Probability and Combinatorics Solutions for Exam 1.1a

IE 111 NAME ___Solutions______________________ 9:00 Section Exam 1.1a Fall 2012 Instructions ● Open book, open notes ● Clearly indicate your answer ● You must show all relevant work and justify your answers appropriately ● Partial credit will be given, but not without sufficient support ● Factorials, permutations, and combinations and other complex formulas do not need to be evaluated. You can leave as 4 C 2 for example. ● Cases where you should compute a final number are noted. ● Each question is worth 25 points ● My guess is that this is a long test.

Question 1 Consider the events shown in the Venn diagram above: Suppose P(A)=0.3 P(C )=0.6 P(E)=0.42. Suppose also that A and C are independent. (Please compute a final probability for each part). a) Find P(C) = 1-P(C ) = 1-0.6 =0.4 S B A D C E

b) Find P((A B) (C D)) = P(A C) = P(A) P(C) = (0.3)(0.4) = 0.12 c) Find P(A C E)) = P(A C)+P(E)) = 0.58+0.42 = 1 =P(A)+P(C)-P(A C) +P(E)) = P(A)+P(C)-P(A)P(C) +P(E)) = 0.3+0.4 – (0.3)(0.4) + 0.42 = 1 d) Find P(C D ) = P(C ) = 0.6

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Question 2 Consider the network of roads and bridges that go from city X to city Y below: Suppose that the probability that each of the bridges is out due to a storm are as follows: P(Bridge A is out) = 0.3 P(Bridge B is out) = 0.2 P(Bridge C is out) = 0.1 P(Bridge D is out) = 0.5 P(Bridge E is out) = 0.4 a) Assuming bridges fail independently, what is the probability I can go from City X to City Y? =P (A B C E ) + P(A D E ) - P(A B C E A D E ) =P (A B C E ) + P(A D E ) - P(A B C D E ) = (0.7)(0.8)(0.9)(0.6) + (0.7)(0.5)(0.6) - (0.7)(0.8)(0.9)(0.5)(0.6) = 0.3024 + 0.21 – 0.1512 = 0.3612 b) Suppose you don’t know which bridges are out in advance, but must choose a route before you leave City X. Which route should you choose? From above, the probability of route A-B-C-E is 0.3024 While the probability of route A-D-E is 0.21 thus choose A-B-C-E Question 2 A group consists of 6 Lehigh students, 4 Moravian students and 3 Muhlenberg students. City Y Bridge C Bridge B Bridge E Bridge D Bridge A City Y City X

a) How many different committees of size 6 can I form? b) If 3 of the students must be from Lehigh, now how many committees of size 6 can I form? c) Given that Joe Blow from Moravian must be on the committee now how many committees of size 6 can I form? d) If each of the committees in part a) is equally likely, what is the probability that Joe Blow is on the committee? e) If each of the 6 positions on the committee is a unique position (e.g. President, VP, Treasurer, secretary, finance committee chair, and membership committee chair), now how many committees can I form from the 13 candidates?

Question 3 ` I am making shish-kebabs out of chunks of chicken, tomatoes, green peppers and mushrooms. Each shish-kebab will have a total of 8 items (chunks) on it. Suppose I have an unlimited supply of each of the four types of items. a) If the order of the chunks on the skewer does matter , and each skewer will have exactly 8 chunks of any type on the skewer, how many shish-kebab configurations can I make? = 4 8 = 65536 b) If the order of the chunks on the skewer doesn’t matter , and each skewer will have exactly 8 chunks of any type, how many shish-kebab configurations can I make? N=4 item types Multi-choose k=8 items: (4-1)+8 Multichoose 8 = 11 C 8 = 165 c) If the only restriction is that each of the 8 items on the skewer must be different to the items on either side of it, how many shish-kebab configurations can I make? 4*3 7 = 8748 d) If I have only 8 chunks as follows: 3 chicken, 2 mushroom, 2 tomato and 1 green pepper, how many shish-kebab configurations can I make? Permutation of like objects 8!/(3!2!2!1!) = 420 Drawing of a Shish-Kebab

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Question 4 I invented a new dice game played with a single fair die with the following rules: ● If you roll a 1 or a 2, then you lose ● If you roll a 5 or a 6, then you win ● If you roll a 3 or a 4, then you roll again ● If you rolled a 3 on the first roll and a 1 or 2 or 3 on the second roll, you win. ● If you rolled a 3 on the first roll and a 4 or 5 or 6 on the second roll, you lose. ● If you rolled a 4 on the first roll and a 1 or 2 or 3 or 4 on the second roll, you win. ● If you rolled a 4 on the first roll and a 5 or 6 on the second roll, you lose. a) Given that you won the game, what is the probability you rolled a 4 on the first roll? P(Win) = (1/3)+(1/6)(1/2)+(1/6)(2/3) = 12/36+7/36=19/36 = 0.53 P(4|Win) = P(4 Win)/P(Win) = (2/18)/(19/36) = 4/19 = 0.2105 b) If I play this game over and over again, what is the probability that I win for the first time on or before my third play of the game? =P(W or LW or LLW) = (19/36)+(17/36)(19/36)+(17/36) 2 (19/36) = 0.8947 c) If I play this game over and over again, what is the probability that I win for the second time on my third play of the game? =P(LWW or WLW) = (2)(19/36) 2 (17/36) = 0.263

IE 111 NAME _________________________ 9:00 Section Exam 1.1b Fall 2012 Instructions ● Open book, open notes ● Clearly indicate your answer ● You must show all relevant work and justify your answers appropriately ● Partial credit will be given, but not without sufficient support ● Factorials, permutations, and combinations and other complex formulas do not need to be evaluated. You can leave as 4 C 2 for example. ● Cases where you should compute a final number are noted. ● Each question is worth 25 points ● My guess is that this is a long test.

Question 1 Consider the events shown in the Venn diagram above: Suppose P(B)=0.3 P(D )=0.6 P(E)=0.42. Suppose also that B and D are independent. (Please compute a final probability for each part). a) Find P(D) = 1-P(D ) = 1-0.6 =0.4 b) Find P((A B) (C D)) = P(B D) = P(B)P(D) = (0.3)(0.4) = 0.12 c) Find P(B D E)) = P(B D)+P(E)) = P(B)+P(D)-P(B D) +P(E) = P(B)+P(D)-P(B)P(D) +P(E) = 0.3+0.4 – (0.3)(0.4) +0.42 = 1 d) Find P(C D ) = P(D ) = 0.6 S B A C D E

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Question 2 Consider the network of roads and bridges that go from city X to city Y below: Suppose that the probability that each of the bridges is out due to a storm are as follows: P(Bridge A is out) = 0.2 P(Bridge B is out) = 0.15 P(Bridge C is out) = 0.15 P(Bridge D is out) = 0.4 P(Bridge E is out) = 0.3 a) Assuming bridges fail independently, what is the probability I can go from City X to City Y? =P (A B C E ) + P(A D E ) - P(A B C E A D E ) =P (A B C E ) + P(A D E ) - P(A B C D E ) = (0.8)(0.85)(0.85)(0.7) + (0.8)(0.6)(0.7) - (0.8)(0.85)(0.85)(0.6)(0.7) = 0.3024 + 0.21 – 0.1512 = 0.3612 = 0.4046 + 0.336 – 0.24276 = 0.49784 b) Suppose you don’t know which bridges are out in advance, but must choose a route before you leave City X. Which route should you choose? From above, the probability of route A-B-C-E is 0.4046 While the probability of route A-D-E is 0.336 thus choose A-B-C-E City Y Bridge C Bridge B Bridge E Bridge D Bridge A City Y City X

Question 2 A group consists of 5 Lehigh students, 7 Moravian students and 4 Muhlenberg students. a) How many different committees of size 5 can I form? b) If 4 of the students must be from Lehigh, now how many committees of size 6 can I form? c) Given that Joe Blow from Moravian must be on the committee, but there are no other restrictions on selecting members, how many committees of size 6 can I form? d) If each of the committees in part a) is equally likely, what is the probability that Joe Blow is on the committee? e) If each of the 6 positions on the committee is a unique position (e.g. President, VP, Treasurer, secretary, finance committee chair, and membership committee chair), now how many committees can I form from the 13 candidates?

Question 3 ` I am making shish-kebabs out of chunks of chicken, tomatoes, and mushrooms. Each shish-kebab will have a total of 9 items (chunks) on it. Suppose I have an unlimited supply of each of the three types of items. a) If the order of the chunks on the skewer does matter , and each skewer will have exactly 9 chunks of any type on the skewer, how many shish-kebab configurations can I make? = 3 9 b) If the order of the chunks on the skewer doesn’t matter , and each skewer will have exactly 9 chunks of any type, how many shish-kebab configurations can I make? N=3 item types Multi-choose k=9 items: (3-1)+9 Multichoose 9 = 11 C 9 c) If the only restriction is that each of the 9 items on the skewer must be different to the items on either side of it, how many shish-kebab configurations can I make? 3*2 8 d) If I have only 9 chunks as follows: 4 chicken, 3 mushroom, and 2 tomato, how many shish-kebab configurations can I make? Permutation of like objects 9!/(4!3!2!) Drawing of a Shish-Kebab

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Question 4 I invented a new dice game played with a single fair die with the following rules: ● If you roll a 1 or a 2, then you lose ● If you roll a 5 or a 6, then you win ● If you roll a 3 or a 4, then you roll again ● If you rolled a 3 on the first roll and a 1 or 2 or 3 on the second roll, you win. ● If you rolled a 3 on the first roll and a 4 or 5 or 6 on the second roll, you lose. ● If you rolled a 4 on the first roll and a 1 or 2 or 3 or 4 on the second roll, you win. ● If you rolled a 4 on the first roll and a 5 or 6 on the second roll, you lose. a) Given that you won the game, what is the probability you rolled a 4 on the first roll? P(Win) = (1/3)+(1/6)(1/2)+(1/6)(2/3) = 12/36+7/36=19/36 = 0.53 P(4|Win) = P(4 Win)/P(Win) = (2/18)/(19/36) = 4/19 = 0.2105 b) If I play this game over and over again, what is the probability that I win for the first time on or before my third play of the game? =P(W or LW or LLW) = (19/36)+(17/36)(19/36)+(17/36) 2 (19/36) = 0.8947 c) If I play this game over and over again, what is the probability that I win for the second time on my third play of the game? =P(LWW or WLW) = (2)(19/36) 2 (17/36) = 0.263

Exam1.1Soln_9am

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