Department of Industrial Engineering IE 202: Engineering Statistics Example Questions Spring 2012

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1 Department of Industrial Engineering IE 202: Engineering Statistics Example Questions Spring 202. Twenty workers are to be assigned to 20 different jobs, one to each job. How many different assignments are possible? The first worker chooses from 20 jobs. There are 9 jobs and 9 workers left. The second worker chooses from 9 jobs. There are 8 jobs and 8 workers left, and so on. So the answer is 20!. 2. John, Jim, Jay and Jack have formed a band consisting of 4 instruments. If each of the boys can play all 4 instruments, how many different arrangements are possible? What if John and Jim can play all 4 instruments but Jay and Jack can each play only piano and drums? Solution If each of the boys can play all 4 instruments, then there are 4! ways for them to play together. If Jay and Jack can only play piano and drum, then John and Jim must play the other 2 instruments. Then there are 2! 2! 4 ways for them to play. 3. In how many ways can 8 people be seated in a row if (a there are no restrictions on the seating arrangement? (b persons A and B must sit next to each other? (c there are 4 men and 4 women and no 2 men or 2 women can sit next to each other? (d there are 5 men and they must sit next to each other? (e there are 4 married couples and each couple must sit together? Solution (a 8! (b First consider A and B as person because they must sit next to each other, then there are 7! ways for them to sit. Next consider that however they sit, A can sit to the left of B or B can sit to the left of A. So the answer is 2 7!. (c Let M represent Man and W represent Woman. Here the arrangement must be M-W-M-W-M- W-M-W OR W-M-W-M-W-M-W-M. In each case there are 4 possible places for the 4 women to sit and 4 possible places for the 4 men to sit. Therefore the answer is: 2 4! 4!. (d First consider the 5 men as person because they must sit next to each other. there are 8 people, so three of them are woment. Therefore there are 4! such seating arrangements. Now consider that no matter where the women are seated, there are 5! ways that the five men can sit next to each other. The answer is 4! 5!. (e Consider every married couple as person, there are 4! ways for the couple to sit. For every couple, the wife can sit to the left of the husand or the husand can sit to the left of the wife, so there are 2 ways for them to sit together. The answer is 4! 2 4.

2 4. Seven different gifts are to be distributed among 0 children. How many distinct results are possible if no child is to receive more than one gift? Solution If the gifts are unique. The first gift can be given to one of 0 children. The second gift goes to one of the remaining 9 children, and so on. The answer is ! 3! If the gifts are the same. The same logic applies, but since the gifts are the same, some of the configurations are the same. We just divide the previous answer by 7!. The answer is 0! 3!7! Another way to solve this is to try picking 7 people to get gifts, out of 0 possible people. So the answer is ( 0 0! 7 3!7! 5. If 8 identical blackboards are to be divided among 4 schools, how many divisions are possible? How many if each school must receive at least blackboard? n Number of blackboards given to school, n 2 Number of blackboards given to school 2, n 3 Number of blackboards given to school 3, n 4 Number of blackboards given to school 4. n + n 2 + n 3 + n 4 8. There is a special formula for this. If each school must receive at least blackboard ( ( 8 7 7! 4 3 3!4! 35ways. If any school may receive no blackboard, ( ( 3! 3!8! 65ways. 6. In how many ways can r objects be selected from a set of n objects if the order of selection is considered relevant? n n n r + ways to choose the st object, ways to choose the 2nd object,... ways to choose the rth object. 7. Determine the number of vectors (x,..., x n such that each x i is either 0 or and n x i k i0 2

3 If we did not have the restriction on the sum, the answer would be 2 n. However, with this restriction, it means we need at least k s in the list, so we may see k ones, k + ones, k + 2 ones,... k ones, ( ( n n i x i k ways this can happen, k k + ones, ( ( n n i x i k + ways this can happen, k +... n ones, ( ( n n i x i n way this can happen. n The answer is n ik ( n i 8. A system is comprised of 5 components, each of which is either working or failed. Consider an experiment that consists of observing the status of each component, and let the outcome of the experiment be given by the vector (x, x 2, x 3, x 4, x 5, where x i is equal to if the component i is working and is equal to 0 if component i is failed. (a How many outcomes are in the sample space of this experiment? (b Suppose that the system will work if components and 2 are both working, or if components 3 and 4 are both working, or if components, 3 and 5 are all working. Let W be the event that the system will work. Specify all the outcomes in W. (c Let A be the event that components 4 and 5 are both failed. How many outcomes are contained in the event A? (d Write out all the outcomes in the event AW. (a I can list the sample space too: (b Let us define the events: (0, 0, 0, 0, 0 (0, 0, 0, 0, (0, 0, 0,, 0 (0, 0, 0,, (0, 0,, 0, 0 (0, 0,, 0, (0, 0,,, 0 (0, 0,,, (0,, 0, 0, 0 (0,, 0, 0, (0,, 0,, 0 (0,, 0,, (0,,, 0, 0 (0,,, 0, (0,,,, 0 (0,,,, (, 0, 0, 0, 0 (, 0, 0, 0, (, 0, 0,, 0 (, 0, 0,, (, 0,, 0, 0 (, 0,, 0, (, 0,,, 0 (, 0,,, (,, 0, 0, 0 (,, 0, 0, (,, 0,, 0 (,, 0,, (,,, 0, 0 (,,, 0, (,,,, 0 (,,,, W event that components and 2 are both working, W 2 event that components 3 and 4 are both working, W 3 event that components, 3 and 5 are all working. P r [W ] P r [W W 2 W 3 ] P r [W ] + P r [W 2 ] + P r [W 3 ] P r [W W 2 ] P r [W W 3 ] P r [W 2 W 3 ] + 2P r [W W 2 W 3 ]. 3

4 To find P r [W ] note that the first two bits must be, but the remaining 3 can be or 0. So there are 2 3 ways to do this. P r [W ] P r [W 2 ] P r [W 3 ] W W 2 event that components, 2, 3 and 4 are all working W W 3 event that components, 2, 3 and 5 are all working W 2 W 3 event that components, 3, 4 and 5 are all working W W 2 W 3 event that components, 2, 3, 4 and 5 are all working P r [W W 2 ] P r [W W 3 ] P r [W 2 W 3 ] P r [W W 2 W 3 ] P r [W ] P r [W W 2 W 3 ] P r [W ] + P r [W 2 ] + P r [W 3 ] P r [W W 2 ] P r [W W 3 ] P r [W 2 W 3 ] + 2P r [W W 2 W 3 ] Easy way: find the elements above where the machine would work: (0, 0, 0, 0, 0 (0, 0, 0, 0, (0, 0, 0,, 0 (0, 0, 0,, (0, 0,, 0, 0 (0, 0,, 0, (0, 0,,, 0 (0, 0,,, (0,, 0, 0, 0 (0,, 0, 0, (0,, 0,, 0 (0,, 0,, (0,,, 0, 0 (0,,, 0, (0,,,, 0 (0,,,, (, 0, 0, 0, 0 (, 0, 0, 0, (, 0, 0,, 0 (, 0, 0,, (, 0,, 0, 0 (, 0,, 0, (, 0,,, 0 (, 0,,, (,, 0, 0, 0 (,, 0, 0, (,, 0,, 0 (,, 0,, (,,, 0, 0 (,,, 0, (,,,, 0 (,,,, P r [W ] (c If 4 and 5 are not working, the corresponding variables are 0. The other variables can be either 0 or. The number of ways this can happen (the number of outcomes in event A is: (d AW is the event that the system is working and elements 4 and 5 are not working. If 4 and 5 are not working, events W 2 and W 3 above can t be true. Thus only W defined above can be true, AW AW event that components and 2 are both working, 4 and 5 are not working. In this case only component 3 may be 0 or. The outcomes in AW are (,, 0, 0, 0 and (,,, 0, 0. 4

5 9. Suppose that A and B are mutually exclusive events for which P (A.3 and P (B.5. What is the probability that (a either A or B occurs? (b A occurs but B does not? (c both A and B occur? P (A 0.3 P (B 0.5 A and B are mutually exclusive means AB, and P r(ab 0. (a P r(a B P (A + P (B P (AB P (A + P (B (b P r (B c A P r (B c A P r (A According to the question, if A occurs, then it is certain that B does not occur. Thus P r (B c A. Finally, P r (B c A P r (B c A P r (A P (A 0.3. (c P r(ab An elementary school is offering 3 language classes: one in Spanish, on in French, and one in German. The classes are open to any of the 00 students in the school. There are 28 students in the Spanish class, 26 in the French class, and 6 in the German class. The 2 students that are in both Spanish and French, 4 that are in both Spanish and German, and 6 that are in both French and German. In addition, there are 2 students taking all 3 classes. (a If a student is chosen randomly, what is the probability that he or she is not in any of the language classes? (b If a student is chosen randomly, what is the probability that he or she is taking exactly one of the language classes? (c If 2 students are chosen randomly, what is the probability that at least is taking a language class? (a Number of students taking at least one language class Spanish French German Spanish + French + German - (Spanish and German but not French - (Spanish and French but not German - (French and German but not French - (Spanish and German and French (2-2 - (4-2 - ( students take at least one language class. Therefore students take no language classes. So the probability that the student is not taking any language class is 48/ (b Number of students taking exactly one language class Spanish + French + German 2 (Spanish and German but not French 2 (Spanish and French but not German 2 (French and German but not French 2 (Spanish and German and French (2 2 2 (4 2 2 ( students take exactly one language class. 5

6 (c Pr(at least one of the two is taking a language class -Pr(none of the two are taking a language class Sixty percent of the students at a certain school wear neither a ring nor a necklace. Twenty percent wear a ring and 30 percent a necklace. If one of the students is chosen randomly, what is the probability that this student is wearing (a a ring or a necklace? (b a ring and a necklace? (a Pr(ring or necklace - Pr(no ring and no necklace (b Pr(ring and necklace Pr(ring + Pr(necklace - Pr(ring and necklace A small community organization consists of 20 families, of which 4 have one child, 8 have two children, 5 have three children, 2 have four children, and has five children. (a If one of these families is chosen at random, what is the probability it has i children, i, 2, 3, 4, 5? (b If one of the children is randomly chosen, what is the probability that child comes from a family having i children, i, 2, 3, 4, 5? (a P r(i P r(i P r(i P r(i P r(i (b Total number of children P r(i P r(i P r(i P r(i P r(i An urn contains 3 red and 7 black balls. Players A and B withdraw balls from the urn consecutively until a red ball is selected. Find the probability that A selects the red ball. (A draws the first ball, then B, and so on. There is now replacement of the balls drawn. 6

7 who chooses A B A B A B A B possible result R 2 B R 3 B B R 4 B B B R 5 B B B B R 6 B B B B B R 7 B B B B B B R 8 B B B B B B B R P r(a gets red ball P r(red on st pull + P r(black on st pull, black on 2nd pull, and red on 3rd pull + P r(black on st pull, black on 2nd pull, black on 3rd pull, black on 4th pulland red on 5th pull + P r (black on st pull, black on 2nd pull, black on 3rd pull, black on 4th pull, black on 5th pull, black on 6th pulland red on 7th pull An urn contains 5 red, 6 blue, and 8 green balls. If a set of 3 balls is randomly selected, what is the probability that each of the balls will be (a of the same color? (b of different colors? Repeat under the assumption that whenever a ball is selected, its color is noted and it is then replaced in the urn before the next selection. This is known as sampling with replacement. (a Set of 3 balls of same color: P r(all same color P r(all red + P r(all blue + P r(all green (b Set of all different colors: P r(all different colors P r(one red + P r(one blue + P r(one green ! (a Set of all 3 balls of same color sampling with replacement P r(all same color P r(all red + P r(all blue + P r(all green (b Set of all 3 balls of different colors sampling with replacement P r(all different colors P r(one red + P r(one blue + P r(one green !

8 5. A forest contains 20 elk, of which 5 are captured, tagged, and then released. A certain time later, 4 of the 20 elk are captured. What is the probability that 2 of these 4 have been tagged? What assumptions are you making? P r(2 of 4 are tagged ( Assumption: any deer, whether tagged or untagged, is equally likely to be caught. 6. Five people designated as A, B, C, D, E, are arranged in linear order. Assuming that each possible order is equally likely, what is the probability that (a there is exactly one person between A and B? (b there are exactly two people between A and B? (c there are exactly three people between A and B? (a Finding the total number of arrangements where there is exactly person between A and B. A C B D E 3! ways A is on and B is on 3. 2 B C A D E 3! ways B is on and A is on 3. 3 D A C B E 3! ways A is on 2 and B is on 4. 4 D B C A E 3! ways B is on 2 and A is on 4. 5 D E A C B 3! ways A is on 3 and B is on 5. 6 D E B C A 3! ways B is on 3 and A is on 5. Total number of arrangements is 3! P r(there is exactly person betwen A and B 36 5! (b Finding the total number of arrangements where there are exactly 2 people between A and B: A C D B E 3! ways A is on and B is on 4. 2 B C D A E 3! ways B is on and A is on 4. 3 D A C E B 3! ways A is on 2 and B is on 5. 4 D B C E A 3! ways B is on 2 and A is on 5. Total number of arrangements is 3! P r(there are exactly 2 people betwen A and B 24 5! (c Finding the total number of arrangements where there are exactly 3 people between A and B: A C D E B 3! ways A is on and B is on 5. 2 B C D E A 3! ways B is on and A is on 5. Total number of arrangements is 3! 2 2. P r(there are exactly 3 people betwen A and B 2 5! Let E, F, and G be three events. Find expressions for the events so that, of E, F, and G, (a only E occurs; (b both E and G but not F occur; 8

9 (c at least one of the events occurs; (d at least two of the events occur; (e all three events occur; (f none of the events occur; (g at most one of the events occurs; (h at most two of the events occur; (i exactly two of the events occur; (j at most three of the events occur. (a only E occurs EF c G c ; (b both E and G but not F occur EF G c ; (c at least one of the events occurs E F G; (d at least two of the events occur EF F G EG; (e all three events occur EF G; (f none of the events occur E c F c G c ; (g at most one of the events occurs EF c G c E c F G c E c F c G E c F c G c (EF F G EG c ; (h at most two of the events occur (EF G c ; (i exactly two of the events occur EF G c EF c G E c F G; (j at most three of the events occur S (sample space. 8. Find the simplest expression for the following events: (a (E F (E F c ; (b (E F (E c F (E F c ; (c (E F (F G. (a (E F (E F c (E F E (E F F c (EE F E (EF c F F c E EF c E; (b (E F (E c F (E F c ; ((E F E c (E F F (E F c (EE c F E c EF F F (E F c (F E c F E F F (E F c F (E F c F E F F c F E (c (E F (F G E (F G F (F G EF EG F F F G F EG 9. Prove that P (E F G P (E + P (F + P (G P (E c F G P (EF c G P (EF G c 2P (EF G. P (E F G P (E F + P (G P ((E F G P (E F + P (G P (EG F G P (E + P (F P (EF + P (G (P (EG + P (F G P (EGF G P (E + P (F + P (G P (EF P (EG P (F G + P (EF G Now we can expand some of these terms: P (EF P (EF G EF G C P (EF G + P (EF G c sincep (EF GEF G c 0, P (EG P (EF G EF c G P (EF G + P (EF c GsinceP (EF GEF c G 0, P (F G P (EF G E c F G P (EF G + P (E c F GsinceP (EF GE c F G 0. 9

10 Finally, P (E F G P (E + P (F + P (G (P (EF G + P (EF G c (P (EF G + P (EF c G (P (EF G + P (E c F G + P (EF G P (E + P (F + P (G P (EF G c P (EF c G P (E c F G 3P (EF G + P (EF G P (E + P (F + P (G P (EF G c P (EF c G P (E c F G 2P (EF G 20. If P (E.9 and P (F.8, show that P (EF.7. In general, prove Bonferroni s inequality, namely, P (EF P (E + P (F We know that P (E F P (E + P (F P (EF. Since all probabilities must be less than, P (E + P (F P (EF P (E + P (F P (EF P (EF P (E + P (F P (EF What is the probability that at least one of a pair of fair dice lands on 6, given that the sum of the dice is i, i 2, 3,..., 2? If i < 7, the answer is zero, since any number (on one dice plus 6 (on the other dice is greater than 6. When we roll two dice, there are 6 6 possible answers. The following pairs of numbers yield a sum of 7: {(, 6, (2, 5, (3, 4, (4, 3, (5, 2, (6, } so the probability of getting at least one 6 when the sum is 7 is 2/6. The following pairs of numbers yield a sum of 8: {(2, 6, (3, 5, (4, 4, (5, 3, (6, 2} so the probability of getting at least one 6 when the sum is 8 is 2/5. The following pairs of numbers yield a sum of 9: {(3, 6, (4, 5, (5, 4, (6, 3} so the probability of getting at least one 6 when the sum is 9 is 2/4. The following pairs of numbers yield a sum of 0: {(4, 6, (5, 5, (6, 4} so the probability of getting at least one 6 when the sum is 0 is 2/3. The following pairs of numbers yield a sum of : {(5, 6, (6, 5} so the probability of getting at least one 6 when the sum is is. Only (6, 6 will yield a sum of 2, so {(5, 6, (6, 5} so the probability of getting at least one 6 when the sum is 2 is. P r(one of the dice is 6 sum of the dice is N 0 ifn 6 3 ifn ifn 8 2 ifn ifn 0 ifn ifn 2 0 otherwise. 0

11 22. An urn contains 6 white and 9 black balls. If 4 balls are to be randomly selected without replacement, what is the probability that the first 2 selected are white and the last 2 black? Consider an urn containing 2 balls, of which 8 are white. A sample of size 4 is to be drawn with replacement (without replacement. What is the conditional probability (in each case that the first and third balls drawn will be white given that the sample contains exactly 3 white balls? There are 8 white and 4 black balls. W : Event that there are exactly 3 white balls W {(BW W W, (W BW W, (W W BW, (W W W B} Z : Event that 3rd and st balls are white Z {(W BW B, (W BW W, (W W W B, (W W W W } W Z {(W BW W, (W W W B} Without replacement: Probabilities of these events happening are: P r[w ] P r[z] P r[w Z] With replacement: P r [Z W ] P r [W Z] P r [W ] Probabilities of these events happening are: P r[w ] 4 4 ( ( 8 2 ( 4 2 ( 8 3 ( 4 ( 8 4 P r[z] ( 8 3 ( 4 P r[w Z] P r [Z W ] P r [W Z] P r [W ]

12 24. An ectopic pregnancy is twice as likely to develop when the pregnant woman is a smoker as it is when she is a nonsmoker. If 32 percent of women of childbearing age are smokers, what percentage of women having ectopic pregnancies are smokers? E : Event that the woman has an ectopic pregnancy. S : Event that the woman is a smoker. We are given: We are asked to find P r[s E]. P r[s] 0.32 P r [E S] 2 P r [E S c ] P r[s E] P r[es] P r[e] P r[e S]P r[s] P r[e] Also known: Therefore: P r[s E] P r[e] P r[e S]P r[s] + P r [E S c ] P r [S c ] P r[e S]P r[s] P r[e S]P r[s] + P r [E S c ] P r [S c ] 2P r[s] 2P r[s] + P r [S c ] ( P r[e S c ]P r[s] 2 P r[e S c ]P r[s] + P r [E S c ] P r [S c ] 25. A total of 500 married working couples were polled about their annual salaries, with the following information resulting: Husband Wife Less than $25,000 More than $25,000 Less than $25, More than $25, For instance, in 36 of the couples, the wife earned more and the husband earned less than $25,000. If one of the couples is randomly chosen, what is (a the probability that the husband earns less than $25,000? (b the conditional probability that the wife earns more than $25,000 given that the husband earns more than this amount? (c the conditional probability that the wife earns more than $25,000 given that the husband earns less than this amount? (a the probability that the husband earns less than $25,000? Total number of couples Number of couples where the husband earns less than $25, P r[husband earns less than $25,000]

13 (b the conditional probability that the wife earns more than $25,000 given that the husband earns more than this amount? P r[w > $25, 000 H > $25, 000] P r[w > $25, 000, H > $25, 000] P r[h > $25, 000] (c the conditional probability that the wife earns more than $25,000 given that the husband earns less than this amount? P r[w > $25, 000 H < $25, 000] P r[w > $25, 000, H < $25, 000] P r[h < $25, 000] In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a the probability that a randomly selected family owns both a dog and a cat? (b the conditional probability that a randomly selected family owns a dog given that it owns a cat? D : Event that the family owns a dog. S : Event that the family owns a cat. Given: P r[d] 0.36, P r[c D] 0.22, P r[c] (a the probability that a randomly selected family owns both a dog and a cat? P r[cd] P r[c D]P r[d] (b the conditional probability that a randomly selected family owns a dog given that it owns a cat? P r[d C] P r[cd] P r[c] Suppose that 5 percent of men and.25 percent of women are color blind. A color blind person is chosen at random. What is the probability of this person being male? Assume that there are an equal number of males and females. What if the population consisted of twice as many males as females? C : Event that someone is color blind. M : Event that someone is male. W : Event that someone is female. If there are an equal number of men and women, P r[m] P r[w ] 0.5. P r[m C] P r[mc] P r[c] P r[c M]P r[m] P r[c M]P r[m] + P r[c W ]P r[w ] If there are twice as many males as females, P r[m] 2P r[w ], and since P r[w ] + P r[m], we find P r[w ] /3, P r[m] 2/3. P r[m C] P r[mc] P r[c] P r[c M]P r[m] P r[c M]P r[m] + P r[c W ]P r[w ]

14 28. Suppose that an insurance company classifies people into one of three classes: good risks, average risks, and bad risks. The company s records indicate that the probabilities that good-, average- and bad-risk persons will be involved in an accident over a -year span are, respectively,.05,.5, and.30. If 20 percent of the population is a good risk, 50 percent an average risk, and 30 percent a bad risk, what proportion of people have accidents in a fixed year? If policy holder A had no accidents in 997, what is the probability that he or she is a good or average risk? G : Event that someone is good risk, V : Event that someone is average risk, B : Event that someone is bad risk, A : Event that someone has an accident in one year. We are given P r[a G] 0.05, P r[a V ] 0.5, P r[a B] P r[g] 0.20, P r[v ] 0.50, P r[b] The probability that a randomly chosen person has an accident is: P r[a] P r[a G]P r[g]+p r[a V ]P r[v ]+P r[a B]P r[b] Therefore the proportion of people who have accidents is 7.5%. Second question asks the probability that a person is good or average risk given that they had no accidents in a given year. In math: P r[g V A c ]. P r[g V A c ] P r[b A c ] P r[bac ] ( A c P r[ac B]P r[b] A c A ball is in any one of n boxes and is in the ith box with probability P i. If the ball is in box i, a search of that box will uncover it with probability α i. Show that the conditional probability that the ball is in box j, given that a search of box i did not uncover it, is P j α i P i ( α i P i α i P i if j i if j i B i : Event that ball is in box i, F i : Event that ball is found in box i, P r [F i B i ] α i. We are asked to find P r [B j F c i ]. P r [B j Fi c ] P r [B jfi c] P r [Fi c P r [Fi c] B j] P r [B j ] n k P r [F i c B k] P r [B k ] Probability of not finding a ball in a box if it isn t in the box is. So, { P r [Fi c if i j B j ] ( α i if i j P r [F c i B j ] P j n k P r [F c i B k] P k and n k P r [F c i B k ] P k ( n P k P i + ( α i P i P i + P i α i P i α i P i k 4

15 if j i: if j i: P r [B j F c i ] P r [F c i B j] P j n k P r [F c i B k] P k P j α i P i. P r [B j F c i ] P r [Fi c B j] P j n k P r [F i c B ( α i P i. k] P k α i P i 30. Three dice are rolled. By assuming that each of the 6 3 possible outcomes is equally likely, find the probabilities attached to the possible values that X can take on, where X is the sum of the 3 dice. If N < 3, P r[n] 0. If N 3, this can only happpen if all 3 dice land on, so Look at all possible sums: P r(n We can see a pattern emerge. Let s put them in one big matrix:

16 Each block has a matrix structure known as a Toeplitz matrix. We can read out the probabilities using 6

17 this matrix: P r[n] if N 3 3 if N if N if N if N if N if N if N if N if N if N if N if N if N if N 7 if N 8 3. Five men and 5 women are ranked according to their scores on an examination. Assume that no two scores are alike and all 0! possible rankings are equally likely. Let X denote the highest ranking achieved by a woman. (For instance, X if the top-ranked person is female. Find P {X i}, i, 2, 3,..., 8, 9, 0. X highest rank achieved by a woman. Without any restrictions, there are 0! ways to rank everyone. First, X with probability 0.5. If X 2, then the highest ranking person must be a man, and the second highest ranking person must be a woman. We can choose man from 5 to be highest ranking, and woman from the five women to be highest ranking. The remaining 8 people will rank among themselves in 8! ways, so: P r(x 2 ( 5 ( 5 0! 8! 5 5 8! 0! If X 3, then the two highest ranking people must be male, and the third highest ranking person must be a woman. We can choose 2 men from 5 to be highest ranking, and then we can arrange them in 2! ways. We choose woman from the five women to be highest ranking. The remaining 7 people will rank among themselves in 7! ways, so: P r(x 3 ( 5 2 ( 5 2! 0! 7! 5! 3!2! 2 5 7! 00 0! If X 4, then the three highest ranking people must be male, and the fourth highest ranking person must be a woman. We can choose 3 men from 5 to be highest ranking, and then we can arrange them in 3! ways. We choose woman from the five women to be highest ranking. The remaining 6 people will rank among themselves in 6! ways, so: P r(x 4 ( 5 3 ( 5 3! 0! 6! 7 5! 3!2! 3! 5 6! 300 0!

18 Similarly, P r(x 5 ( 5 4 ( 5 4! 0! 5! 5! 4!! 4! 5 5! 600 0! P r(x 6 5!5! 0! Four independent flips of a coin are made. Let X denote the number of heads obtained. Plot the probability mass function of the random variable X 2. The pmf of X 2 is: P r(x ( 4 P r(x ( 4 P r(x 2 2 ( 4 P r(x 3 3 ( 4 P r(x

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20 33. If the distribution function of X is given by F (b calculate the probability mass function of X. f(x 0 b < b < 3 5 b < b < b < 3.5 b b 0 0 b 5 b 2 0 b 3 0 b otherwise. 34. Four busses carrying 48 students from the same school arrive at a football stadium. The buses carry, respectively, 40, 33, 25, and 50 students. One of the students is randomly selected. Let X denote the number of students that were on the bus carrying the randomly selected student. One of the 4 bus drivers is also randomly selected. Let Y denote the number of students on her bus. (a Which of E [X] or E [Y ] do you think is larger? Why? (b Compute E [X] and E [Y ]. (a Which of E [X] or E [Y ] do you think is larger? Why? E[X] should be larger because of sampling bias. (b Compute E [X] and E [Y ]. E[X] E[Y ] A box contains 5 red and 5 blue marbles. Two marbles are withdrawn randomly. If they are the same color, then you win $.0; if they are different colors then you win -$.00 (That is, you lose $.00. Calculate (a the expected value of the amount you win; (b the variance of the amount you win. (a the expected value of the amount you win; 20

21 E[X] 2P r(red, red 0. (P r(red, blue + P r(blue, red 2.2P r(blue, blue (b the variance of the amount you win. V ar(x (2 ( 0. 2 P r(red, red + ( 0. ( 0. 2 (P r(red, blue + P r(blue, red +( 2.2 ( 0. 2 P r(blue, blue ( Suppose that, in flight, airplane engines will fail with probability p, independently from engine to engine. If an airplane needs a majority of its engines operative to complete a successful flight, for what values of p is a 5-engine plan preferable to a 3-engine plane? 5 engine plane: 3 engine plane: The 5 engine plane is less likely to crash if P r(all engines fail ( p 5 P r(all engines fail ( p 3 ( p 5 < ( p 3 ( p 2 < Since p is always less than, its square will be less than also. So the 5 engine plane is always safer. 37. A communications channel transmits the digits 0 and. However, due to static, the digit transmitted is incorrectly received with probability.2. Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we will transmit instead of 0 and instead of. If the receiver of the message uses majority decoding, what is the probability that the message will be wrong when decoded? What independence assumptions are you making? P r[wrong] P r[between 3 and 5 incorrect bits ( ( ( ! 3!2! ! 4!! We are assuming that the probability that the bit is wrong does not depend on the probability that the previous bit is wrong. 2

22 38. It is known that diskettes produced by a certain company will be defective with probability.0, independently of each other. The company sells the diskettes in packages of size 0 and offers a money-back guarantee that at most of the 0 diskettes in the package will be defective. The guarantee is that the customer can return the entire package of diskettes if he or she finds more than one defective diskette in it. If someone buys 3 packages, what is the probability that he or she will return exactly of them? Pr(defective diskette 0.0. P r(exactly return ( 3 P r(more than defective diskettep r( or fewer defective diskettes 2 ( P r( or fewer defective diskettes P r(2 or more defective diskettes P r(exactly return When coin is flipped, it lands on heads with probability.4; when coin 2 is flipped, it lands on heads with probability.7. One of these coins is randomly chosen and flipped 0 times. (a What is the probability that the coin lands on heads on exactly 7 of the 0 flips? (b Given that the first of these ten flips lands heads, what is the conditional probability that exactly 7 of the 0 flips lands on heads? 7H : Event that coin lands on heads exactly 7 times, C : Event that coin is picked, C2 : Event that coin 2 is picked. (a What is the probability that the coin lands on heads on exactly 7 of the 0 flips? P r[7h] P r[7h C]P r[c] + P r[7h C2]P r[c2] ( ( ! 7!3! 0.5 ( (b Given that the first of these ten flips lands heads, what is the conditional probability that exactly 7 of the 0 flips lands on heads? The fact that the first flip lands heads changes the probability of which coin was chosen. H : Event that the coin lands heads the first time. 22

23 P r[c H] P r[c2 H] P r(h CP r(c P r(h P r(h C2P r(c2 P r(h P r(h CP r(c P r(h CP r(c + P r(h C2P r(c2 P r(h C2P r(c2 P r(h CP r(c + P r(h C2P r(c2 Therefore, P r[7h] P r[7h C, H]P r[c H] + P r[7h C2, H]P r[c2 H] ( ( ! 7!3! 0.5 ( If you buy a lottery ticket in 50 lotteries, in each of which your chance of winning a prize is 00, what is the (approximate probability that you will win a prize (a at least once? (b exactly once? (c at least twice? (a at least once? (b exactly once? ( P r[at least one win] P r[no wins] P r[exactly one win] ( 50 ( (c at least twice? P r[at least two wins] P r[exactly one win] P r[no wins] Let X be a random variable having expected value µ and variance σ 2. Find the expected value and variance of Y X µ σ [ ] X µ E[Y ] E σ E[X] µ σ 0 µ Y. 23

24 [ V ar(y E (Y µ Y 2] E [(Y 0 2] E [ Y 2] [ (X ] [ ] µ 2 (X µ 2 E [(X µ 2] E E σ σ 2 σ 2 V ar(x σ 2 V ar(x σ 2 σ2 σ Find V ar(x if P (X a p P (X b E[X] pa + ( pb µ V ar(x E [(X µ 2] (a pa ( pb 2 p + (b pa ( pb 2 ( p (a b 2 ( p 2 p + p 2 (b a 2 ( p (a b 2 p( p(( p + p (a b 2 p( p 24