make if we knew that the die was loaded and that a 3 always comes up. In addition to being acceptable, we want the probability assignment to accurately reflect the experiment; we want the assignment of probabilities to be reasonable. For example, suppose we flip a fair coin. The sample space is S { H, T}, and the simple events are H the coin comes up heads and T the coin comes up tails. While the assignment 52

11 PH ( ) 3/ 4, PT ( ) / 4 is acceptable, it is not reasonable. Since the coin is fair, we reasonably assume that a head is just as likely to turn up as a tail. This implies that we should set H) = T) from which it follows that is the assignment we should make. PH ( ) / 2, PT ( ) / 2 Examples 2.2:. A box contains marbles: 6 red, 3 blue, and 2 white. After shaking the box thoroughly, you draw a marble out of the box. What is the sample space for this experiment and what probability would you assign to the simple events. 2. A coin is biased in such a way that a head is twice as likely to come up as a tails. Determine a probability assignment for the simple events of the sample space S { H, T} that reflects this bias. 3. A die is rolled 000 times with the results given in the table. Outcome Number of times Give two probability assignments for the simple events. Solutions:. The marble will either be red (R), blue (B), or white (W). Therefore, S { R, B, W}. Since 6 of the marbles are red, 3 are blue and 2 are white, we assign 6 3 2,, PR ( ) PB ( ) PW ( ). 53

12 2. We ll let PT ( ) x, 0 x, be the probability that a tail comes up. Then the probability that a head comes up is PH ( ) 2 x. Now, by condition (ii), x 2x 3x x. 3 Thus, 2, PH 3 3 PT ( ) ( ). 3. For the first assignment we ll use the results given in the table: ) 0.66, 2) 0.70, 3) ) 0.70, 5) 0.65, 6) These numbers are close to 6 assumption, then we set so it appears that the die is fair. If we make that ), 2), 3), 4), 5), 6). Examples 2. and 2.2 illustrate two typical ways in which acceptable and reasonable probability assignments are made for the simple events in a sample space S.. Theoretical. We use assumptions and analytical reasoning (the die is fair so the outcomes have equal probability; the coin is biased in a certain way), and impose the two conditions that the probability assignment must satisfy. No experiments are actually performed. 2. Empirical. For one of our assignments in Examples we used the results of an actual experiment. Empirical probability can be defined more precisely as follows: Suppose we perform an experiment n times and the simple event e i occurs with frequency f ( e i ). That is, the outcome e i occurs f ( e i ) times. Then the ratio f ( ei ) / n is called the relative frequency or empirical probability of e i. 54

13 frequency of occurence of ei f( ei) Pe ( i ). number of trials n We leave it as an exercise to verify that this assignment is acceptable. The connection between theoretical and empirical probability assignments for a given random experiment is contained in a theorem which states that the empirical probabilities approach the theoretical probabilities as n gets larger and larger, provided the theoretical probability assignment is correct. Look at Example again. The empirical probability assignments to the outcomes are all close to /6. Based on this evidence, if we rolled the die 0,000 times or a 00,000 times, we would expect that a would occur essentially /6 of the time, a 2 would occur essentially /6 of the time, and so on. So, we assume that the die is fair and assign (theoretically) the probabilities )=/6, 2)=/6,, 6)=/6. The Equally Likely Assumption: If we toss an unbiased coin, we would expect that a heads is as likely to turn up as a tails. There are two possible outcomes, they are equally likely to occur, and so we assign the value, /2, to each outcome; P ( H ) T ) / 2. To say this another way, if we were to toss the coin a large number of times, we would expect that half (or at least very close to half) the tosses would result in heads, and half in tails. If we roll a fair die, there are six possible outcomes and any one number is as likely to turn up as any other. Thus, we assign the values P ( ) )... 6) / 6. If we draw a card from a well-shuffled, standard deck, then any one card is a likely to be drawn as any other and so we assign the value P ( e) / 52 to each of the 52 simple events. In general, if a sample spaces S has n elements, S e, e2,..., e n and if we determine that each of the outcomes e i is as likely to occur as any other, then we assign the probability value / n to each. That is, Examples 2.3: e i ), i,2,3,..., n n. A red die and a green die are rolled. The dice are fair. Let e ij denote the outcome i on the red die, j on the green die. What is P e )? ( ij 55

14 2. Two identical die are rolled and the sum of the numbers that turn up is noted. What is the sample space? Are the outcomes equally likely? 3. A combination lock has 3 dials, each labeled with the 9 digits from 0 to 9. An opening combination is a sequence of 3 digits with no digit repeated. How many opening combinations are there? What is the probability of a person guessing the right opening combination? 4. A 5-card hand is dealt from a standard, well shuffled, 52-card deck. How many simple events are in the sample space? Are they equally likely? If so, what are the probability assignments for the simple events? Solutions:. The dice are fair; each outcome is as likely to occur as any other; there are 36 possible outcomes (6 possibilities on the red die, 6 possibilities on the green die, 6 x 6 =36). Therefore, Pe ( ). i, j The sample space is S = {2, 3, 4, 5,,, 2}. There are outcomes. If we assume that the outcomes are equally likely, then we would assign to each outcome. While this assignment is acceptable, is it reasonable? Note that there is only one way to get a 2, both dice have to come up. Similarly, there is only one way to get a 2, both dice have to come up 6. On the other hand, there are several ways to get a 7 a and a 6, a 2 and a 5, etc. Therefore a 7 is more likely to occur than a 2 or a 2. We conclude that the outcomes are not equally likely and so the equally likely assignment is not appropriate. We will finish this example below. 3. The first digit can be any one of the 0 numbers. Since no digit can be repeated, the second digit can be any one of the remaining 9 and the third any one of the 8 that remain after the first 2 numbers have been chosen. Thus, by the Multiplication Principle, the number of opening combinations is 0 x 9 x 8 = 720 Note that the number of opening combinations is the number of permutations of 0 things taken 3 at a time, P 0, 3. The sample space is the set of possible opening combinations. The probability of guessing the correct one is / The number of 5-card hands that can be dealt from a standard deck is the number of combinations of 52 things taken 5 at a time; C 52, 5. Each hand is a likely to occur as any other so each hand has probability /C 52, 5. 56

15 of an Event: Suppose that we have been given a random experiment, determined the sample space S, and devised an acceptable probability assignment for the simple events. How do we define the probability of an arbitrary event E?, 2 3 e n be the sample space for an experiment, and suppose E is an arbitrary subset of S. We define the probability of E as follows: Let S e e, e,..., (i) If E is the empty set, then P ( E) 0. (ii) If E is a simple event, then E) has already been assigned. (iii) E { e, e,..., e } is a compound event, then i i i 2 k E) e i ) e i )... e i ). 2 k That is, E) is the sum of the probabilities of the simple events that comprise E. (iv) If E = S, then P ( E).This follows from (iii) and the definition of the probability of simple event. Note: An event E such that E) = 0 is called an impossible event. An event E such that E) = is called a certain event. Examples 2.4:. Toss a fair coin twice. What is the probability that at least one tail appears? 2. Roll a red die and a green die. (a) What is the probability that the sum of the numbers that turn up is 7? (b) What is the probability that the sum of the numbers that turn up is less than 4? (c) What is the probability that the sum of the numbers is 3? 57

16 3. A company receives orders from women in the following age groups: Age Number of Orders Under and over 3 00 A customer is selected at random from the set of 00 customers. Solutions: (a) Give the sample space for the experiment and determine the probabilities of the simple events. (b) What is the probability that the customer selected is under 40?. The sample space is S { HH, HT, TH, TT}. The outcomes are equally likely and so we assign probability to each. Let E = at least one tail. Then E { HT, TH, TT} and 4 3 PE ( ) The sample space is illustrated in the figure. GREEN DIE (,) (,2) (,3) (,4) (,5) (,6) RED DIE (2,) (2,2) (2,3) (2,4) (2,5) (2,6) (3,) (3,2) (3,3) (3,4) (3,5) (3,6) (4,) (4,2) (4,3) (4,4) (4,5) (4,6) (5,) (5,2) (5,3) (5,4) (5,5) (5,6) (6,) (6,2) (6,3) (6,4) (6,5) (6,6) 58

17 As we discussed previously, the outcomes are equally likely to occur and so we assign the value to each. 36 (a) Let A = the sum of the numbers is 7. Then and A {(,6),(2,5),(3,4),(4,3),(5,2),(6,)} 6 PA ( ) (b) Let B = the sum of the numbers is less than 4. Then and B {(,), (, 2), (2,)} 3 PB ( ) (c) Let C = the sum is 3. Since the sum of the numbers on the dice is a number between 2 and 2. C and PC ( ) 0; C is an impossible event. 3. (a) Based on this data, 20) 22 /00, 20-29) 22 /00, 30-39) 47 /00, 40-49) 8/00, 49) 3/00 (b) The probability that she is under 40 is: P ( 20) 20 29) 30 39) It is clear that the critical step in determining the probabilities of events associated with a given experiment is the assignment of acceptable, reasonable probabilities to the simple 59

18 events in the sample space S. Once that has been done, we can find the probability of any event connected with the experiment. The equally likely case revisited: As illustrated above, probability experiments that result in equally likely simple events is the easiest case; each of the simple events is assigned probability / n where n is the total number of possible outcomes. While not all probability experiments have equally likely outcomes, many can be restated or refined so that the outcomes are equally likely and this is an effective approach to solving probability problems. So now we consider the special case of the probability of an event E when the outcomes of the experiment are equally likely. Suppose that S e e, e,...,, 2 3 is the sample space of an experiment and that the outcomes are equally likely. Then e n e i ), i,2,3,..., n. n If E { e, e,..., e }, (E consists of k simple events), then i i i 2 k E) e i ) e i 2 )... e That is, the probability of the event E is simply i k )... n n n k terms k n. the number of elementsin E n( E) E). the number of elementsin S n( S) Examples 2.5:. Toss two identical coins. (a) What is the probability that the coins match (both heads or both tails)? (b) What is the probability that at least one head turns up? 2. Roll two identical dice (the completion of Example 2.3.2). 60

19 (a) What is the probability that the sum of the numbers that turn up is 7? (b) What is the probability that the sum of the numbers that turn up is less than 4? 3. A committee has 6 female and 5 male members. A 5-member subcommittee is to be selected at random from the committee members. (a) What is the probability that the subcommittee will consist of 3 females and 2 males? (b) What is the probability that the subcommittee will have at least 4 females? Solutions:. If we toss two identical coins the set of possible outcomes is S { HH, HT, TT}. Stated in terms of the number of heads that appear, S {2,,0}. However, these simple events are not equally likely; there is only one way to get 2 heads and only one way to get 0 heads, but exactly head can occur in two ways. If the coins were distinguishable, then the sample space would be R { HH, HT, TH, TT} and these events are equally likely with each having the probability /4. Note that the compound event { HT, TH } in R corresponds to the simple event HT (or exactly head) in S. Since the coins don t know whether or not they are distinguishable, we will assume that they are and then go on to determine the probability assignment for S:,, PHH ( ) PHT ( ) PTT ( ). (a) Let A = the coins match. Then A { HH, TT} and 2 PA ( ). (b) Let B = at least head. Then B { HH, HT, TH} and 3 4 PB ( ). 2. The sample space is: S {2, 3, 4,..., 2}. As we have seen, these outcomes are not equally likely; the assignment 2) 3) 4)... 2) is acceptable, but it is not reasonable. 6

20 As in part, we handle this situation by defining a sample space for the experiment in which the outcomes are equally likely. Assume that one of the dice is red and the other is green. There is no loss in generality here; the dice don t know what color they are. Now see Example (a) If A = the sum of the numbers is 7, then 6 PA ( ). (b) If B = the sum of the numbers is less than 4, then 2 PB ( ). 3. The sample space is the set of 5-member subcommittees and, since the subcommittee will be selected at random, any particular 5-member subcommittee is as likely to be chosen as any other. (a) Let E = the subcommittee has 3 female and 2 male members. Then no. of subcommittees with 3 females and 2 males PE ( ). no. of 5-member subcommittees The number of 5-member subcommittees is the combinations of things taken 5 at a time, C,5. View the selection of a subcommittee with 3 female members and 2 male members as a 2-step process: Step: Select the 3 female members. This can be done in C 6,3 ways. Step2: Select the 2 male member. This can be done in C 5,2 ways. Therefore, 6! 5! C6,3C 5,2 3!3! 2!3! PE ( ). C!, !6! (b) Let F = at least 4 females. The number of subcommittees with at least 4 female members equals the number with exactly 4 female members plus the number with exactly 5 female members. Proceeding as above, the number of subcommittees with 4 female and male member is given by C6,4C5, The number of subcommittees with 5 female members is given by C6,5 6. Thus, PF ( )

22 (a) 6 (b) 3 (c) An odd number. (d) 7 (e) Greater than Among 00 students in a college fraternity, 50 are majoring in the humanities majors, 30 are majoring in a science and 20 are engineering majors. If a student is selected at random, what is the probability that he is: (a) An engineering major? (b) Neither a science nor an engineering major? 0. A roulette wheel has 38 spaces numbered, 2,, 36, 0 and 00. Find the probability of getting: (a) An odd number. (b) A number greater than 24. (c) A number less than 5. (d) A prime number.. An experiment consists of dealing 5 cards from a standard deck of 52 cards. What is the probability of being dealt: (a) 5 hearts? (b) 5 black cards? (c) 4 of a kind (i.e., 4 aces, or 4 kings, or 4 queens, and so on)? (d) 3 aces and 2 queens? (e) 3 of one kind and two of another (a full house )? 2. A 4-person committee is to be composed of employees from two departments, department A with 5 employees and department B with 20 employees. If the 4 people are selected at random from the pool of 35 people, what is the probability that: (a) All are from department B? (b) 3 are from B and is from A? (c) 2 are from A and 2 are from B? (d) At least 3 are from department A? 64

23 III. UNION AND INTERSECTION OF EVENTS; COMPLEMENT OF AN EVENT; ODDS Unions and Intersections: Suppose we are given an experiment with sample space S. Let A and B be events in S and let E be the event either A occurs or B occurs. Then E occurs if the outcome of the experiment is either in A, or in B, or in both A and B. Therefore we conclude that E A B. Similarly, if we let F be the event both A and B occur, then F A B. Recall that if A and B are sets, then A B x x A or x B and A B x x A x B and. Example 3.: Roll a fair die. The sample space of equally likely simple events is: S, 2, 3, 4, 5, 6. Let A be the event an odd number turns up and let B be the event the number that turns up is divisible by 3. (a) Find the probability of the event E = the number that turns up is odd or is divisible by 3. (b) Find the probability of the event F = the number that turns up is odd and is divisible by 3. Solutions: (a) A {, 3, 5}, B {3, 6}, E A B {, 3, 5, 6} Since the simple events are equally likely, 65

24 n( A B) 4 2 P ( E) A B), n( S) 6 3 (b) F A B {3} and n( A B) P ( F) A B). n( S) 6 Suppose that A and B are events, and that E A B. We want to find E) in terms of A) and B)? There are two cases to consider: (i) A and B are disjoint: A B Ø. In this case, the events A and B are said to be mutually exclusive (if event A occurs, then event B cannot occur and vice versa). For mutually exclusive events A and B, the elements in A B are the elements in A plus the elements in B, and () A B) A) B). (ii) A and B are not disjoint: A B Ø. In this case, the elements in A B are the elements in A plus the elements in B minus the elements in A B. Therefore, (2) A B) A) B) A B). Note that the general formula (2) actually holds for both cases: since (2) reduces to () when A B Ø [ P ( A B) Ø) 0 ]. Also note the connection with addition principle of counting: Examples 3.2:. Roll a pair of fair dice. n(ab) = n(a) + n(b) n(ab). (a) What is the probability that the sum of the numbers is 7 or? (b) What is the probability that both dice either turn up the same number or that the sum of the numbers is less than 5? 66

25 2. What is the probability that a number selected at random from the first 50 positive integers is exactly divisible by 3 or 4? 3. A certain city has 2 daily newspapers, the Chronicle and the Times. A survey of 00 residents of the city was conducted to determine the readership of the two newspapers. It was found that 80 took the Chronicle, 65 took the Times, and 5 took neither paper. Assuming that the survey reflects the actual readership of the city, what is the probability that a resident selected at random: (a) Reads at least one of two papers? (b) Reads both papers? Solutions:. (a) Let A = sum is 7 and B = sum is. Then, A {(,6),(2,5),(3,4),(4,3),(5,2),(6,)}, B {(5,6),(6,5)}, and AB. Therefore, PA ( B) (b) Let A = both dice turn up the same number and B = the sum is less than 5. Then, A {(,),(2,5),(3,3),(4,4),(5,5),(6,6)}, B {(,),(,2),(,3),(2,),(2,2),(3,)}, and A and B are not disjoint. Therefore, PA ( B) PA ( ) PB ( ) PA ( B) S = {, 2, 3,, 50} and each number is as likely to be chosen as any other. Let A = the number is divisible by 3, B = the number is divisible by 4, and E A B. There are 6 numbers between and 50 that are divisible by 3; ; 3 6 PA ( ). 50 There are 2 numbers between and 50 that are divisible by 4; ; 2 2 PB ( )

26 Some of these numbers are divisible by both 3 and 4; for example, 2 and 24. The numbers that are divisible by both 3 and 4 are divisible by 2; there are 4 of these; ; 4 PA ( B) Thus, PE ( ) (a) Let C = reads Chronicle and T = reads Times. Since 5% take neither paper, we conclude that 85% read at least one of the two papers. Thus, PC ( T) (b) From the given information we get the Venn diagram; C T 5 and conclude that PC ( T) Complement of an Event: Suppose that E is an event in a sample space S. Since with E satisfies c E, the complement of E, together it follows that E E c S and E E c Ø, c c S) E E ) E) E ). Since P ( S) (recall the definition of probability of an event ), we have which implies c P ( E) E ) (3) E c ) E). 68

27 Examples 3.3:. The probability of an event E is PE ( ) 0.63, what is the probability of the complement of E? 2. A coin is tossed 4 times. What is the probability of getting at least one tail? 3. In a class of 0 students, 6 are female and 4 are male. If 3 of the students are selected at random, find the probability that at least one female is selected? Solutions: c. PE ( ) PE ( ) There are 6 equally likely outcomes and S { HHHH, HHHT, HHTH,..., TTTT}. c Let E = at least one tail. Then E = no tails = all heads. Thus, c PE ( ) and 6 5 PE ( ) Let E = at least one female is selected. Then E = no female is selected = all males are selected. There are C0,3 20 combinations of 0 students taken 3 at a time and there are C4,3 4 ways to select 3 male students. Therefore, c 4 29 PE ( ) and PE ( ) Odds It is common in gaming situations to quote the probability of winning (or losing) in terms of odds. The relationship between the odds for (or against) event E and the probability that event E occurs is given by: c (i) odds for E) E) E, provided P ( E) ; c E) E ) (ii) odds against c E ) E, provided P ( E) 0. E) 69

28 When possible, odds are expressed as a ratio of positive integers p / q,(also written p : q ). In this context, the ratio p / q (or p : q) is usually expressed as p to q. Examples 3.4:. What are the odds for rolling a number divisible by 3 in a single roll of a fair die? 2. What are the odds against rolling a 7 or an in one roll of a pair of fair dice? 3. What are the odds that there is at least one boy in a family of 4 children? Solutions:. Let E = the number is divisible by 3. Then E {3,6} ; 2 PE ( ) 6 3 and c 2 PE ( ). 3 The odds for E : 3 2 3, to 2, or : Let E = 7 or. Then, Pe (), and c 7 PE ( ). 9 The odds against E : , 7 to 2, or 7: Let E = at least one boy. Then, c E = all girls, and c PE ( ) ; 6 5 PE ( ). 6 The odds for E : , 5 to, or 5:. Odds to probability 70

29 Suppose we know that the odds for an event E are p to q. Then what is the probability that E occurs? What is E)? If the odds for E are p to q, then by the definition of odds given above, Therefore, E) E) p q q E) p[ E)] q E) p p E). ( p q) E) p and p E). p q Examples 3.5:. Calculate the probability of an event E if the odds for E are 5:2. 2. Calculate the probability of an event F if the odds against F are 9:5. Solutions:. 5 5 PE ( ) The odds for F are 5:9 ; 5 5 PF ( ) Exercises 3.3:. A standard die is rolled twice. The outcomes are equally likely. What is the probability that both rolls result in the same number of spots? 2. In the experiment described in Problem, what is the probability that the results 7

30 of the two rolls are different? 3. What is the probability that out of three people selected randomly, at least two will have the same birth month? Assume that all sequences of three birth months are equally likely. 4. In a certain city, 60% of the residents listen either to the opera or the Opry on the radio. Thirty percent of them listen to the opera. Forty four percent of them listen to the Opry. What percentage of the residents listens to both? 5. Suppose that PA ( ) 0.4, PB ( ) 0.6 and PA ( B) 0.. Find the probabilities of c c c the events B A and A B. A Venn diagram may be helpful. 6. The probability that a certain baseball team will win its division is 2/. What are the odds on its winning the division? 7. In a three-horse race, the odds on Blue Boy are 3:7 and the odds on Showgirl are :. What are the odds on Slim Jim? IV. CONDITIONAL PROBABILITY ; INTERSECTION; INDEPENDENCE Conditional : Let A and B be events in a sample space S for a given experiment. Suppose we are told that the experiment has been performed and that the event A has occurred. What effect does this additional knowledge have on the occurrence of event B? Does the probability that B will also occur increase? Decrease? Stay the same? The probability that B will occur given that A has already occurred is called a conditional probability and is denoted by P ( B A) (read the probability of B given A ). Example 4.: An experiment consists of tossing a penny, a nickel and a dime. Events A, B, C, and D are defined as follows: A = the penny comes up heads, B = at least two heads come up, C = exactly two tails come up, 72

31 D = the dime comes up tails. Solution: (a) Calculate P ( B A) and compare it with P (B). (b) Calculate P ( C A) and compare it with P (C). (c) Calculate P ( D A) and compare it with P (D). We know from previous examples that the sample space is: S HHH, HHT, HTH, HTT, THH, THT, TTH, TTT where, for example, HTH means that the penny came up heads, the nickel came up tails and the dime came up heads. Also, the outcomes are equally likely and so the probability of each of the simple events is /8. P ( B) 4/8 / 2. (a) Event B HHH, HHT, HTH, THH Now suppose that we are told that event A has occurred; that is, the penny came up heads. Since we know that the penny came up heads, the sample space has to be changed. It is now S ' HHH, HHT, HTH, HTT These are equally likely events and the probability of getting at least two heads is now 3 / 4 ; P ( B A) 3/ 4. Knowing that A has occurred has increased the probability B will occur. (b) Event C HTT, THT, TTH occurs is P ( C) 3/ 8.. In absence of any information, the probability that C Now assume that event A has occurred. Then S' is the new sample space of equally likely events and the probability of getting exactly two tails is / 4 ; P ( C A) / 4. Knowing that A has occurred decreased the probability of C occurring. and P ( D) 4/8 / 2 (c) Event D HHT, HTT, THT, TTT 73

32 If we assume that event A has occurred, then the sample space is S ' and P ( D A) 2 / 4 / 2. In this case, knowing that A has occurred did not change the probability of D occurring. Example 4. illustrates that in a given experiment, the occurrence of some event A may or may not influence the occurrence of some other event B. Examples 4.2:. A box contains 0 disks numbered, 2, 3,, 0. A person draws a disk from the box. (a) What is the probability that the number is a prime? (b) Suppose the person states that the number on the disk is odd. Now what is the probability that the number is a prime? 2. One hundred people responded to a survey that asked whether they thought women in the armed forces should be permitted to participate in combat. The results of the survey were: Gender Yes No Totals Male Female Totals Solutions: (a) What is the probability that a respondent answered yes given the respondent was a female? (b) What is the probability that a respondent was a male, given that the respondent answered no?. The sample space S {,2,3,...,0} consists of 0 equally likely outcomes. Let A = the number on the disk is odd and let B = the number on the disk is a prime (a) B {2,3,5,7} and na ( ) 4 2 PB ( ). ns ( )

33 (b) Given that the disk drawn has an odd number on it, the sample space becomes S' A {,3,5,7,9}. Now AB {3,5,7} and na ( B) 3 PBA ( ). na ( ) 5 Knowing that the number on the disk is odd improved the probability that the number is a prime. 2. (a) Given that the respondent was a female, the sample space becomes the set of 50 female respondents. The probability that she answered yes (Y) is 8 4 PY ( ) (b) Given that the respondent answered no, the sample space becomes the set of 74 respondents who answered no. The probability that the respondent was a male (M) is 32 6 P ( M ) We want to obtain a formula for the conditional probability of B given A in terms of the probabilities of the events A and B themselves. We will use a sample space S of equally likely outcomes to motivate the formula. Recall that in the equally likely case, the probability of an event E is given by n( E) E). n( S) Knowing that the event A has occurred means that we must change the sample space from S to the subset S' A. Now, to determine the probability that B will occur, given that A has already occurred, we need the number of elements in B that are also in A; that is, the number of elements in B A. Thus, n( B A) B A). n( A) Dividing numerator and denominator by n(s), we obtain 75

34 n( B A) n( S) B A) B A). n( A) A) n( S) This formula holds in general. For events A and B in an arbitrary sample space S, the conditional probability of B given A is given by (4) B A) B A), provided P ( A) 0. A) Examples 4.3:. A box contains black chips and white chips. A person draws two chips in succession without replacement. If the probability of selecting a black chip and a white chip is 5/56 and the probability of selecting a black chip on the first draw is 3/8, find the probability of selecting a white chip on the second draw given that first chip selected was a black chip. 2. In a certain housing subdivision, 35% of the homes have a family room and a fireplace. If 70% of the homes have a family room, find the probability that a home has a fireplace given that it has a family room. Solutions:. Let B = black chip and W = white chip. Then 5 PB ( W) PW ( B). PB ( ) Let R = family room and let F = fireplace. Then PR ( F) 0.35 PFR ( ). PR ( ) Intersection of Events; the Product Rule: 76

35 Let A and B be two events in a sample space S, and assume A) 0 and P ( B) 0. Then, by the formula (4), B A) B A) and A) A B) A B). B) Solving the first equation for B A) and the second equation for A B), we get B A) A) B A) and A B) B) A B). Since A B B A, we have the product rule: (5) A B) A) B A) B) A B). We can use either of these equations to calculate A B). Examples 4.4:. 70% of the shoppers at a certain grocery store are female and 80% of the female shoppers have a charge account at the store. Find the probability that a shopper selected at random is a female and has a charge account. 2. Suppose 60% of the male shoppers at the same grocery store have a charge account at the store. What is the probability that a shopper selected at random is a male who does not have a charge account. Solutions: Let C = shoppers with charge account, F = female shoppers, and M = male shoppers.. Since 70% of the shoppers are female, the probability that a shopper selected at random is a female is PF ( ) Since 80% of the female shoppers have charge accounts at the store, the probability that a shopper has a charge account, given that the shopper is a female, is PCF ( )

36 Therefore, the probability that a shopper selected at random is a female and has a charge account is PC ( F) PF ( ) PCF ( ) We conclude that 30% of the shoppers at the grocery store are male. Since 60% of the males have charge accounts, 40% do not. Therefore, PM ( ) 0.30 and c PC ( M) The probability that a shopper selected at random is a male and does not have a charge account is c c PC ( M) PM ( ) PC ( M) Trees: We used tree diagrams in the section on Counting to help us count the number of outcomes in an experiment that involved several steps. Here we use tree diagrams, called probability trees, to help us calculate the probabilities of events for experiments that consist of a sequence of steps. Example 4.5. A box contains 9 balls; 6 of them are red and 3 are white. A ball is drawn from the box and, without replacing it, a second ball is drawn. (a) Use a tree diagram to determine the sample space S. (b) Assign probabilities to the simple events. (c) Let A be the event the second ball is red. Calculate A). (d) What is the probability that the first ball is red, given that the second ball is white? Solutions: (a) R 2 R W 2 R 2 W W 2 78

37 The sample space is S { R R2, R W2, W R2, W W2} where R R2 represents the outcome red on the first ball and red on the second ball, R W2 represents red on the first ball and white on the second, and so on. (b) The probability of drawing a red ball on the first draw is 6/ 9 2/ 3 so we assign the probability 2/ 3 to the branch SR. The probability that the first ball drawn is white is 3/ 9 / 3 so we assign that value to the branch SW. Now to the next level: What probability should we assign to the branch R R 2? This is the conditional probability P R R 2 ), that is, the probability that the second ball is red ( given that the first ball was red. Since the box now contains 8 balls and 5 of them are red, P ( R R ) 5/8. Continuing in this manner, we assign probabilities to the other branches 2 of the tree to obtain the result in Figure A. R, R 2 2/3 R 5/8 3/8 R 2 W 2 5/2 /4 R, W 2 /3 3/4 R W /4 W 2 /2 /4 W, R 2 W, W 2 Figure A ) ( R W2 ) R ) W2 R ) Now, P ( R R ) R ) R R, P, and so on. The results are shown in Figure B. (c) A R R, W } and { 2 R2 5 2 ( A) R R2 ) W R2 ) P. 79

38 (d) By the conditional probability formula, W2 ) 4 R W2 ) R W2 ). We get W ) P ( R from the tree diagram. Now, the event W 2 = the second ball is white is given by W R W W 2 2, W 2 and so 2 ) 4 2 P ( W 3 2. Therefore, R W2 ) 3 P ( R W2 ). W ) 4 Independent/Dependent Events: As we discovered in Example 4., the occurrence of some event A may or may not change the probability that some other event B occurs. Events A and B are independent if the occurrence of one does not change the probability of the occurrence of the other. In symbols, A and B are independent if A B) A) or if B A) B). To be precise, we should say that if A B) A), then A is independent of B, or if B A) B), then B is independent of A. However, it can be shown that if A and B are events with nonzero probabilities, then A is independent of B if and only if B is independent of A. This is one of the problems in the Exercise Set. If A B) A) or if B A) B), then A and B are dependent events. Formula (5) provides another way to view independence. In particular, if A and B are independent in the sense that A B) A) or B A) B), then by formula (5), (6) A B) A) B). Moreover, if A and B are events with nonzero probabilities, then (6) implies that A and B are independent. There are some obvious examples of the concept of independence/dependence. If we toss a fair coin twice, the outcome on the second toss is independent of the outcome on the first toss, a coin has no memory. If we draw a card from a standard deck, put it back, and then draw a second card, the outcome on the second card is independent of the 80

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