Exam 1
1. 13(a) Indicate the appropriate symbol <,>,,, or = and explain your choice. P(E [ F) ? P(E) + P(F): (b) Indicate if the following statement is true or false, explain your answer. Any event is independent with itself. (c) Consider two events, E and F. If P(E) = :6 and P(F) = :6 are E and F mutually exclusive? Explain your answer.
(a) Indicate the appropriate symbol <,>,,, or = and explain your choice. P(E [ F) ? P(E) + P(F): Solution: The formula should say P(E [ F) P(E) + P(F): Recall that we know the following formula relating E [ F and E \ F: P(E [ F) = P(E) + P(F) P(E \ F): But P(E \ F) 0, so P(E \ F) 0. This says that P(E [ F) = P(E) + P(F) P(E \ F) P(E) + P(F): Thus the correct symbol is \". A more intuitive and less algebraic approach: the sum P(E) + P(F) accounts for outcomes in the intersection twice!! So the probability of the union, P(E [ F) is equal to this sum if the intersection is empty (the events W and F are mutually exclusive) and less than this sum otherwise. (b) Indicate if the following statement is true or false, explain your answer. Any event is independent with itself. Solution: This is false. for any event E, P(E) P(E) = P(E)2, and P(E \ E) = P(E). These two quantities are equal only if P(E) = 0 or 1, which doesn't cover \any event." (c) Consider two events, E and F. If P(E) = :6 and P(F) = :6 are E and F mutually exclusive? Explain your answer. Solution: No, E and F are not mutually exclusive. To be mutually exclusive means that P(E \ F) = 0. But look what this would mean for P(E [ F): P(E [ F) = P(E) + P(F) P(E \ F) = 0:6 + 0:6 0 = 1:2: But we can't have a probability over 1! Thus P(E \ F) must be positive, so E and F are not mutually exclusive. What we're saying, in short,
1.01: Let A and B be events with P(A) = 0:5, P(B0) = 0:4, and P(A [ B) = 0:9. (a) Find P(B). (b) Find P(A \ B). (c) Are A and B independent? Explain your reasoning. (d) Are A and B mutually exclusive? Explain your reasoning.
1. Let A and B be events with P(A) = 0:5, P(B0) = 0:4, and P(A [ B) = 0:9. (a) Find P(B). Solution: P(B) = 1 P(B0) = 1 0:4 = 0:6. (b) Find P(A \ B). Solution: We know the other three quantities in the formula P(A [ B) = P(A) + P(B) P(A \ B): Substitution gives 0:9 = 0:5 + 0:6 P(A \ B); so P(A \ B) = 0:2. (c) Are A and B independent? Explain your reasoning. Solution: No, A and B are not independent. Recall that A and B are independent when P(A \ B) = P(A) P(B). Here P(A \ B) = 0:2 but P(A) P(B) = (0:5)(0:6) = 0:3. (d) Are A and B mutually exclusive? Explain your reasoning. Solution: No. Recall that two events A and B are mutually exclusive if they have no events in common. But here P(A \ B) 6= 0, so A \ B 6=. Thus they are not mutually exclusive.
1.02: On a problem set consisting of 10 true-false questions, a student guesses all his answers. For each problem, the probability that he answers correctly is the same as the probability that he gets it wrong. (a) What is the probability that he gets at least 7 problems correct? (b) What are the odds that he gets at least 7 problems correct?
2. On a problem set consisting of 10 true-false questions, a student guesses all his answers. For each problem, the probability that he answers correctly is the same as the probability that he gets it wrong. (a) What is the probability that he gets at least 7 problems correct? Solution: Let E be the event that he gets at least 7 problems correct (7 or 8 or 9 or 10 correct) in the sample space S of all possible ways to answer the ten questions. P(E) = n(E) n(S) = # of ways to get at least 7 correct # of ways to answer the 10 questions : The denominator is n(S) = 210 = 1024; the numerator is slightly more complicated. The number of ways to get exactly 7 correct is the number of ways to select 7 questions out of 10; this is C(10; 7), or \10 choose 7." Similarly, there are C(10; 8) ways to get exactly 8 questions correct, C(10; 9) = 10 ways to get exactly 9 questions correct, and C(10; 10) = 1 way to get all 10 questions right. Thus n(E) = C(10; 7) + C(10; 8) + C(10; 9) + C(10; 10) = 176; And the probability that the student will guess 7 or more correct is P(E) = n(E) n(S) = 176 1024 = 11 64 = 0:171875: (b) What are the odds that he gets at least 7 problems correct? Solution: Odds are a ratio: probability of success to probability of failure, or P(E) P(E0) : Here P(E) = 11 64 , so P(E0) = 1 11 64 = 53 64 . Hence the odds for getting at least 7 answers correct by guessing are 11=64 53=64 = 11 53; or 11 to 53. (The odds against getting 7 or more correct by guessing are 53 to 11.)
1.10: A high tech electronics research rm has a great need to protect its secret projects. Their lab is protected by a security fence, door alarms on the entrance, pressure sensitive switches in the hallways and motion detectors in the lab. When challenged by a spy, these systems are known to be 75%, 85%, 90% and 96% e ective respectively. (a) What is the probability that a spy will be detected trying to steal secrets? (b) The security rm hired to do the job runs independent tests from time to time to determine if the systems are still being operated correctly by the sta . If 1000 such tests are run, what is the probability of a successful detection every time? (Assume that the system continues to work as reliably as in the description above.)
A high tech electronics research rm has a great need to protect its secret projects. Their lab is protected by a security fence, door alarms on the entrance, pressure sensitive switches in the hallways and motion detectors in the lab. When challenged by a spy, these systems are known to be 75%, 85%, 90% and 96% e ective respectively. (a) What is the probability that a spy will be detected trying to steal secrets? Solution: The spy is detected if at least one of the systems works. Sounds like maybe we should look at the complement. P(at least one system works) = 1 P(no system works) OR 1 P(all systems fail!): The probability that the spy successfully avoids detection. . . that all systems fail to detect him. . . is P(all fail) = (1 0:75)(1 0:85)(1 0:90)(1 0:96) = (0:25)(0:15)(0:10)(0:04) = 0:00015: Thus, the spy will be detected with probability 1 :00015 = 0:99985. (b) The security rm hired to do the job runs independent tests from time to time to determine if the systems are still being operated correctly by the sta . If 1000 such tests are run, what is the probability of a successful detection every time? (Assume that the system continues to work as reliably as in the description above.) Solution: The probability of a test \spy" being caught in one of these tests is 0:99985, as in part (a). If we run 1; 000 such tests, and we assume they are all independent, then the probability that the spy will be caught each time is (0:99985)1000 0:86070: That is, the spy will be detected in all 1; 000 tests with probability 0:86070.
1.12: A pair of fair dice are tossed. (a) What is the probability that the sum of the tosses is even, given that at least one of the dice comes up odd? Show your work! (b) If F is the event that the sum of the dice is odd, and E is the event that at least one of the tosses is even, determine whether or not E and F are independent events. Justify your conclusion.
A pair of fair dice are tossed. (a) What is the probability that the sum of the tosses is even, given that at least one of the dice comes up odd? Show your work! Solution: If we let A be the event \the sum of the two dice is even" and B be the event \at least one of the two dice is odd," then the question asks for P(AjB). Since P(AjB) = P(A \ B) P(B) = n(A \ B)=n(S) n(B)=n(S) = n(A \ B) n(B) ; it is enough to count outcomes in A \ B and in B. The event \at least one die is odd" is complicated{ we need to consider those outcomes in which only the rst die is odd, only the second is odd, or both are odd. The complement contains those outcomes in which no dice are odd{ or both dice are even. This is easier to count. B0 = n (2; 2); (2; 4); (2; 6); (4; 2); (4; 4); (4; 6); (6; 2); (6; 4); (6; 6) o : Thus n(B0) = 9; since n(S) = 62 = 36, we must have n(B) = n(S) n(B0) = 36 9 = 27. Now we turn to the intersection A \ B. In words, this event is: the sum of the two dice is even and at least one die is odd. This means that both dice must be odd (the sum of an odd and an even is odd { that won't work!). Thus A \ B is the event: both dice are odd, which is simply A \ B = n (1; 1); (1; 3); (1; 5); (3; 1); (3; 3); (3; 5); (5; 1); (5; 3); (5; 5) o : Thus n(A \ B) = 9. Now we can nally compute our conditional probability: P(AjB) = n(A \ B) n(B) = 9 27 = 1 3: Thus the probability that the sum of the tosses is even, given that at least one of the dice comes up odd, is 1 3 . (b) If F is the event that the sum of the dice is odd, and E is the event that at least one of the tosses is even, determine whether or not E and F are independent events. Justify your conclusion. Solution: In this part, E is the event \at least one toss is even" and F is the event \the sum of the two dice is odd." We're asked if E and F are independent; this is equivalent to the question, does P(E \ F) = P(E) P(F)? Just as we counted B (and S) in part (a), we can count that n(E) = 27 (and n(S) = 36). Thus P(E) = 27 36 = 3 4 . The event F is slightly more complicated. That the sum is odd means that one die is even and the other odd. We can list these possibilities: F = n (1; 2); (1; 4); (1; 6); (2; 1); (2; 3); (2; 5); (3; 2); : : : o One way to think of this is that the rst die can be any number, but there are only three possibilities for the second die (namely the other parity { if the rst is odd, the second must be even). Thus there are n(F) = 6 3 = 18 elements of F, and so P(F) = n(F) n(S) = 18 36 = 1 2 . Finally, we nd the probability of E \F. Notice that the event E \F is simply F! This is because F E. We've already seen this: if the sum is odd, one of the two dice must be even. That is, if the toss is in F, then it must also be in E. Since E \ F = F, we know P(E \ F) = P(F) = 1 2 . So: is P(E \ F) = P(E) P(F)? No: P(E \ F) = 1 2 , but P(E) P(F) = 3 4 1 2 = 3 8 . Thus E and F are not independent.
1.07: A special deck of cards consists of cards numbered 1; 2; 3; : : : ; 9; each value appears in the three suits fRed; White; Blueg. A card is selected at random. Let E = the event that the selected card is a 5, and F = the event that the selected card is Red. Are events E and F independent? Verify your answer mathematically.
A special deck of cards consists of cards numbered 1; 2; 3; : : : ; 9; each value appears in the three suits fRed; White; Blueg. A card is selected at random. Let E = the event that the selected card is a 5, and F = the event that the selected card is Red. Are events E and F independent? Verify your answer mathematically. Solution: Yes, E and F are independent. Let's compute probabilities: P(E) = P(5) = 3 27 = 1 9 (there are 3 ves out of the 27 cards), P(F) = P(Red) = 9 27 = 1 3 (9 of the 27 cards are red), and so P(E) P(F) = 1 9 1 3 = 1 27 . But P(E \ F) = P(5 \ Red) = 1 27 , too! (There is only one Red 5 in the deck.) Thus P(E \ F) = P(E) P(F), so E and F are independent.
1.11: A student goes bobbing for apples. In the tub there are 4 Granny Smiths, 3 Empire, 3 Macintosh, and 1 Gala apple. The student is allowed two turns and retrieves one apple for each turn. For sanitary reasons, once an apple is retrieved it is not placed back in the tub. (a) What is the probability that the second apple is a Granny Smith? (b) What is the probability that the rst apple was an Empire, given that the second apple was a Granny Smith?
A student goes bobbing for apples. In the tub there are 4 Granny Smiths, 3 Empire, 3 Macintosh, and 1 Gala apple. The student is allowed two turns and retrieves one apple for each turn. For sanitary reasons, once an apple is retrieved it is not placed back in the tub. (a) What is the probability that the second apple is a Granny Smith? Solution: One approach is to draw a tree diagram. Write S for Granny Smith, E for Empire, M for Macintosh, and G for Gala. We'll also write SE for the event \ rst apple selected is a Granny Smith, second apple is an Empire" and similarly for other possible events. Summing all paths of the tree that end in Smith, we have P(\S is second") = P(SS) + P(ES) + P(MS) + P(GS) = 4 11 3 10 + 3 11 4 10 + 3 11 4 10 + 1 11 4 10 = 40 110 = 4 11: (b) What is the probability that the rst apple was an Empire, given that the second apple was a Granny Smith? Solution: From the tree diagram and the usual formula for conditional probability, we compute: P(\E is rst" j \S is second") = P(\E is rst" \ \S is second") P(\S is second") = P(ES) 4=11 = 12=110 4=11 = 0:3 Thus the probability that the rst apple was an Empire given that the second was a Granny Smith is 0:3.
1.06: In a SAPS (Student Activity Participation Study) survey, 20 engineering majors, 15 business majors, 30 science majors, 25 liberal arts majors, and 10 human development majors were selected. Eight of the engineering, 10 of the science, 7 of the human development, 6 of the business, and 9 of the liberal arts majors selected for the study were female. (a) What is the probability of selecting a female if one person from this group is chosen randomly? (b) What is the probability that a female student selected randomly from this group is a science major? (c) What is the probability that a male student selected randomly from this group is a business major?
In a SAPS (Student Activity Participation Study) survey, 20 engineering majors, 15 business majors, 30 science majors, 25 liberal arts majors, and 10 human development majors were selected. Eight of the engineering, 10 of the science, 7 of the human development, 6 of the business, and 9 of the liberal arts majors selected for the study were female. Remarks: It will help us in this problem to make a table of students, by major and sex: Eng. Bus. Sci. L.A. H.D. Total Total 20 15 30 25 10 100 Female 8 6 10 9 7 40 Male 12 9 20 16 3 60 (a) What is the probability of selecting a female if one person from this group is chosen randomly? Solution: From the table, P(F) = 40 100 = 0:4: (b) What is the probability that a female student selected randomly from this group is a science major? Solution: Again, from the table: P(Sci j F) = P(Sci \ F) P(F) = 10=100 40=100 = 1 4 = 0:25: (c) What is the probability that a male student selected randomly from this group is a business major? Solution: Again, from the table: P(Bus jM) = P(Bus \M) P(M) = 9=100 60=100 = 3 20 = 0:15: Things to think about: how else could this be phrased? what phrase would indicate that you should look for the probability of the intersection, P(Bus \M)?
1.04: In an international language school, every student takes at least one of three foreign language classes. There are 29 students taking French, 40 students taking German, and 31 students taking Chinese. It is also known that there are 18 students taking both German and French, 12 students taking both Chinese and French, and 15 students taking German and Chinese. Only 5 students manage to take all three. (a) How many students are in this school? (b) If a student is chosen at random, what is the probability that he or she is not taking German? (c) If a student is chosen at random, what is the probability that he or she is taking exactly one language class?
In an international language school, every student takes at least one of three foreign language classes. There are 29 students taking French, 40 students taking German, and 31 students taking Chinese. It is also known that there are 18 students taking both German and French, 12 students taking both Chinese and French, and 15 students taking German and Chinese. Only 5 students manage to take all three. Remarks: This problem is really an exercise in counting. A Venn diagram would be useful; you should make one before diving into the questions (and answers). (a) How many students are in this school? Solution: We use our handy dandy formula: n(G [ F [ C) = n(G) + n(F) + n(C) n(G \ F) n(G \ C) n(F \ C) + n(G \ F \ C) = 29 + 40 + 31 18 12 15 + 5 = 60: Alternatively, from a correct Venn diagram, there are 9 + 7 + 4 + 5 + 10 + 13 + 12 = 60 students in this school. (b) If a student is chosen at random, what is the probability that he or she is not taking German? Solution: We compute: P(G0) = 1 P(G) = 1 40 60 = 1 3: Thus 1=3 of the students in this school are not taking German. (c) If a student is chosen at random, what is the probability that he or she is taking exactly one language class? Solution: The numbers 9, 4, 12 should appear in the \exactly one" regions of your Venn diagram, so the probability that a student is in only one language class is 9 + 4 + 12 60 = 25 60 = 5 12 0:4167:
1.03: Let E and F be events with P(E) = 0:7, P(F) = 0:2, and P(E [ F) = 0:8. (a) Find P(E j F). (b) Find P(E0 \ F0). (c) Find P(E0 \ F).
Let E and F be events with P(E) = 0:7, P(F) = 0:2, and P(E [ F) = 0:8. Picture: It's worth noting that you could also draw a Venn diagram for this problem. (a) Find P(E j F). Solution: We may compute P(E j F) via the formula P(E j F) = P(E \ F) P(F) : For this, we must rst compute P(E \ F); we know the other three quantities in the formula P(E [ F) = P(E) + P(F) P(E \ F): Substitution gives 0:8 = 0:7 + 0:2 P(E \ F), so P(E \ F) = 0:1. Hence P(E j F) = P(E \ F) P(F) = 0:1 0:2 = 0:5: (b) Find P(E0 \ F0). Solution: Here we use DeMorgan's Law: E0 \ F0 = (E [ F)0: P(E0 \ F0) = P((E [ F)0) = 1 P(E [ F) = 1 0:8 = 0:2: (c) Find P(E0 \ F). Solution: Recall that P(F) = P(E \ F) + P(E0 \ F). We know two of these numbers and are asked for the third. Easy! P(E0 \ F) = P(F) P(E \ F) = 0:2 0:1 = 0:1:
1.09: Long, long ago in the 1960s, at a faraway college, a class on popular culture took an informal poll. Out of 140 students, results indicated that 87 enjoyed rock music, 102 enjoyed country music and 37 enjoyed other forms of music. In addition, 2 students indicated they did not enjoy music of any kind. Furthermore, 55 liked both rock and country, 24 enjoyed country and other and 28 enjoyed rock and other. If a student from the class is selected at random, what is the probability that they like. . . (a) . . . other forms of music? (b) . . . all three forms listed? (c) . . . only one of the forms listed?
Long, long ago in the 1960s, at a faraway college, a class on popular culture took an informal poll. Out of 140 students, results indicated that 87 enjoyed rock music, 102 enjoyed country music and 37 enjoyed other forms of music. In addition, 2 students indicated they did not enjoy music of any kind. Furthermore, 55 liked both rock and country, 24 enjoyed country and other and 28 enjoyed rock and other. Three set version of a familiar equation: P(A [ B [ C) = P(A) + P(B) + P(C) P(A \ B) P(B \ C) P(C \ A) + P(A \ B \ C): If a student from the class is selected at random, what is the probability that they like. . . (a) . . . other forms of music? Solution: We are told that there are 140 students polled, 37 of whom enjoy \other forms" of music. Thus, a student picked at random will enjoy other forms of music with probability P(\Other") = 37 140 0:2643: (b) . . . all three forms listed? Solution: Now let A be the students that like rock music, B be the students that like country music, and C be the students that enjoy other music. We're asked to nd P(A \ B \ C). We'll use the equation given, noting that we are told all the probabilities except this last one. We have (substituting in): P(A [ B [ C) = P(A) + P(B) + P(C) P(A \ B) P(B \ C) P(C \ A) + P(A \ B \ C) or 138 140 = 87 140 + 102 140 + 37 140 55 140 24 140 28 140 + P(A \ B \ C): Solving, we get P(A \ B \ C) = 138 87 102 37 + 55 + 24 + 28 140 = 19 140 0:1357: (c) . . . only one of the forms listed? Solution: For this part it is perhaps simplest to draw a Venn diagram. From this counting it should be clear that the probability that a randomly selected student likes only one kind of music is P(\only one") = 23 + 42 + 4 140 = 69 140 0:4929:
2.2: Records of a midwestern university show that in one semester, 32% of students received an \A" in math, 34% received an \A" in chemistry, and 12% received an \A" in both. (a) If a student received an \A" in chemistry, what is the probability that he or she received an \A" in math as well? (b) What is the probability that a randomly selected student received an \A" in either math or chemistry (or both)? (c) Are the events \a student received an `A' in math" and \a student received an `A' in chemistry" independent events?
Records of a midwestern university show that in one semester, 32% of students received an \A" in math, 34% received an \A" in chemistry, and 12% received an \A" in both. (a) If a student received an \A" in chemistry, what is the probability that he or she received an \A" in math as well? Let C be the event \a student receives an `A' in chemistry" and M be the event \a student receives an `A' in math." We are given P(M) = 0:32, P(C) = 0:34, and P(M \C) = 0:12. This question asks us for P(M jC), which we nd in the usual way: P(M jC) = P(M \ C) P(C) = 0:12 0:34 = 12 34 = 6 17 0.3529 : (b) What is the probability that a randomly selected student received an \A" in either math or chemistry (or both)? In the notation of the previous part, this question asks for P(M[C). Using inclusion/exclusion: P(M [ C) = P(M) + P(C) P(M \ C) = 0:32 + 0:34 0:12 = 0.54 : (c) Are the events \a student received an `A' in math" and \a student received an `A' in chemistry" independent events? We check: Does P(M \ C) = P(M) P(C)? P(M \ C) = 0:12 but P(M) P(C) = (0:32) (0:34) = 0:1088. Since these are not equal, the answer is: no, they are not independent events . An easier alternative: we use our computation from part (a): P(MjC) 6= P(M), therefore, M and C are not independent events.
1.05: Suppose that an insurance company classi es people into one of three categories { good risks, average risks, and bad risks. Their records indicate that the probabilities that good, average, and bad risk persons will be involved in an accident over a 1-year span are, respectively, 0:05, 0:20, and 0:40. Moreover, the data indicates that 20 percent of the population are \good risks," 50 percent are \average risks," and the rest are \bad risks." (a) What proportion of people have accidents in a given year? (b) If a policy holder had no accidents in 1998, what is the probability that he or she is an average risk?
Suppose that an insurance company classi es people into one of three categories { good risks, average risks, and bad risks. Their records indicate that the probabilities that good, average, and bad risk persons will be involved in an accident over a 1-year span are, respectively, 0:05, 0:20, and 0:40. Moreover, the data indicates that 20 percent of the population are \good risks," 50 percent are \average risks," and the rest are \bad risks." (a) What proportion of people have accidents in a given year? Solution: The simplest way to do this is using a tree diagram. Your tree should have the risk category rst, followed by the accident category. From this information we can compute: P(\Accident") = P(\Good Risk" \ \Acc.") + P(\Average Risk" \ \Acc.") + P(\Bad Risk" \ \Acc.") = (:2)(:05) + (:5)(:2) + (:3)(:4) = 0:01 + 0:10 + 0:12 = 0:23 or 23%: That is, 23% of drivers will be in an accident in a given year. (b) If a policy holder had no accidents in 1998, what is the probability that he or she is an average risk? Solution: We use the tree diagram we drew in part (a) and the usual formula for conditional probability. We're asked to nd P(`average risk' j `no accident'). In part (a) we found that P(\Accident") = 0:23, so P(\No Accident") = 1 0:23 = 0:77. Thus P(`average risk' j `no accident') = P(average risk \ no accident) P(no accident) = 0:40 0:77 = 40 77 0:5195: note: we can arrive at the same value for P(\no accident") by summing the three paths of the tree that end in no accident.
1.08: The following table describes the voters in a certain district. Democrat Republican Independent Male 400 450 150 Female 600 550 350 Find the probability that . . . (a) . . . a randomly selected voter is a Republican. (b) . . . a randomly selected voter is female. (c) . . . a randomly selected male voter is Independent. (d) . . . a randomly selected Democrat is female. (e) . . . a randomly selected voter is male, given that the voter is not a Republican. (f) . . . a randomly selected voter is a Democrat, given that the voter is not an Independent.
The following table describes the voters in a certain district. Democrat Republican Independent Male 400 450 150 Female 600 550 350 Remarks: Before we turn to the questions, we expand the above table to include totals by sex and political party: Democrat Republican Independent Total Male 400 450 150 1,000 Female 600 550 350 1,500 Total 1,000 1,000 500 2,500 Notice that now the answers will follow quickly! Find the probability that . . . (a) . . . a randomly selected voter is a Republican. Solution: P(R) = 1; 000 2; 500 = 0:4. (b) . . . a randomly selected voter is female. Solution: P(F) = 1; 500 2; 500 = 0:6. (c) . . . a randomly selected male voter is Independent. Solution: P(I jM) = P(I \M) P(M) = 150=2; 500 1; 000=2; 500 = 0:15. (d) . . . a randomly selected Democrat is female. Solution: P(F jD) = P(F \ D) P(D) = 600=2; 500 1; 000=2; 500 = 0:6. (e) . . . a randomly selected voter is male, given that the voter is not a Republican. Solution: P(M jR0) = P(M \ R0) P(R0) = (400 + 150)=2; 500 (1; 000 + 500)=2; 500 = 550 1; 500 = 11 30 0:3667. (f) . . . a randomly selected voter is a Democrat, given that the voter is not an Independent. Solution: P(Dj I0) = P(D \ I0) P(I0) = P(D) P(I0) = 1; 000=2; 500 (1; 000 + 1; 000)=2; 500 = 0:5. Again, things to think about: how else could this be phrased? what phrase would indicate that you should look for the probability of an intersection (Female \ Democrat, for instance)?
