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MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: since A and B are events, then P ( neither ) = P (( A ∪ B ) c ) = P ( A c ∩ B c ) . Since A and B are independent, so are A c and B c (why?). Thus, P ( neither ) = P ( A c ) P ( B c ) = (1 − P ( A ))(1 − P ( B )) = (1 − 0 . 2)(1 − 0 . 2) = 0 . 64 . University of Ottawa 2

MAT 2377 – Probability and Statistics for Engineers Practice Set Q3 . A smoke-detector system consists of two parts A and B . If smoke occurs then the item A detects it with probability 0 . 95 , the item B detects it with probability 0 . 98 whereas both of them detect it with probability 0 . 94 . What is the probability that the smoke will not be detected? 0 . 01 a) 0 . 99 b) 0 . 04 c) 0 . 96 d) none of the preceding e) University of Ottawa 5

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: we have P ( only player 1 succeeds ) = P ( A 1 ∩ A c 2 ∩ A c 3 ) = P ( A 1 ) P ( A c 2 ) P ( A c 3 ) = 0 . 5 × 0 . 3 × 0 . 4 = 0 . 06 P ( only player 2 succeeds ) = P ( A c 1 ∩ A 2 ∩ A c 3 ) = P ( A c 1 ) P ( A 2 ) P ( A c 3 ) = 0 . 5 × 0 . 7 × 0 . 4 = 0 . 14 P ( only player 3 succeeds ) = P ( A c 1 ∩ A c 2 ∩ A 3 ) = P ( A c 1 ) P ( A c 2 ) P ( A 3 ) = 0 . 5 × 0 . 3 × 0 . 6 = 0 . 09 But these three events are mutually exclusive, so P ( exactly one succeeds ) = P (1) ∪ P (2) ∪ P (3) = P (1)+ P (2)+ P (3) = 0 . 29 . University of Ottawa 8

MAT 2377 – Probability and Statistics for Engineers Practice Set Q6 . There is a theorem of combinatorics that states that the number of permutations of n objects in which n 1 are alike of kind 1 , n 2 are alike of kind 2 , ..., and n r are alike of kind r (that is, n = n 1 + n 2 + · · · + n r ) is n ! n 1 ! · n 2 ! · · · · · n r ! . Find the number of different words that can be formed by rearranging the letters in the following words (include the given word in the count): NORMAL a) HHTTTT b) ILLINI c) MISSISSIPPI d) University of Ottawa 11

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: there are 1000 students in total. Let A and E represent the events that the student passed and that the student is an engineer, respectively. Then P ( A c | E ) = P ( A c ∩ E ) P ( E ) ) = 60 / 100 490 / 1000 = 0 . 12 . University of Ottawa 14

MAT 2377 – Probability and Statistics for Engineers Practice Set Q9 . A medical research team wished to evaluate a proposed screening test for Alzheimer’s disease. The test was given to a random sample of 450 patients with Alzheimer’s disease; in 436 cases the test result was positive. The test was also given to a random sample of 500 patients without the disease; only in 5 cases was the result was positive. It is known that in Canada 11 . 3% of the population aged 65+ have Alzheimer’s disease. Find the probability that a person has the disease given that their test was positive (choose the closest answer). 0 . 97 a) 0 . 93 b) 0 . 99 c) 0 . 07 d) none of the preceding e) University of Ottawa 17

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let p = 0 . 1 denote the probability that an item is defective. Then 1 − p = 0 . 9 is the probability that an item is not defective. Let X denote the number of defective items in the sample. The probability that none of the items is defective is P ( X = 0) = P ( item 1 is not defective ) × · · · × P ( item 12 is not defective ) = (1 − p ) 12 = 0 . 9 12 ≈ 0 . 2824 The probability that exactly one of the items is defective is P ( X = 1) = P ( only item 1 is defective ) + · · · + P ( only item 12 is defective ) . University of Ottawa 20

MAT 2377 – Probability and Statistics for Engineers Practice Set The event that only item j is defective (for j = 1 , . . . , 12 ) occurs when item j is defective (with probability p ) AND the remaining 11 items are not defective (each with probability 1 − p ). Since the items are sampled independently, P ( only item j is defective ) = p (1 − p ) 11 ≈ 0 . 0314 . But there are ( 12 1 ) = 12! 1!11! = 12 ways to chose which of the 12 items will be defective (sampling without replacement), so P ( X = 1) = p (1 − p ) 11 + · · · + p (1 − p ) 11 | {z } ( 12 1 ) =12 times = 12 1 p (1 − p ) 11 ≈ 12(0 . 0314) ≈ 0 . 3766 . University of Ottawa 21

MAT 2377 – Probability and Statistics for Engineers Practice Set The probability that exactly two of the items are defective is P ( X = 2) = P ( only items 1 , 2 are defective ) + · · · + P ( only item 11 , 12 are defective ) . The event that only items j, k are defective (for j, k = 1 , . . . , 12 , j ̸ = k ) occurs when items j ̸ = k are defective (each with probability p ) AND the remaining 10 items are not defective (each with probability 1 − p ). Since the items are sampled independently, P ( only items j, k are defective ) = p 2 (1 − p ) 10 ≈ 0 . 0031 . But there are ( 12 2 ) = 12! 2!10! = 66 ways to chose which 2 of the items will be defective, so University of Ottawa 22

MAT 2377 – Probability and Statistics for Engineers Practice Set P ( X = 2) = p 2 (1 − p ) 10 + · · · + p 2 (1 − p ) 10 | {z } ( 12 2 ) =66 times = 12 2 p 2 (1 − p ) 10 ≈ 66(0 . 0031) = 0 . 2301 . The probability of at most two defective items in the sample is thus P ( X ≤ 2) = P ( X = 0) + P ( X = 1) + P ( X = 2) = 12 1 p 0 (1 − p ) 12 + 12 2 p 1 (1 − p ) 11 + 12 2 p 2 (1 − p ) 10 ≈ 0 . 2824 + 0 . 3766 + 0 . 2071 = 0 . 8891 University of Ottawa 23

MAT 2377 – Probability and Statistics for Engineers Practice Set Q11 . A student can solve 6 problems from a list of 10 . For an exam 8 questions are selected at random from the list. What is the probability that the student will solve exactly 5 problems? 0 . 98 a) 0 . 02 b) 0 . 28 c) 0 . 53 d) none of the preceding e) University of Ottawa 24

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let X and 8 − X be the number of questions on the exam that the student can and cannot solve, respectively. There are ( 6 X ) ways to randomly chose X exam questions that the student can solve, and ( 10 − 6 8 − X ) = ( 4 8 − X ) ways to randomly chose 8 − X questions that the student cannot solve: there are thus ( 6 X )( 4 8 − X ) ways to randomly chose X questions that the student can solve AND 8 − X questions that the student cannot solve. Since there are ( 10 8 ) total ways to chose 8 exam questions randomly from the 10 problems, P ( X ) = ( 6 X )( 4 8 − X ) ( 10 8 ) . For X = 5 , P ( X = 5) = ( 6 5 )( 4 3 ) ( 10 8 ) = 6 · 4 45 ≈ 0 . 53 . University of Ottawa 25

MAT 2377 – Probability and Statistics for Engineers Practice Set Q12 . Consider the following system with six components. We say that it is functional if there exists a path of functional components from left to right. The probability of each component functions is shown. Assume that the components function or fail independently. What is the probability that the system operates? 0 . 1815 a) 0 . 8185 b) 0 . 6370 c) 0 . 2046 d) none of the preceding e) University of Ottawa 26

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let Box A consist of components 2 , 3 , 4 , 5 , Box B consist of components 2 , 3 , 4 , and Box C consist of components 2 , 3 . We will denote the probability that Box j operates by P ( j ) , j ∈ { A, B, C } . We are interested in the probability P ( S ) that the system operates. Because the components function or fail independently, P ( S ) = P ( component 1 and Box A and component 6 operate ) = P (1 operates ) × P ( A ) × P (6 operates ) = 0 . 5 · P ( A ) · 0 . 5 = 0 . 5 2 P ( A ) . There are two ways for Box A to operate: either component 5 operates (with probability 0 . 5 ) or Box B operates: University of Ottawa 27

MAT 2377 – Probability and Statistics for Engineers Practice Set P ( A ) = P ( component 5 operates or Box B operates ) = P (5 operates ) + P ( B ) − P ( component 5 operates and Box B operates ) = P (5 operates ) + P ( B ) − P (5 operates ) P ( B ) = 0 . 5 + P ( B ) − 0 . 5 P ( B ) = 0 . 5(1 + P ( B )) . Thus, P ( S ) = 0 . 5 2 P ( A ) = 0 . 5 2 · 0 . 5(1 + P ( B ) = 0 . 5 3 (1 + P ( B )) . In order for Box B to operate, we need both Box C and component 4 to operate, so that University of Ottawa 28

MAT 2377 – Probability and Statistics for Engineers Practice Set P ( B ) = P ( Box C operates and component 4 operates ) = P (4 operates ) P ( C ) = 0 . 7 P ( C ) . Thus, P ( S ) = 0 . 5 3 (1 + P ( B )) = 0 . 5 3 (1 + 0 . 7 P ( C )) . Finally, there are two ways for Box C to operate: either component 2 operates (with probability 0 . 7 ) or component 3 operates (also with probability 0 . 7 ): University of Ottawa 29

MAT 2377 – Probability and Statistics for Engineers Practice Set P ( C ) = P ( component 2 operates or component 3 operates ) = P (2 operates ) + P (3 operates ) − P ( components 2 operates and component 3 operates ) = P (2 operates ) + P (3 operates ) − P (2 operates ) P (3 operates ) = 0 . 7 + 0 . 7 − 0 . 7 · 0 . 7 = 0 . 7(2 − 0 . 7) = 0 . 7 · 1 . 3 . Thus, P ( S ) = 0 . 5 3 (1 + 0 . 7 P ( C )) = 0 . 5 3 (1 + 0 . 7 · 0 . 7 · 1 . 3) = 0 . 5 3 (1 + 0 . 7 2 · 1 . 3) = 0 . 24046 . University of Ottawa 30

MAT 2377 – Probability and Statistics for Engineers Practice Set Q13 . Three events are shown in the Venn diagram below. Shade the region corresponding to the following events: A c a) ( A ∩ B ) ∪ ( A ∩ B c ) b) ( A ∩ B ) ∪ C c) ( B ∪ C ) c d) ( A ∩ B ) c ∪ C e) University of Ottawa 31

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) This is the set of all outcomes not in A : b) We can rewrite ( A ∩ B ) ∪ ( A ∩ B c ) = A ∩ ( B ∪ B c ) = A ∩ S = A , so this is the set of all outcomes in A : University of Ottawa 32

MAT 2377 – Probability and Statistics for Engineers Practice Set c) This is the set all outcomes either in C or in the intersection of A and B : d) This is the set of all outcomes that are not in either B or C (or that are neither in B nor in C ): University of Ottawa 33

MAT 2377 – Probability and Statistics for Engineers Practice Set e) Since C and A ∪ B are mutually exclusive (disjoint), C ⊆ ( A ∩ B ) c and ( A ∩ B ) c ∪ C = ( A ∩ B ) c , so this is the set of all outcomes not in A ∩ B : University of Ottawa 34

MAT 2377 – Probability and Statistics for Engineers Practice Set Q14 . Pieces of aluminum are classified according to the finishing of the surface and according to the finishing of edge. The results from 85 samples are summarized as follows: Edge Surface excellent good excellent 60 5 good 16 4 Let A denote the event that a selected piece has ”excellent” surface, and let B denote the event that a selected piece has “excellent” edge. If samples are elected randomly, determine the following probabilities: P ( A ) a) P ( B ) b) P ( A c ) c) P ( A ∩ B ) d) P ( A ∪ B ) e) P ( A c ∪ B ) f) University of Ottawa 35

MAT 2377 – Probability and Statistics for Engineers Practice Set g) If the selected piece has excellent edge finishing, what is the probability that it has excellent surface finishing? h) If the selected piece has good surface finishing, what is the probability that it has excellent edge finishing? i) Are A and B independent? University of Ottawa 36

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) P ( A ) = 65 / 85 b) P ( B ) = 76 / 85 c) P ( A c ) = 1 − 65 / 85 = 20 / 85 d) P ( A ∩ B ) = 60 / 85 e) P ( A ∪ B ) = P ( A )+ P ( B ) − P ( A ∩ B ) = 65 / 85+76 / 85 − 60 / 85 = 81 / 85 f) P ( A c ∪ B ) = P ( A c ) + P ( B ) − P ( A c ∩ B ) = 20 / 85 + 76 / 85 − 16 / 85 = 80 / 85 g) P ( A | B ) = P ( A ∩ B ) /P ( B ) = 60 / 76 h) P ( B | A c ) = P ( B ∩ A c ) /P ( A c ) = 16 / 20 i) P ( A | B ) ̸ = P ( A ) – events are not independent. University of Ottawa 37

MAT 2377 – Probability and Statistics for Engineers Practice Set Q15 . If P ( A ) = 0 . 1 , P ( B ) = 0 . 3 , P ( C ) = 0 . 3 , and events A, B, C are mutually exclusive, determine the following probabilities: P ( A ∪ B ∪ C ) a) P ( A ∩ B ∩ C ) b) P ( A ∩ B ) c) P (( A ∪ B ) ∩ C ) d) P ( A c ∩ B c ∩ C c ) e) P [( A ∪ B ∪ C ) c ] f) University of Ottawa 38

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) P ( A ∪ B ∪ C ) = P ( A ) + P ( B ) + P ( C ) = 0 . 7 , since the events are mutually exclusive. b) P ( A ∩ B ∩ C ) = 0 c) P ( A ∩ B ) = 0 d) P (( A ∪ B ) ∩ C ) = 0 e) P ( A c ∩ B c ∩ C c ) = P [( A ∪ B ∪ C ) c ] = 1 − P ( A ∪ B ∪ C ) = 0 . 3 (draw Venn diagram) f) P [( A ∪ B ∪ C ) c ] = 1 − P ( A ∪ B ∪ C ) = 0 . 3 University of Ottawa 39

MAT 2377 – Probability and Statistics for Engineers Practice Set Q16 . The probability that an electrical switch, which is kept in dryness, fails during the guarantee period, is 1 %. If the switch is humid, the failure probability is 8 %. Assume that 90 % of switches are kept in dry conditions, whereas remaining 10 % are kept in humid conditions. a) What is the probability that the switch fails during the guarantee period? b) If the switch failed during the guarantee period, what is the probability that it was kept in humid conditions? University of Ottawa 40

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: Let F, H, D represent the event that the switch fails, that it is humid, and that it is dry, respectively. Note that H and D are mutually exclusive and exhaustive. Given: P ( F | D ) = 0 . 01 , P ( F | H ) = 0 . 08 , P ( D ) = 0 . 9 , P ( H ) = 0 . 1 . a) According to the Law of Total Probability, P ( F ) = P ( F | D ) P ( D ) + P ( F | H ) P ( H ) = 0 . 01 · 0 . 9 + 0 . 08 · 0 . 1 = 0 . 009 + 0 . 008 = 0 . 017 . b) According to Bayes’ Theorem, P ( H | F ) = P ( H ∩ F ) P ( F ) = P ( F | H ) P ( H ) P ( F ) = 0 . 4706 . University of Ottawa 41

MAT 2377 – Probability and Statistics for Engineers Practice Set Q17 . The following system operates only if there is a path of functional device from left to the right. The probability that each device functions is as shown. What is the probability that the circuit operates? Assume independence. 0 . 98 0 . 90 1 2 0 . 97 0 . 90 3 4 0 . 95 0 . 90 5 6 0 . 99 7 University of Ottawa 42

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let Box A consist of components 1 , 2 , 3 , 4 ; Box B of components 5 , 6 ; Box C of component 7 . Since all components are independent, P ( system works ) = P ( A works ) P ( B works ) P (7 works ) . Now, B is just a parallel system, so that it works if any of its two components work: P ( B works ) = 0 . 9 + 0 . 95 − 0 . 9 · 0 . 95 = 1 . 85 − 0 . 855 = 0 . 995 . Furthermore, since all the components are independent of one another, P (2 and 4 work ) = P (2 works ) P (4 works ) = 0 . 9 · 0 . 9 = 0 . 81 P (1 and 3 work ) = P (1 works ) P (3 works ) = 0 . 98 · 0 . 97 = 0 . 9506 . University of Ottawa 43

MAT 2377 – Probability and Statistics for Engineers Practice Set Now, A is a parallel system consisting of two independent sub-systems: 2 , 4 and 1 , 3 , so that P ( A works ) = P (2 and 4 work ) + P (1 and 3 work ) − P (2 and 4 work ) P (1 and 3 work ) = 0 . 81 + 0 . 9506 − 0 . 81 · 0 . 9506 = 0 . 9906 . Thus P ( system works ) = 0 . 9906 · 0 . 995 · 0 . 99 = 0 . 9758 . University of Ottawa 44

MAT 2377 – Probability and Statistics for Engineers Practice Set Q18 . An inspector working for a manufacturing company has a 95 % chance of correctly identifying defective items and 2 % chance of incorrectly classifying a good item as defective. The company has evidence that 1 % of the items it produces are nonconforming (defective). 1. What is the probability that an item selected for inspection is classified as defective? 2. If an item selected at random is classified as non defective, what is the probability that it is indeed good? University of Ottawa 45

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: let A be the event that an item is classified as defective and D be the event that an item is defective; so that D c is the event that an item is ’good’. What is known is that P ( D ) = 0 . 01 , P ( A | D ) = 0 . 95 , and P ( A | D c ) = 0 . 02 . a) According to the the Law of Total Probability, P ( A ) = P ( A ∩ D )+ P ( A ∩ D c ) = P ( A | D ) P ( D )+ P ( A | D c ) P ( D c ) ≈ 0 . 0293 . b) According to Bayes’ Theorem, P ( D c | A c ) = P ( A c | D c ) P ( D c ) P ( A c ) = (1 − P ( A | D c )) P ( D c ) 1 − P ( A ) ≈ 0 . 999 . University of Ottawa 46

MAT 2377 – Probability and Statistics for Engineers Practice Set Q19 . Consider an ordinary 52-card North American playing deck ( 4 suits, 13 cards in each suit). a) How many different 5 − card poker hands can be drawn from the deck? b) How many different 13 − card bridge hands can be drawn from the deck? c) What is the probability of an all-spade 5 − card poker hand? d) What is the probability of a flush ( 5 − cards from the same suit)? e) What is the probability that a 5 − card poker hand contains exactly 3 Kings and 2 Queens? f) What is the probability that a 5 − card poker hand contains exactly 2 Kings, 2 Queens, and 1 Jack? University of Ottawa 47

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) The number of possible 5 − card poker hands drawn from a deck of 52 playing cards is 52 C 5 = 52 5 = 52! 5!47! = 2 , 598 , 960 . b) The number of possible 13 − card poker hands drawn from a deck of 52 playing cards is 52 C 13 = 52 13 = 52! 13!39! = 635 , 013 , 559 , 600 . University of Ottawa 48

MAT 2377 – Probability and Statistics for Engineers Practice Set c) The number of possible 5 − card hands that are all-spades is N = ( 13 5 )( 39 0 ) because the 5 spades in the hand can be selected from the 13 spades in the deck in ( 13 5 ) ways, after which zero non-spade can be selected in ( 39 0 ) = 1 way, and so N = ( 13 5 ) · 1 = 13! 5!8! = 1287 . The probability of obtaining such a hand is thus 1287 2 , 598 , 960 ≈ 0 . 000495 . d) For any of the suits S, H, D, C , the probability of a 5 − card hand with all cards in the same suit has been computed above to be 0 . 000495 . Thus, P ( flush ) = P ( S flush ) + P ( H flush ) + P ( D flush ) + P ( C flush ) ≈ 4(0 . 000495) ≈ 0 . 00198 . University of Ottawa 49

MAT 2377 – Probability and Statistics for Engineers Practice Set e) Let A be the outcome which the hand consists of exactly 3 Ks and 2 Qs. We can select the 3 Ks in any one of ( 4 3 ) ways and the 2 Qs in any one of ( 4 2 ) ways, so that the number of such hands is N B = 4 3 4 2 = 4! 1!3! · 4! 2!2! = 4 · (24 / 4) = 24 , and P ( B ) = 24 2 , 598 , 960 ≈ 0 . 0000092 . f) Same idea, but this time P ( C ) = ( 4 2 )( 4 2 )( 4 1 ) 2 , 598 , 960 = 144 2 , 598 , 960 ≈ 0 . 000055 . University of Ottawa 50

MAT 2377 – Probability and Statistics for Engineers Practice Set Q20 . Students on a boat send messages back to shore by arranging seven coloured flags on a vertical flagpole. a) If they have 4 orange flags and 3 blue flags, how many messages can they send? b) If they have 7 flags of different colours, how many messages can they send? c) If they have 3 purple flags, 2 red flags, and 2 yellow flags, how many messages can they send? University of Ottawa 51

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) The question is to find the number of distinguishable permutations of 4 Os and 3 Bs, which is to say 7 C 4 = 7 4 = 7! 4!3! = 35 . b) If the flags were of different colours, we would be looking at the number of (distinguishable) permutations of 7 different objects, which is to say 7! = 5040 . c) The number of distinguishable combinations of 3 Ps, 2 Rs, and 2 Ys is 7! 3!2!2! University of Ottawa 52

MAT 2377 – Probability and Statistics for Engineers Practice Set Q21 . The Stanley Cup Finals in hockey or the NBA Finals in basketball continue until either the representative team form the Western Conference or from the Eastern Conference wins 4 games. How many different orders are possible ( WWEEEE means that the Eastern team won in 6 games) if the series goes 4 games? a) 5 games? b) 6 games? c) 7 games? d) University of Ottawa 53

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) There are only 2 orders: EEEE and WWWW . b) Let’s start with Finals won by the Eastern team, i.e.: 4 out of 5 of the wins were by E , with the 5th (and last) win by E . Note that this is equal to ( 3 0 ) + ( 3 3 ) . In other words, only the order of the 4 first games matters, from which 3 must be wins by E (in order for the 5 th game’s win by E to end the Final): there are ( 4 3 ) = 4 ways of picking 3 wins by E in the first 4 games. The same reasoning applies to Finals won by the Western team. There are thus 8 ways to end the Finals in 5 games. Note that this is equal to ( 4 1 ) + ( 4 3 ) . University of Ottawa 54

MAT 2377 – Probability and Statistics for Engineers Practice Set c) A similar reasoning shows that there are ( 5 3 ) = 5 · 4 2 = 10 ways for either the Eastern or the Western team to win in 6 games, so 20 ways in total. Note that this is equal to ( 5 2 ) + ( 5 3 ) . d) A similar reasoning shows that there are ( 6 3 ) = 6 · 5 · 4 6 = 20 ways for either the Eastern or the Western team to win in 7 games, so 40 ways in total. Note that this is equal to ( 6 3 ) + ( 6 3 ) . University of Ottawa 55

MAT 2377 – Probability and Statistics for Engineers Practice Set Q22 . Consider an ordinary 52-card North American playing deck ( 4 suits, 13 cards in each suit), from which cards are drawn at random and without replacement, until 3 spades are drawn. a) What is the probability that there are 2 spades in the first 5 draws? b) What is the probability that a spade is drawn on the 6 th draw given that there were 2 spades in the first 5 draws? c) What is the probability that 6 cards need to be drawn in order to obtain 3 spades? d) All the cards are placed back into the deck, and the deck is shuffled. 4 cards are then drawn from. What is the probability of having drawn a spade, a heart, a diamond, and a club, in that order? University of Ottawa 56

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) Let A be the event that there are 2 spades in the first 5 draws. Since the cards are drawn without replacement, P ( A ) = ( 13 2 )( 39 3 ) ( 52 5 ) = 13! 2!11! · 39! 3!36! · 5!47! 52! = 13 · 12 2 · 39 · 38 · 37 6 · 120 52 · 51 · 50 · 49 · 48 = 1 , 026 , 492 , 480 3 , 742 , 502 , 400 ≈ 0 . 274 . b) Let B be the event that the 6 th draw was a spade. If A has already occurred, there are only 47 cards left for the 6 th draw, out of which only 11 are spades. Thus P ( B | A ) = 11 47 ≈ 0 . 234 . University of Ottawa 57

MAT 2377 – Probability and Statistics for Engineers Practice Set c) The event A ∩ B is the event that 6 cards need to be drawn in order to obtain 3 spades: first, there must be 2 spades in the first 5 draws ( A ), and once A has happened, there must be a spade on the 6 th draw ( B ): P ( A ∩ B ) = P ( B | A ) P ( A ) ≈ 0 . 234 · 0 . 274 ≈ 0 . 064 . d) If we assume that the first is a spade, the second is a heart, the third is a diamond, and the fourth is a club, then its probability is P ( S 1 ∩ H 2 ∩ D 3 ∩ C 4 ) = P ( S 1 ) P ( H 2 | S 1 ) P ( D 3 | S 1 ∩ H 2 ) P ( C 4 | S 1 ∩ H 2 ∩ D 3 ) = 13 52 · 13 51 · 13 50 · 13 49 ≈ 0 . 0044 . If we take all 4 cards at a time and order is not important, it is: ( 13 1 )( 13 1 )( 13 1 )( 13 1 ) ( 52 4 ) = 13 × 13 × 13 × 13 (52 × 51 × 50 × 49) / 4! ≈ 4! × 0 . 0044 . University of Ottawa 58

MAT 2377 – Probability and Statistics for Engineers Practice Set Q23 . A student has 5 blue marbles and 4 white marbles in his left pocket, and 4 blue marbles and 5 white marbles in his right pocket. If they transfer one marble at random from their left pocket to his right pocket, what is the probability of them then drawing a blue marble from their right pocket? University of Ottawa 59

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: for notation, let BL , BR , and WL denote drawing a blue marble from the left pocket, a blue marble from the right pocket, and a white marble from the left pocket, respectively. By the Law of Total Probability, P ( BR ) = P ( BL ∩ BR ) + P ( WL ∩ BR ) = P ( BL ) P ( BR | BL ) + P ( WL ) P ( BR | WL ) = 5 9 · 5 10 + 4 9 · 4 10 = 41 90 ≈ 0 . 456 . University of Ottawa 60

MAT 2377 – Probability and Statistics for Engineers Practice Set Q24 . An insurance company sells a number of different policies; among these, 60 % are for cars, 40 % are for homes, and 20 % are for both. Let A 1 , A 2 , A 3 , A 4 represent people with only a car policy, only a home policy, both, or neither, respectively. Let B represent the event that a policyholder renews at least one of the car or home policies. a) Compute P ( A 1 ) , P ( A 2 ) , P ( A 3 ) , and P ( A 4 ) . b) From past data, we know that P ( B | A 1 ) = 0 . 6 , P ( B | A 2 ) = 0 . 7 , P ( B | A 3 ) = 0 . 8 . Given that a client selected at random has a car or a home policy, what is the probability that they will renew one of these policies? University of Ottawa 61

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) P ( A 1 ) = 0 . 4 , P ( A 2 ) = 0 . 2 , P ( A 3 ) = 0 . 2 . Because the events are mutually excl. & exhaustive, P ( A 4 ) = 1 − P ( A 1 ) − P ( A 2 ) − P ( A 3 ) = 0 . 2 . b) The event A 1 ∪ A 2 ∪ A 3 represent those clients who have a car policy, a home policy, or both. Thus, P ( B | A 1 ∪ A 2 ∪ A 3 ) = P ( B ∩ ( A 1 ∪ A 2 ∪ A 3 )) P ( A 1 ∪ A 2 ∪ A 3 ) = P (( B ∩ A 1 ) ∪ ( B ∩ A 2 ) ∪ ( B ∩ A 3 )) P ( A 1 ∪ A 2 ∪ A 3 ) . University of Ottawa 62

MAT 2377 – Probability and Statistics for Engineers Practice Set But A 1 , A 2 , A 3 are mutually exclusive, so that B ∩ A 1 , B ∩ A 2 , B ∩ A 3 are also mutually exclusive, and P ( B | A 1 ∪ A 2 ∪ A 3 ) = P ( B ∩ A 1 ) + P ( B ∩ A 2 ) + P ( B ∩ A 3 ) P ( A 1 ) + P ( A 2 ) + P ( A 3 ) = P ( B | A 1 ) P ( A 1 ) + P ( B | A 2 ) P ( A 2 ) + P ( B | A 3 ) P ( A 3 ) P ( A 1 ) + P ( A 2 ) + P ( A 3 ) = 0 . 6 · 0 . 4 + 0 . 7 · 0 . 2 + 0 . 8 · 0 . 2 0 . 4 + 0 . 2 + 0 . 2 = 0 . 54 0 . 80 = 0 . 675 . University of Ottawa 63

MAT 2377 – Probability and Statistics for Engineers Practice Set Q25 . An urn contains four balls numbered 1 through 4 . The balls are selected one at a time, without replacement. A match occurs if ball m is the m th ball selected. Let the event A i denote a match on the i th draw, i = 1 , 2 , 3 , 4 . a) Compute P ( A i ) , i = 1 , 2 , 3 , 4 . b) Compute P ( A i ∩ A j ) , i, j = 1 , 2 , 3 , 4 , i ̸ = j . c) Compute P ( A i ∩ A j ∩ A k ) , i, j, k = 1 , 2 , 3 , 4 , i ̸ = j, i ̸ = k, j ̸ = k . d) What is the probability of at least 1 match? University of Ottawa 64

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: (difficult!) a) P ( A i ) = 3! 4! . b) P ( A i ∩ A j ) = 2! 4! . c) P ( A i ∩ A j ∩ A k ) = 1! 4! . d) P ( A 1 ∪ A 2 ∪ A 3 ∪ A 4 ) = 1 1! − 1 2! + 1 3! − 1 4! . University of Ottawa 65

MAT 2377 – Probability and Statistics for Engineers Practice Set Q26 . The probability that a company’s workforce has at least one accident in a given month is (0 . 01) k , where k is the number of days in the month. Assume that the number of accidents is independent from month to month. If the company’s year starts on January 1, what is the probability that the first accident occurs in April? University of Ottawa 66

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: for any month X , let X represent the event that an accident takes place during month X . If the monthly probabilities are independent of one another, then P ( J c ∩ F c ∩ M c ∩ A ) = P ( J c ) P ( F c ) P ( M c ) P ( A ) = (1 − 31(0 . 01)) · (1 − 28(0 . 01)) · (1 − 31(0 . 01)) · 30(0 . 01) = 0 . 69 · 0 . 72 · 0 . 69 · 0 . 30 ≈ 0 . 103 . University of Ottawa 67

MAT 2377 – Probability and Statistics for Engineers Practice Set Q27 . A Pap smear is a screening procedure used to detect cervical cancer. Let T − and T + represent the events that the test is negative and positive, respectively, and let C represent the event that the person tested has cancer. Among patients that do not have the disease, the test reports a ‘positive’ with probability 0.19; Among patients that have the disease, the test reports a ’negative’ with prob 0.16. In North America, the rate of incidence for this cancer is roughly 8 out of 100,000 women. Based on these numbers, do you think that the Pap smear is an effective procedure? What factors influence your conclusion? University of Ottawa 68

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: from the statement of the problem, we have P ( T − | C ) = 0 . 16 , P ( T + | C ) = 0 . 84 , P ( T + | C c ) = 0 . 19 , P ( T − | C c ) = 0 . 81 , P ( C ) = 0 . 00008 , P ( C c ) = 0 . 99992 . According to Bayes’ Theorem and the Law of Total Probability, P ( C | T + ) = P ( T + | C ) P ( C ) P ( T + ) = P ( T + | C ) P ( C ) P ( T + | C ) P ( C ) + P ( T + | C c ) P ( C c ) = 0 . 84 · 0 . 00008 0 . 84 · 0 . 00008 + 0 . 19 · 0 . 99992 ≈ 0 . 0000354 . For every million positive Pap smears, only 354 represent true cases of cervical cancer. The procedure is ineffective because the cancer rate is small, and because the error rates of the procedure are relatively high. University of Ottawa 69

MAT 2377 – Probability and Statistics for Engineers Practice Set Q28 . Of three different fair dice, one each is given to Elowyn, Llewellyn, and Gwynneth. They each roll the die they received. Let E = { Elowyn rolls a 1 or a 2 } , LL = { Llewellyn rolls a 3 or a 4 } , and G = { Gwynneth rolls a 5 or a 6 } be 3 events of interest. a) What are the probabilities of each of E , LL , and G occurring? b) What are the probabilities of any two of E , LL , and G occurring simultaneously? c) What is the probability of all three of the events occurring simultaneously? d) What is the probability of at least one of E , LL , or G occurring? University of Ottawa 70

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) P ( E ) = P ( LL ) = P ( G ) = 1 / 3 . b) P ( E ∩ LL ) = P ( LL ∩ G ) = P ( G ∩ E ) = 1 / 9 . c) P ( E ∩ LL ∩ G ) = 1 / 27 . d) Using De Morgan’s Law and assuming that the events are independent, P ( E ∪ LL ∪ G ) = 1 − P (( E ∪ LL ∪ G ) c ) = 1 − P ( E c ∩ LL c ∩ G c ) = 1 − P ( E c ) P ( LL c ) P ( G c ) = 1 − (1 − P ( E ))(1 − P ( LL ))(1 − P ( G )) = 1 − (1 − 1 / 3) 3 = 1 − 8 / 27 ≈ 0 . 704 . University of Ottawa 71

MAT 2377 – Probability and Statistics for Engineers Practice Set Q29 . Over the course of two baseball seasons, player A obtained 126 hits in 500 at-bats in Season 1, and 90 hits in 300 at-bats in Season 2; player B , on the other hand, obtained 75 hits in 300 at-bats in Season 1 , and 145 hits in 500 at-bats in Season 2 . A player’s batting average is the number of hits they obtain divided by the number of at-bats. a) Which player has the best batting average in Season 1? In Season 2? b) Which player has the best batting average over the 2-year period? c) What is happening here? University of Ottawa 72

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) In season 1 , player A ’s batting average is 126 500 = 0 . 252 ; player B ’s is 75 300 = 0 . 250 . In season 2 , player A ’s batting average is 90 300 = 0 . 300 ; player B ’s is 145 500 = 0 . 290 . In both seasons, player A has a stronger batting average. b) Over the 2 seasons, player A ’s batting average is 126+90 500+300 = 0 . 270 , while player B ’s batting average is 75+145 300+500 = 0 . 275 ; player B has a stronger battin average. c) Simpson’s paradox! University of Ottawa 73

MAT 2377 – Probability and Statistics for Engineers Practice Set Q30 . A stranger comes to you and shows you what appears to be a normal coin, with two distinct sides: Heads ( H ) and Tails ( T ). They flip the coin 4 times and record the following sequence of tosses: HHHH . a) What is the probability of obtaining this specific sequence of tosses? What assumptions do you make along the way in order to compute the probability? What is the probability that the next toss will be a T . b) The stranger offers you a bet: they will toss the coin another time; if the toss is T , they give you 100$ , but if it is H , you give them 10$ . Would you accept the bet (if you are not morally opposed to gambling)? c) Now the stranger tosses the coin 60 times and records 60 × H in a row: H · · · H . They offer you the same bet. Do you accept it? d) What if they offered 1000$ instead? 1 , 000 , 000$ ? University of Ottawa 74

MAT 2377 – Probability and Statistics for Engineers Practice Set Solution: a) (1 / 2) 4 = 0 . 0625 ; independence of tosses and fairness of coin; 1 / 2 = 0 . 5 . b) The bet is to your advantage: half the time you will receive 100$ , half the time you will lose 10$ ; on average, you receive more than you lose if you take the bet. c) You know what? Even though it’s theoretically possible for a fair coin to be tossed independently 60 times and to come up H 60 times in a row, the probability of this happening is so vanishingly small at (1 / 2) 60 = 8 . 7 × 10 − 19 that I don’t actually believe that my assumptions were warranted: either the coin is not fair, or the tosses are not independent. At any rate, I smell a rat and I do not take the bet. University of Ottawa 75