STAT 418 MID-TERM EXAMINATION 1 FALL, 2008NAME (in BLOCK letters):SIGNATURE:Student Number:Instructions:1. Please verify that your exam paper contains all seven questions.2. You may use a calculator and one sheet of notes, but you may not use any books.3. Illegible handwriting will not be graded!4. To earn partial credit, explain your arguments carefully and show all your work. If youneed extra paper then please writ e on the back of your paper.5. Your instructor or teaching ass istant will answer no questions during the examinationperi od. Simply explain any error you may find in a question and proceed to the nextquestion.DO NOT WRITE BELOW THIS LINEQuestion Marks1234567TotalProblem 1. [This problem is based on E xercise 7, p. 16, in the textbook.]In how many way s can 4 boys and 4 girls sit in a row if:(a) There are no restrictions on the seating arrangement?(b) The boys must sit together and the girls must sit together?(c) The girls must si t together and there are no restrictions on where the boys may sit?(d) No two people of the same sex may sit together?(e) Jill (one o f the girls) must sit beside Jack (one of the boys)?(a) 8!(b) 2! · 4! · 4!(c) 5! · 4!(d) 2 · 4 · 4 · 3 · 3 · 2 · 2 · 1 · 1 = 2 · 4! · 4!(e) 2! · 7!Problem 2. [This problem is based on Self-Test Exercise 3, p. 22, in the textbook.]A president, treasurer, and secretary, all different, are to be chosen from a club consistingof 1 5 people ( whose names are Ann, Bob, Cate, Don, Eda, Fran, Gail, Hal, Iona, Jane,Kate, Leo, Mary, Nan, and Oscar). How many different choices of officers are possible if:(a) There are no restrictions?(b) Ann and Bob wi ll not serve tog et her?(c) Cate and Don wi ll serve together or not at all?(d) Eda must be an officer?(e) Oscar will serve only if he is president?(a)!153"· 3!, or 15 · 14 · 13(b) (15 · 14 · 13) − (13 · 3!(c) (13 · 3!) + (13 · 12 · 11)(d) 3! ·!142"(e) (14 · 13 · 12) +!142"· 2!Problem 3. [This problem is based on Self-Test Exercise 2, p. 63 in the textbook.]A customer visiting a department store will buy a suit with probability .22, a shirt withprobability .30, and a tie with probability .28. The customer will purchase both a suit anda shirt with probability .11, both a suit and a tie with probability .14, and both a shirt anda tie with probability . 10. Also, a customer will purchase all three items with probability.06. What is the probability that a customer purchases:(a) None of t he three items?(b) Exactly one o f the three items?(c) Exactly two of the three items?A simple way to solve this problem is to draw the corresponding Venn diagram, insert theprobabilities in the various regions, and then deduce the answers.Problem 4. [This problem is based on E xercise 34, p. 59 and Self-Test Exercise 3, p. 64.]Thirteen cards are dealt one by one from a well-shuffled, standard deck of 52 playingcards. Find the probability that:(a) The thirteenth card is an ace.(b) The first ace occurs on the thirteenth card.(c) A ll thirteen cards are of the same suit (i.e., a ll hearts, or a ll clubs, or all spades, or alldiamonds).(d) The set of thirteen cards is a Yarborough, i.e., does not contain a ten, jack, queen,king, or ace.(a)452(b)485247514650454944484347424641454044394338423741440(c)(41)(1313)(5213)(d)(3213)(5213)Problem 5. [This problem is based on E xercises 1, 2, and 4, p. 111 in the textbook.]Two fair dice are rolled. What is the conditi onal probabili ty that:(a) At least one die lands on 6 given that the dice land on different numbers?(b) The first di e lands on 6 given that the sum of the dice is 9?(c) At least one die lands on 6 g iven that the sum of the dice is 9?(a) Let A be the event that at least one die lands on 6, and let B be the event that the dicelands on different numbers. There are 30 elementary events in B, therefore P (B) = 30/36.Also, there are 1 0 elementary events in A ∩ B; therefore, P (A ∩ B) = 1 0 /36. BecauseP (B) #= 0, we obtainP (A|B)=P (A ∩ B)P (B)=10/3630/36=13.(b) Given that the sum of the dice is 9, the reduced sample space is {(3, 6), (6, 3), (4, 5), (5, 4)}.Among the elements of the reduced sample space, only the outcome (6,3) satisfies the eventthat the first die lands on 6. Therefore, given that the sum of the dice is 9, the conditio nalprobability that the first die lands on 6 is 1/4.Alternatively, let A = {The first die lands on 6} and B = {The sum of the dice is 9}.Then, A ∩ B = {(6, 3)}, so that P (A ∩ B) = 1/36. Also, B = {(3, 6), (6, 3), (4, 5), (5, 4)},so that P (B) = 4/36. Therefore, because P (B) #= 0, we obtainP (A|B)=P (A ∩ B)P (B)=1/364/36=14.(c) This problem i s similar to part (b). Given that the sum of the dice is 9, the reducedsample space is {(3, 6), (6, 3), (4, 5), (5, 4)}. Among the elements of the reduced samplespace, only the out comes (6,3) and (3,6) satisfy the condition that at least one die landson 6. Therefore, given that the sum of t he dice is 9, the conditional probability that atleast one die lands on 6 is 2/4, or 1/2.Problem 6. [This problem is based on E xercises 1,2 and 4, p. 111 in the textbook.](a) Let {A, B} be a partition of a sample space and let E be any event. State Bayes’Theorem as it applies to the conditional probability, P (A|E).Consider two boxes, t he first containing 1 gold and 1 silver coins, and the secondcontaining 2 gold coins and 1 silver coins. A box is selected at random, and a coin isdrawn at random from it.(b) What is the probability that the coin drawn is a gold coin? (Hint : The Law of Tot alProbability.)(c) Given that the coin drawn was a gold coin, what is the conditional probability thatthe box selected was the first box?(d) Given that the coin drawn was not gold, what is the probabili ty that the box selectedwas not the first box?(a)P (A|E)=P (E|A)P (A)P (E|A)P (A)+P (E|B)P (B)(b) Let A = {Box 1 is drawn}, B = {Box 2 is drawn}, and E = {A gol d coi n is drawnfrom the box selected}. Then, P (A)=P (B) = 1/2, P (E|A) = 1/2, and P (E|B) = 2 /3.Therefore, by the Law of Total Pro babi lity,P (E)=P (E|A)P (A)+P (E|B)P (B)=12·12+23·12.(c) Using the same notation as in part (b), it follows from Bayes’ Theorem thatP (A|E)=P (E|A)P (A)P (E|A)P (A)+P (E|B)P (B)=12·1212·12+23·12.(d) Let A = {Box 1 is drawn}, B = {Box 2 is drawn}, and E = {A silver coin is drawnfrom the box selected}. Then, P (A)=P (B) = 1/2, P (E|A) = 1/2, and P (E|B) = 1 /3.By Bayes’ Theorem,P (B|E)=P (E|B)P (B)P (E|A)P (A)+P (E|B)P (B)=13·1212·12+23·12.Bonus Problem 7. …