MA 162 - Section 6.3 and 7.3 Completed Section 6.3 1. Four commuter trains and three express buses depart from City A to City B in the morning, and three commuter trains and three express buses operate on the return trip in the evening. In how many ways can a commuter from City A to City B complete a daily round trip via bus and/or train? Solution: There are 7 ways to travel from City A to City B (4 trains and 3 buses) and there are 6 ways to travel from City B to City A (3 trains and 3 buses), so there are 7 6 = 42 ways for a commuter to make a round trip. 2. In a survey conducted by a union, member were asked to rate the importance of the following issues: (1) job security, (2) increased fringe benefits, and (3) health benefits. Five different responses were allowed for each issue. Among completed surveys, how many different responses to this survey were possible? Solution: There are 5 responses to the first issues, 5 responses to the second, and 5 responses to the third, so in total there are 5 5 5 = 125 different responses to the survey. round trip. 3. A new Shanghai Restaurant offers a choice of three appetizers, two choices of soups, and ten choices of entrees for its lunch special. How many different complete lunches can be ordered from the restaurant s lunch special menu? Solution: Using the multiplication principle, there are 3 2 10 = 60 different complete lunches. 4. Lugano Leather Company makes an executive attache case equipped with two rolling combination locks, each one with a provision for a three-digit number. The attache case is unlocked by setting each lock to produce the correct combination. Each roller in each of the two locks has ten digits. 1
a. How many possible combinations are there? Solution: There are 10 options for each of the 3 digits on the left lock, and 10 options for each of the 3 digits on the right lock, so there are 10 10 10 = 1000 different options for the left lock and 10 10 10 = 1000 different options for the right lock. So there are 1000 1000 = 1, 000, 000 total options. b. How many combinations are possible if the lock on the left-hand side of the attache case must end with an even number and the lock on the right-hand side of the attache case must end with an odd number? Solution: There are 10 options for the first 2 digits on the left lock, and 5 options for the third digit on the left lock, so there are 10 10 5 = 500 different options for the left lock. There are 10 options for first two 2 digits on the right lock, and 5 options for the third digit on the right lock, so there are 10 10 5 = 500 different options for the right lock. So there are 500 500 = 250, 000 total options. 5. The 2014 Toyota Camry comes with six grades of models, two sizes of engines, four choices of transmissions, six exterior colors, and two interior colors. How many choices of the Camry are available for a prospective buyer? Solution: Using the multiplication principle, there are 6 2 4 6 2 = 576 total choices of the Camry available. 6. Over the years, the state of California has used different combinations of letters of the alphabet and digits on its automobile license plates. a. At one time, license plates were issued that consisted of three letters followed by three digits. How many different license plates can be issued under this arrangement? Solution: The 3 letters have 26 options each, and the three digits have 10 options each. So in total, there are 26 3 10 3 = 17, 576, 000 total license plates. b. Later on, license plates were issued that consisted of three digits followed by three letters. How many different license plates can be issued under this arrangement? Solution: Since the order of the letters and digits does not matter, there are the same number of license plates as before, so 26 3 10 3. c. Recently, license plates were issued using a combination of one letter, followed by three digits, followed by another three letters. How many different license plates can be issued under this arrangement? Page 2
Solution: The first letter has 26 options, and the following pattern of digits and letters is the same as in part b, so there are 26 3 10 3 options for those. Thus in total, there are 26 26 3 10 3 = 26 4 10 3 = 26 26 3 10 3 = 456, 976, 000 total license plates. 7. An exam consists of ten true-or-false questions. Assuming that every question is answered, in how many different ways can a student complete the exam? In how many ways can the exam be completed if a student can leave some questions unanswered because a penalty is assessed for each incorrect answer? Solution: There are 2 choices for each of the 10 problems, so there are 2 10 = 1024 total options to answer every question. If the student can leave a question blank, this is the same as having a third option for each question, so there are 3 10 total options to allow questions to be unanswered. 8. (a) How many seven-digit telephone numbers are possible if the first digit must be nonzero? Solution: There are 9 options for the first digit, and 10 options for each of the following six digits, so there are 9 10 6 = 9, 000, 000 possible numbers. (b) How many direct-dialing numbers for calls within the United States and Canada are possible if each number consists of a 1 plus a three-digit area code (the first digit of which must be nonzero) and a number of the type described in part (a)? Solution: There are 9 options for the first area code digit, and then 10 options for the following two digits. Then the remaining digits, we know there are 9 10 6 possible options. So there are 9 10 9 10 6 = 9 2 10 7 = 810, 000, 000 total numbers. Page 3
Section 7.3 1. An experiment consists of selecting a card at random from a 52-card deck. Refer to this experiment for the following and find the probability of the event. (a) A king of diamonds is drawn. Solution: There is exactly one king of diamonds. 1 probability of of drawing the king of diamonds. 52 Out of 52 cards, you have a (b) A diamond or a king is drawn. Solution: There are 13 diamonds, 4 kings, and 1 diamond that is a king. So P ( Diamond or king ) = P (Diamond)+P (king) P (Diamond and King) = 13 16. 52 (c) A face card is drawn. + 4 1 = 52 52 52 Solution: There are 12 face cards in a deck. So the probability of drawing a face card is 12 52. (d) A red face card is drawn. Solution: There are 6 red face cards. So our probability is 6 52 (e) An ace is not drawn. Solution: There are four aces. So the probability of not drawing one of these cards is P (not an ace) = 1 P (ace) = 1 4 = 48. 52 52 (f) A black face card is not drawn. Solution: There are 6 black face cards. So we get 1 6 = 46 52 52 2. Explain why the following statement is incorrect. Joanne, a high school senior, has applied for admission to four colleges, A, B, C, and D. She has estimated that the probability that she will be accepted for admission by college A, B, C, and D is.5,.3,.1, and.08, respectively. Thus the probability that she will be accepted for admission by at least one college is P (A) + P (B) + P (C) + P (D) =.5 +.3 +.1 +.08 =.98. Solution: The probability cannot be computed this way because the event of her being accepted into each college is not mutually exclusive from accptance into each of the the other colleges (although their likely hoods are independent). Page 4
3. In a survey of 1089 adults conducted by Duke University between January 16, 2013 and January 22, 2013, the following question was asked: How serious a threat is climate change? The results of the survey are summarized below: Answer Very Serious VS Somewhat Serious SS Not that much NM Not at all N Respondents 38% 46% 15% 1% On the basis of the results of the survey, what is the probability that a person chosen at random from the survey said that climate change posed: (a) A very serious threat or a somewhat serious threat? Solution: P (VS or SS) = P (VS) + P (SS) =.38 +.46 =.84 (b) No threat? Solution: P (N) =.01 (c) A somewhat serious threat or not that much of a threat? Solution: P (SS or NM) = P (SS) + P (NM) =.46 +.15 =.61 4. Let E and F be two events of an experiment with sample space S. Suppose P (E) =.4, P (F ) =.5, and P (E F ) =.1. Compute: (a) P (E F ) Solution: Given our formula P (E F ) = P (E)+P (F ) P (E F ) =.5+.4.1 =.8 (b) P (E c F c ) Solution: Using De Morgan s Law, we get P (E c F c ) = P ((E F ) c ) = 1 P (E F ) = 1.8 =.2 (c) P (E c F ) Solution: Here, we can interpret directly from a venn diagramm that P (E c F ) = P (F ) P (E F ) =.4.1 =.3 Page 5