Section 4.3 Solutions

Section 4.3 Solutions. 1) If two dates are selected at random from the 365 days of the year and are listed in the order they were selected. a) Use the counting principle to determine the number of possible outcomes if the dates are selected with replacement. _____365____*_____365_____= 133225 First day second day Answer: 133,225 possible outcomes b) Use the counting principle to determine the number of possible outcomes if the dates are selected without replacement. _____365____*_____364_____= 132860 First day second day Answer: 132,860 possible outcomes 3) Joe has 5 shirts, 3 pairs of pants and 4 pairs of shoes. He needs to make an outfit containing one of each item. How many different outfits are possible? (We assume that one different item makes a different outfit) __5_____*____3____*___4_____ = 60 shirts pants shoes Answer: 60 outfits 5) A television remote has buttons for the digits 0 – 9. If you press two buttons, how many numbers are possible if a) The same button may be pressed twice? This is like a with replacement problem. _____10____*_____10_____= 100 First number second number Answer: 100 possible outcomes b) The same button may not be pressed twice? This is like a without replacement problem. _____10____*_____9_____= 90 First number second number Answer: 90 possible outcomes 7) A movie theater sells 3 sizes of popcorn (small, medium, and large) with 3 choices of toppings (no butter, butter, extra butter). How many possible ways can a bag of popcorn be purchased? ______3______*_____3______ = 9 Size topping Answer: 9 9) A state issues license plates consisting of letters and numbers. There are 26 letters and the letters may be repeated. There are 10 digits and the digits may be repeated. How many possible license plates can be issued with two letters followed by three numbers? ____26____*______26______*____10______*____10_______*____10______= 676,000 First letter second letter 1st number 2nd number 3rd number Answer: 676,000 11) Burger Queen offers 4 types of burgers, 5 types of beverages, and 3 types of desserts. If a meal consists of 1 burger, one beverage and one dessert, how many possible meals can be chosen? ____4________*_____5_______*_____3_____= 60 Burger beverage dessert Answer: 60 13) Reid is ordering a birthday cake for a friend. There are 5 cake flavors and 4 frosting flavors to choose from. How many different cakes can Reid order? _____5_____*____4_____ = 20 Cake frosting Answer: 20 15) Answer question 14 but assume the license plates must be of the form 2 letters followed by 4 numbers and there are no other restrictions. ____26___*______26___*____10______*____10_______*____10___*__10_____= 6,760,000 First letter second letter 1st number 2nd number 3rd number 4th number Answer: 6,760,000 17) The standard New York state license plate has three letters followed by four digits. How many different license plates are possible if neither the digits nor the letters can be repeated? ____26___*______25___*____24_____*___10____*____9___*____8___*__7_____ 1st letter 2nd letter 3rd letter 1st # 2nd # 3rd # 4th # Answer: 78,624,000 19) You are assigned a computer generated 4-digit password to access your new voice mail account. If the digits can be repeated, how many passwords are possible? ___10____*___10__*____10___*___10____ 1st # 2nd # 3rd # 4th # Answer: 10,000 #20 – 30: Counting principle problems with sample space construction and probability questions. 21) A couple plans to have two children. a) Determine the number of points in the sample space of the possible arrangements of boys and girls. _______2_____*______2________ = 4 First child second child Answer: 4 b) Construct a tree diagram and list the sample space. Answer: Sample Space: {BB BG GB GG} Find the probability the couple has: c) Two girls The numerator will be the one element in the sample space that has two girls. The denominator will be the 4 elements in the sample space. Answer: 1/4 d) At least one girl The numerator counts BG GB and GG, it should be 3. The denominator consists of all 4 elements in the sample space. Answer: 3/4 e) A girl then a boy The numerator is 1, as GB is the only element that I count here. The denominator is still the 4 elements in the sample space. Answer: ¼ 23) A coin is tossed three times and a sequence of heads and tails is recorded. a) Determine the number of points in the sample space ____2_____*_____2_____*_____2______= 8 1st flip 2nd flip 3rd flip Answer: 8 b) Construct a tree diagram and list the sample space Answer: Sample Space {HHH HHT HTH HTT THH THT TTH TTT} Find the probability that: c) No heads are tossed The numerator is 1 as the TTT is the only event with no heads. The denominator is the 8 items in the sample space. Answer: 1/8 d) Exactly one head is tossed The numerator consists of these 3 elements: TTH THT HTT The denominator is the 8 items in the sample space. Answer: 3/8 e) Three heads are tossed The numerator is 1 as the HH is the only event with three heads. The denominator is the 8 items in the sample space. Answer: 1/8 25) Two six sided dice are rolled and the number on each face is recorded. a) Determine the number of points in the sample space _______6_______*_______6______= 36 first dice second dice Answer: 36 b) Construct a tree diagram and list the sample space 25b) Sample Space: (1,1) sum 2 (2,1) sum 3 (3,1) sum 4 (4,1) sum 5 (5,1) sum 6 (6,1) sum 7 (1,2) sum 3 (1,3) sum 4 (1,4) sum 5 (1,5) sum 6 (1,6) sum 7 (2,2) sum 4 (2,3) sum 5 (2,4) sum 6 (2,5) sum 7 (2,6) sum 8 (3,2) sum 5 (3,3) sum 6 (3,4) sum 7 (3,5) sum 8 (3,6) sum 9 (4,2) sum 6 (4,3) sum 7 (4,4) sum 8 (4,5) sum 9 (4,6) sum 10 (5,2) sum 7 (5,3) sum 8 (5,4) sum 9 (5,5) sum 10 (5,6) sum 11 (6,2) sum 8 (6,3) sum 9 (6,4) sum 10 (6,5) sum 11 (6,6) sum 12 c) Find the probability that a double is rolled (both dice have the same number) The numerator is 6. I need to count these: (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) The denominator is the 36 items in the sample space. Answer: 6/36 = 1/6 d) Find the probability that a sum of 7 is rolled The numerator is 6. I need to count: (1,6) (2,5) (3,4) (4,3) (5,2) (6,1) The denominator is the 36 items in the sample space. Answer: 6/36 = 1/6 e) Find the probability a sum of 2 is rolled The numerator is a 1 as the only item I can count is (1,1) The denominator is the 36 items in the sample space. Answer: 1/36 f) Are you as likely to roll a sum of 2 as you are of rolling a sum or 7 Answer: It is 6 times more likely to roll the sum of a 7. 27) A coin is flipped then a 6 sided dice is rolled. a) Determine the number of points in the sample space _____2______*______6______= 12 Coin dice b) Construct a tree diagram and list the sample space Sample Space: { H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6} c) Find the probability that a head is flipped and an even number is rolled The numerator counts: H2 H4 H6, The numerator will be 3. The denominator counts the 12 items in the sample space. Answer: 3/12 = ¼ d) Find the probability that a tail is flipped and a number less than 3 is rolled The numerator counts T1 T2, the numerator is 2. The denominator counts the 12 items in the sample space. Answer: 2/12 = 1/6 29) A coin is flipped then a number is picked out of a hat containing the numbers 1,2 and 3. a) Determine the number of points in the sample space ____2_______*_____3______ = 6 Coin number Answer: 6 b) Construct a tree diagram and list the sample space Sample Space: {H1 H2 H3 T1 T2 T3} c) Find the probability that a head is flipped and an even number is selected The numerator is 1, as I can only count H1. The denominator is the 6 elements in the sample space. Answer: 1/6 d) Find the probability that a tail is flipped and a number less than 3 is selected The numerator is 2. I can count T1 and T2 The denominator is the 6 elements in the sample space. Answer: 2/6 = 1/3