MAT 326 Probability and statistics for Engineers Chapter 4: Probability 4.1: Sample Space and relationships among events • A sample space is a set of all possible outcomes of an experiment. • Notation: S • An event is any subset of a sample space. • Notation: A, B, C Examples • Sample space associated with a die toss: S={1,2,3,4,5,6} • Consider the event A: observing an even outcome. A={2,4,6} • Sample space associated with Flipping a coin: S={T,H} • T: tail, H: head Examples • Flip two coins: S={HH,HT,TH,TT} • B: observing at least one head; B={HH,HT,TH} • C: Observing at most one tail; C={HH,HT,TH} • D: Observing exactly one head; D={HT,TH} Examples • Roll two dice: S={(1,1),(1,2),…,(1,6),(2,1),(2,2),…,(6,6)} • A: sum of outcomes is equal to 11; A={(5,6),(6,5)} • B: sum of outcomes is at least 11; B={(5,6),(6,5),(6,6)} Relationships among events • Let ๐ be a sample space from an experiment and A and B two events. • Union: ๐ด ∪ ๐ต = {๐ฅ|๐ฅ ∈ ๐ด ๐๐ ๐ฅ ∈ ๐ต} • Intersection: ๐ด ∩ ๐ต = {๐ฅ|๐ฅ ∈ ๐ด ๐๐๐ ๐ฅ ∈ ๐ต} • Complement of an event: ๐ดาง = {๐ฅ ∈ ๐ ๐๐๐ ๐ฅ ∉ ๐ด} Relationships among events • Mutually exclusive events: ๐ด ∩ ๐ต = ∅ (empty set) • ๐ด ∪ ๐ดาง = ๐ • De Morgan’s laws: • ๐ด ∪ ๐ต = ๐ดาง ∩ ๐ตเดค • ๐ด ∩ ๐ต = ๐ดาง ∪ ๐ตเดค Venn diagram • Mutually exclusive events B A Venn diagram • Complement of an event Venn diagram • Union of two events Venn diagram • Intersection of 2 events Example • Flip a coin 3 times: S={HHT,HTH,HTT,HHH,THH,THT,TTH,TTT} • A: at least one head; A={HHT,HTH,HTT,HHH,THH,THT,TTH} • B: exactly one head; B={HTT,THT,TTH} • ๐ด∩๐ต =๐ต • ๐ด∪๐ต =๐ด • ๐ด:าง No head; ๐ดาง = {๐๐๐} • ๐ด ∪ ๐ดาง = ๐ • ๐ดาง ∩ ๐ต = ∅ H T H H T T H T H H T H T T 4.2-Definition of probability • Suppose S is a sample space associated with an experiment. A probability is a numerically valued function that assigns a number p(A) to every event A such that •๐ ๐ด ≥0 •๐ ๐ =1 • If ๐ด1 , ๐ด2 , … , ๐ด๐ is a sequence of mutually exclusive events, then • ๐ ๐ด1 ∪ ๐ด2 ∪ โฏ ∪ ๐ด๐ = σ๐๐=1 ๐ ๐ด๐ Equally likely outcomes • Roll a die. S={1,2,3,4,5,6} • Probability of each outcome is equal to 1/6 • Roll two dice: S={(1,1),(1,2),…,(6,6)} • Probability of each outcome is equal to 1/36 Equally likely outcomes • In general, if S is a finite sample space of some experiment and A is any event, under the assumption of equally likely outcomes, the probability that event A occurs is given by •๐ ๐ด = ๐ด ๐ • ๐ denotes the cardinality of the set S, i.e., the number of elements Examples • Flip three coins: S={HHT,HTH,HTT,HHH,THH,THT,TTH,TTT} • What’s the probability of obtaining at least one head. • A: at least one head; A={HHT,HTH,HTT,HHH,THH,THT,TTH} • p=7/8 • Roll two dice. What’s the probability that the sum of the outcomes is 8. • A: sum of outcomes is 8: {(5,3),(3,5),(4,4),(6,2),(2,6)} • p=5/36 Probability of the union of two events • Let S be the sample space of some experiment. Suppose A and B are two given events. • ๐ ๐ด∪๐ต =๐ ๐ด +๐ ๐ต −๐ ๐ด∩๐ต • If the events A and B are mutually exclusive then • ๐ ๐ด∪๐ต =๐ ๐ด +๐ ๐ต Example • Consider the national percentages for HIV-related risk groups (Science, 1992) given in Table 4.3. A firm hires a new worker after a national advertising campaign. • A) What is the probability that the worker falls in the “risky partner” category? • B) What is the probability that the worker is in at least one of the risk groups? • C) If the firm hires 1,000 workers, how many are expected to be at risk if the 1,000 came from the population at large? • D) If the firm hires 1,000 workers, how many are expected to be at risk if the 1,000 come from high-risk cities? Risk Group • Multiple partners • Risky partner • Transfusion recipient • Multiple partner and risky partner • Multiple partner and transfusion recipient • Risky partner and transfusion recipient • All others • No risk national Percent Number 7.0 170 3.2 76 2.3 55 high risk cities Percent Number 9.5 651 3.7 258 2.1 144 1.7 41 3.0 209 0.0 1 0.3 20 0.2 0.7 84.9 4 1 2,045 0.3 0.7 80.4 19 51 5,539 • A) Solutions to practical problems of this type always involve assumptions. To answer parts (a) and (b) with the data given, we must assume that the new worker is randomly selected from the national population, which implies P(risky partner) = 0.032 • b Let’s label the seven risk groups, in the order listed in the table, E1, E2, , E7 Then the event of being in at least one of the groups can be written as the union of these seven, that is, E1∪ E2 ∪ … E7 • The event “at least one” is the same as the event “E1 or E2 or E3 or … or E7” Because these seven groups are listed in mutually exclusive fashion, P(E1∪ ๐ธ2 ∪ ๐ธ3 … ∪ ๐ธ7) = P(E1) + P(E2) + โฏ P(E7) =0.070+0.032+0.023+0.017+0.0+0.002+0.007=0.151 • Note that the mutually exclusive property is essential here; otherwise, we could not simply add the probabilities • C) We assume that all 1,000 new hires are randomly selected from the national population (or a subpopulation of the same makeup). Then 15.1% of the 1,000, or 151 workers, are expected to be at risk. • D) We assume that the 1,000 workers are randomly selected from high-risk cities. Then 19.6%, or 196 workers, are expected to be at risk 4.3-Counting techniques • For many problems, it’s not straightforward to count the possible outcomes. • Roll a dice 5 times • All possible outcomes in a lottery. • All possible phone numbers in a region. • All possible car plates in a region. Multiplication rule • If ๐ธ1 is an experiment with ๐1 outcomes and ๐ธ2 is an experiment with ๐2 outcomes, then the experiment which consists of performing ๐ธ1 first and then ๐ธ2 consists of ๐1 โ ๐2 outcomes. • Generalization for an arbitrary number of experiments. • If ๐ธ1 is an experiment with ๐1 outcomes, ๐ธ2 is an experiment with ๐2 outcomes, and so on up to ๐ธ๐ which is an experiment with ๐๐ outcomes, then the experiment which consists of performing ๐ธ1 first and then ๐ธ2 up to ๐ธ๐ consists of ๐1 โ ๐2 โ โฏ โ ๐๐ outcomes. Example • How many different license plates are possible if Kentucky uses three letters followed by three digits. • ๐ธ1 : choosing first letter, ๐ธ2 : choosing second letter, ๐ธ3 : choosing third letter • ๐น1 : choosing first digit, ๐น2 : choosing second digit, ๐น3 : choosing third digit • 263 โ 103 Permutations • The number of arrangements or permutations of r objects taken from n distinct objects is given by • ๐๐๐ = ๐! ๐−๐ ! • 0! = 1, ๐! = 1.2. … ๐ • Proof: By the multiplication rule, The number of arrangements is ๐! given by ๐. ๐ − 1 . ๐ − 2 … ๐ − ๐ − 1 = ๐−๐ ! Example • From among 10 employees, three are to be selected for travel to three outof-town plants, A, B, and C, with one employee traveling to each plant. Because the plants are in different cities, the order of assigning the employees to the plants is an important consideration. The first person selected might, for instance, go to plant A and the second to plant B. In how many ways can the assignments be made? • Solution : Because order is important, the number of possible distinct assignments is 10! ๐310 = =10(9)(8)=720 7! In other words, there are 10 choices for plant A but only nine for plant B and eight for plant C. This gives a total of 10(9)(8) ways of assigning employees to the plants Example • An assembly operation in a manufacturing plant involves four steps, which can be performed in any order. If the manufacturer wishes to experimentally compare the assembly times for each possible ordering of the steps, how many orderings will the experiment involve? • Solution: The number of orderings is the permutation of n 4 things taken r 4 at a time. (All steps must be accomplished each time.) This turns out to be 4 4! ๐4 = =24 0! because 0! 1 by definition. Example • Four names are drawn from the 24 members of a club for the offices of President, Vice-President, Treasurer, and Secretary. In how many different ways can this be done? Combinations • In permutation, order is important. But in many problems the order of selection is not important and interest centers only on the set of r objects. • The number of subsets or combinations of size r that can be selected from n different objects is given by • ๐ ๐ = ๐! ๐! ๐−๐ ! Example • How many committees of two chemists and one physicist can be formed from 4 chemists and 3 physicists? • Number of ways of choosing two chemists: • Number of ways of choosing one physicist: • By multiplication rule, the answer is: 3.6=18 4 2 3 1 = = 4! 2!2! 3! 1!2! =6 =3 Exercise 14 • Two vehicles in succession are observed moving through the intersection of two streets. • A) List the possible outcomes, assuming each vehicle can go straight, turn left, or turn right. • B) Assuming the outcomes to be equally likely, find the probability that at least one vehicle turns left. • C) Find the probability that at most one vehicle turns. • A) Sample space; S={SS,SL,SR,LS,LL,LR,RS,RL,RR} • B) Let A be the event: at least one vehicle turns left. Then A={SL,LS,LL,LR,RL}. Hence p=5/9 • C) Let B be the event: at most one vehicle makes a turn. B={SS,SL,SR,LS,RS}. Hence p=5/9 Exercise 18 • Seven applicants have applied for two jobs. How many ways can the jobs be filled if • A) the first person chosen receives a higher salary than the second? • B) There are no differences between the jobs? • A)7.6=42 (permutations) • B) 7 2 = 7! 2!5! = 21 (combinations) Exercise 19 • A package of six light bulbs contains two defective bulbs. If three bulbs are selected for use, find the probability that none is defective. • Sample space: 6 3 = 6! 3!3! = 20 • Event: None of the three selected bulbs are defective: • Probability: 4 3 = 4! 1!3! =4 • The number of ways of partitioning ๐ distinct objects into ๐ groups containing ๐1 , ๐2 , … ๐๐ objects, respectively, is • ๐! ๐1 !๐2 !…๐๐ ! where • σ๐๐=1 ๐๐ = ๐ Exercise 21 • A flee of eight taxis is to be randomly assigned to three airports A, B, and C, with two going to A, five to B, and one to C. • A) In how many ways can this be done? • B) What is the probability that the specific cab driven by Jones is assigned to airport C? • In how many ways can this be done? 8! 2!5!1! = 6.7.8 2 = 168 • B) What is the probability that the specific cab driven by Jones is assigned to airport C? Exercises 1. How many different words can be created by rearranging the letters in SELFIESTICK? 2. A multiple choice test has 5 questions and each question has 4 choices. How many ways are there to answer the questions? What’s the probability that a student get all answers correctly? • 1) 11! 2!2!2! • 2) 45 • 1/45 • 3)A manufacturing company has two retail outlets. It is known that 30% of the potential customers buy products from outlet I alone, 50% buy from outlet II alone, 10% buy from both I and II, and 10% buy from neither. Let A denote the event that a potential customer, randomly chosen, buys from I and B denote the event that the customer buys from II. Find the following probabilities. • p(A), p(๐ด ∪ ๐ต), p ๐ด ∩ ๐ต , ๐ ๐ตเดค , ๐ ๐ด ∪ ๐ต , 3)p(A)=0.4, • p(๐ด ∪ ๐ต)=0.9, • p ๐ด ∩ ๐ต = 0.1, • ๐ ๐ตเดค = 0.4, • ๐ ๐ด ∪ ๐ต = 0.1, • 4)Suppose that three employees are to be selected from ten to visit a new plant. In how many ways can the selection be made? • If two of the ten employees are female and eight are male, what is the probability that exactly one female gets selected among the three? • 10 3 • 5)A firm is placing three orders for supplies among five different distributers. Each order is randomly assigned to one of the distributors, and a distributor can receive multiple orders. Find the probability that all orders go to different distributors. • 5.4.3 53 Conditional probability • ๐ ๐ด|๐ต = ๐ ๐ด∩๐ต • ๐ ๐ต|๐ด = ๐ ๐ด∩๐ต • ๐ ๐ด|๐ต = ๐ ๐ต|๐ด ๐ ๐ด ๐ ๐ต ๐ ๐ด ๐ ๐ต Examples • If we randomly pick two television sets in succession from a shipment of 240 television sets of which 15 are defective, what is the probability that they will be both defective? • A: first television picked was defective • B: second tv picked was defective • ๐ด ∩ ๐ต: both tv picked were defective • ๐ ๐ด ∩ ๐ต = ๐ ๐ด ๐ ๐ต|๐ด = 15 240 โ 14 239 = 7 1912 Examples • A drawer contains 4 black, 6 brown, and 8 red socks. Two socks are selected at random from the drawer. • A) What is the probability that both socks are of the same color? • B) What is the probability that both socks are red if it is known that they are of the same color? • There are 18 2 = 18! 2!16! = 153 ways to select two socks • Let A be the event that two socks selected at random are of the same color. Then the cardinality of A is given by 42 + 62 + 82 = 6 + 15 + 28 = 49 • Hence p(A)=49/153 • B: two socks selected are red. The cardinality of B is • P(B|A)=p(A∩B)/p(A)=p(B)/p(A)= 28/153 49/153 =28/49=4/7 8 2 = 28 Independent events • Two events A and B are independent iff p(A|B)=p(A) • ๐ ๐ด|๐ต = ๐ ๐ด∩๐ต ๐ ๐ต • This is equivalent to ๐ ๐ด ∩ ๐ต = ๐ ๐ด ๐ ๐ต . • EX33) Suppose that p(A)=0.6, p(B)=0.3, and ๐ ๐ด ∩ ๐ต =0.15. Are events A and B independent? No. Example: Classification of manufactured pens Assembly line 1 Assembly line 2 Total Defective-trash 8 2 10 Defective-to be fixed 13 27 40 Non defective 59 91 150 Total 80 120 200 • Suppose a pen is selected at random. What is the probability that it is a defective-trash pen manufactured by assembly line 1? • Suppose a pen selected at random was manufactured by assembly line 1. then what is the probability that it is defective-trash? • Are the events, defective-trash and produced by assembly line 2 dependent or independent? • Suppose a pen is selected at random. What is the probability that it is a defective-trash pen manufactured by assembly line 1?8/200 • Suppose a pen selected at random was manufactured by assembly line 1. then what is the probability that it is defective-trash?8/80 • Are the events, defective-trash and produced by assembly line 2 dependent or independent? dependent • P(defective trash)=10/200, p(produced by assembly line 2)=120/200 • P(defective trash manufactured by assembly line 2)=2/200 • 10/200x120/200 is different from 2/200. Probability of complement of an event • If ๐ดาง is the complement of an event A in a sample space S, then ๐ ๐ดาง = 1 − ๐(๐ด). Example • A quality-control inspector has ten assembly lines from which to choose products for testing. Each morning of a five-day week, she randomly selects one of the lines to work on for the day. Find the probability that a line is chosen more than once during the week • Solution: It is easier here to think in terms of complements and first find the probability that no line is chosen more than once. If no line is repeated, five different lines must be chosen on successive days, which can be done in ways. The total number of possible outcomes for the selection of five lines without restriction is , by an extension of the multiplication rule. Thus, P(no line is chosen more than once)= P(a line is chosen more than once) =1-0.3=0.7 Exercise Solution • Design A: combine relays 1 and 2 into the equivalent relay 5 which • opens with prob = 0.12 and closes with prob =1 − 0.12. Similarly, • combine the relays 3 and 4 into the equivalent relay 6 which opens • with prob= 0.12 and closes with prob = 1 − 0.12. The new circuit is in series. Hence, the prob that the current will flow is 1 − 0.12 2 = 0.9801 • Design B: combine relays 1 and 3 into the equivalent relay 5 which closes with prob = 0.92 and open with =prob 1 − 0.92. Similarly, combine relays 2 and 4 into the equivalent relay 6 which closes with prob = 0.92 and open with prob = 1 − 0.92. The new circuit is in parallel. Hence, the prob that the current will flow is 1 − (1 − 0.92 )2 = 0.9639 Bayes’ rule • If ๐ต1 , ๐ต2 form a partition of S (their union is equal to S and they are mutually exclusive), and A is any event in S, then • ๐ ๐ต1 |๐ด = ๐ ๐ต1 ๐ ๐ด|๐ต1 ๐ ๐ต1 ๐ ๐ด|๐ต1 +๐ ๐ต2 ๐ ๐ด|๐ต2 • Proof: • ๐ ๐ต1 |๐ด = ๐ ๐ด∩๐ต1 ๐ ๐ด = ๐ ๐ด∩๐ต1 ๐ ๐ด∩๐ต1 +๐ ๐ด∩๐ต2 = ๐ ๐ต1 ๐ ๐ด|๐ต1 ๐ ๐ต1 ๐ ๐ด|๐ต1 +๐ ๐ต2 ๐ ๐ด|๐ต2 • If ๐ต1 , ๐ต2 , … , ๐ต๐ form a partition of S (their union is equal to S and they are mutually exclusive), and A is any event in S, then • ๐ ๐ต1 |๐ด = ๐ ๐ต1 ๐ ๐ด|๐ต1 ๐ ๐ต1 ๐ ๐ด|๐ต1 +๐ ๐ต2 ๐ ๐ด|๐ต2 +โฏ+๐ ๐ต๐ ๐ ๐ด|๐ต๐ 28 p. 185 • Electric motors coming off two assembly lines are pooled for storage in a common stockroom, and the room contains an equal number of motors from each line. Motors are periodically sampled from that room and tested. It is known that 10% of the motors from line I are defective and 15% of the motors from line II are defective. • i) If a motor is randomly selected from the stockroom and found to be defective, find the probability that it came from line I. • ii) Find the probability that a randomly selected motor is defective. Solution • i) ๐ต1 : motor came from line I. • ๐ต2 : motor came from line II. • A: motor is defective • ๐ ๐ต1 = ๐ ๐ต2 = 0.5; ๐ ๐ด|๐ต1 = 0.1; ๐ ๐ด|๐ต2 = 0.15 • ๐ ๐ต1 |๐ด = ๐ ๐ต1 ๐ ๐ด|๐ต1 ๐ ๐ต1 ๐ ๐ด|๐ต1 +๐ ๐ต2 ๐ ๐ด|๐ต2 = 0.1 0.1+0.15 = 0.4 • ii) p(A)=๐ ๐ต1 ๐ ๐ด|๐ต1 + ๐ ๐ต2 ๐ ๐ด|๐ต2 =0.1+0.15=0.25 Nb 47 p.202 • Two methods A and B are available for teaching a certain industrial skill. The failure rate is 20% for A and 10% for B. However, B is more expensive and hence is used only 30% of the time. (A is used the other 70%) A worker is taught the skill by one of the methods but fails to learn it correctly. What is the probability that he was taught by method A? Solution • A: worker is taught by method A • B: worker is taught by method B • C: Worker fails to learn • ๐ ๐ด|๐ถ = ๐ ๐ด ๐ ๐ถ|๐ด ๐ ๐ด ๐ ๐ถ|๐ด +๐ ๐ต ๐ ๐ถ|๐ต = 0.7โ0.2 0.7โ0.2+0.1โ0.3 = 0.82 Nb 40 p.201 • An incoming lot of silicon wafers is to be inspected for defectives by an engineer in a microchip manufacturing plant. In a tray containing 20 wafers, assume four are defective. Two wafers are to be randomly selected for inspection. Find the probability that • A) Both are nondefective • B) At least one of the two is nondefective • C) Both are nondefective given that at least one is nondefective Solution • Let A: Both are nondefective and B: at least one is nondefective • A) • B) • C) ๐ ๐ด|๐ต = ๐ ๐ต∩๐ด ๐ ๐ต = ๐ ๐ด ๐ ๐ต Nb 37 p.200 • A purchasing office is to assign a contract for computer paper and a contract for microcomputer disks to any one of the three firms bidding for these contracts. (Any one firm could receive both contracts.) Find the probability that • A) Firm I receives a contract given that both contracts do not go to the same firm. • B) Firm I receives both contracts • C) Firm I receives the contract for paper given that it does not receive the contract for disks. Solution • Sample space: 9 outcomes • ๐1 ๐ท1 , ๐1 ๐ท2, ๐1 ๐ท3 , ๐2 ๐ท1 , ๐2 ๐ท2 , ๐2 ๐ท3 , ๐3 ๐ท1 , ๐3 ๐ท2 , ๐3 ๐ท3 • A: Firm I receives a contract • B: Both contracts do not go to the same firm. • A) 1 • B) p(Firm I receives both contracts)= 9 Continue • C: Firm I receives the contract for paper • D: Firm I does not receive the contract for disks • C) Ex • Suppose A and B are independent events. Are ๐ดาง and ๐ตเดค also independent? • ๐ ๐ดาง ∩ ๐ตเดค = ๐ ๐ด ∪ ๐ต = 1 − ๐ ๐ด ∪ ๐ต = • 1− ๐ ๐ด +๐ ๐ต −๐ ๐ด ∩ ๐ต = 1−๐ ๐ด −๐ ๐ต −๐ ๐ด ๐ ๐ต = 1−๐ ๐ด −๐ ๐ต 1−๐ ๐ด = 1−๐ ๐ด 1−๐ ๐ต • = ๐ ๐ดาง ๐ ๐ตเดค • Hence ๐ดาง and ๐ตเดค are also independent. Ex • Vehicles coming into an intersection can turn left or right or go straight ahead. Two vehicles enter an intersection in succession. Find the probability that at least one of the two vehicles turns left given that at least one of the two vehicles turns. • S={SS, SR,SL, RS,RR,RL, LS,LR,LL} • 5/8 Ex • Suppose a certain disease has an incidence rate of 0.1% (that is, it afflicts 0.1% of the population). A test has been devised to detect this disease. The test does not produce false negatives (that is, anyone who has the disease will test positive for it), but the false positive rate is 5% (that is, about 5% of people who take the test will test positive, even though they do not have the disease). Suppose a randomly selected person takes the test and tests positive. What is the probability that this person actually has the disease? • A: A person has the disease • ๐ด:าง A person does not have the disease • B: A person test positive • ๐ ๐ด|๐ต = ๐ ๐ด ๐ ๐ต|๐ด ๐ ๐ด ๐ ๐ต|๐ด +๐ ๐ดาง ๐ ๐ต|๐ดาง = 0.001×1 0.001×1+0.999×0.05 = 0.0196 Ex 15 • A commercial building is designed with two entrances, door I and door II. Two customers arrive and enter the building. • A) List the elements of a sample space for this observational experiment. • B) If all elements in part (a) are equally likely, find the probability that both customers use door I; then find the probability that both customers use the same door. • a. S {(C1d1, C2d1), (C1d1, C2d2), (C1d2, C2d1), (C1d2, C2d2)} • b. P(both customers use door I) =1/4 P(both customers use the same door)=1/2 Example • Five motors (numbered 1 through 5) are available for use, and motor 2 is defective. Motors 1 and 2 come from supplier I, and motors 3, 4, and 5 come from supplier II. Suppose two motors are randomly selected for use on a particular day. • Let A denote the event that the defective motor is selected and B the event that at least one motor comes from supplier I. • Find P(A) and P(B) First motor selected Second motor selected 1 2 3 4 5 2 1 3 4 5 3 1 2 4 5 4 1 2 3 5 5 1 2 3 4 • The tree diagram shows that there are 20 possible outcomes for this experiment, which agrees with our calculation using the product rule. That is, there are 20 events of the form {1, 2}, {1, 3}, and so forth. Because the motors are randomly selected, each of the 20 outcomes has probability 1/20. Thus, ๐ ๐ด = ๐ 1,2 ∪ ๐ 2,1 ∪ ๐ 2,3 ∪ ๐ 2,4 ∪ ๐ 2,5 ∪ ๐ 3,2 ∪ ๐ 4,2 • Similarly, we can show that B contains 14 of the 20 outcomes and that P(B)=14/20=0.7 Example • Five applicants for a job are ranked according to ability, with applicant number 1 being the best, number 2 second best, and so on. These rankings are unknown to an employer, who simply hires two applicants at random. • What is the probability that this employer hires exactly one of the two best applicants? • The number of possible outcomes for the process of selecting two applicants from five is • If one of the two best is selected, the selection can be done in =2 • The other selected applicant must come from among the three lowest-ranking applicants, which can be done in =3 • Thus, the event of interest (hiring one of the two best applicants) can 2×3 occur in =0.6 10 Example • A section of an electrical circuit has two relays in parallel, as shown in the Figure. The relays operate independently, and when a switch is thrown, each will close properly with probability only 0.8. • If the relays are both open, find the probability that current will flow from s to t when the switch is thrown 1 t s 2 • Let O denote an open relay and C a closed relay. The four outcomes for this experiment are shown by the following: • E1 (O,O) E2(O,C) E3(C,O) E4(C,C) • Because the relays operate independently, we can find the probability for these outcomes as follows: P(E4) = P(C)P(C) = (0.8)(0.8) = 0.64 P(E3) = P(C)P(O) = (0.8)(0.2) = 0.16 P(E2) = P(O)P(C) = (0.2)(0.8) = 0.16 P(E1) = P(O)P(O) = (0.2)(0.2) = 0.04 • If A denotes the event that current will flow from s to t, then ๐ด = ๐ธ2 ∪ ๐ธ3 ∪ ๐ธ4 ๐๐ ๐ด = ๐ธ1 That is, at least one of the relays must close for current to flow. Thus P(A)=1-P(A)=1-P(E1)=1-0.04=0.96 which is the same as P(E2)+P(E3)+P(E4) 1. If A and B are two independent events, P(A) = 0.6 and P(B) = 0.7, เดค is equal to then ๐(๐ดาง ∪ ๐ต) (a) 0.88 (b) 0.58 (c) 0.7 (d) None of these b