Section 4.1 Solutions

Section 4.1 solutions 1) In a given week, a veterinarian treated the following animals: Animal Dog Cat Bird iguana Total Number treated 40 35 15 5 95 Determine the empirical probability that the next animal she treats is a) A dog: (just divide the number of dogs by the total number of pets treated, and reduce.) 40/95 Answer: 8/19 b) A cat: (just divide the number of cats by the total number of pets treated, and reduce.) 35/95 Answer: 7/19 c) An iguana: (just divide the number of dogs by the total number of pets treated, and reduce.) 5/95 Answer: 1/19 3) A survey was given to 250 people at a Sun’s game. 30 of them said Charles Barkley was their favorite Sun of all time. a) If another person at the game is asked to name their favorite Sun of all time what is the probability they will select Charles Barkley? Just divide the number of people that selected Charles Barkley by the number of people surveyed. 30/250 Answer: 3/25 b) Estimate the number of people of the 12,000 in attendance will say Charles Barkley is their favorite Sun’s player of all time? (Round to the nearest whole number) 3 25 ∗ 12000 = 1440 𝐴𝑛𝑠𝑤𝑒𝑟 1440 𝑝𝑒𝑜𝑝𝑙𝑒 5) John has taught 500 MAT 142 students. 220 of his MAT 142 students received a grade of an A. a) Determine the empirical probability that a student in John’s MAT 142 class receives a grade of an A. 220/500 Answer: 11/25 b) Estimate the number of A’s John will give in a MAT 142 class that has 30 students. (Round to the nearest whole number) 11 25 ∗ 30 = 13.2 Answer: about 13 students 7) A survey was conducted to determine students’ favorite breeds of dogs. Each student chose one breed. Dog Collie Spaniel Lab Boxer PitBull Other # 10 15 35 8 5 12 a) What is the empirical probability that a students’ favorite dog is a boxer? (Hint 85 students’ were surveyed) Answer: 8/85 b) 200 students are asked to name their favorite dog. Estimate how many of them will choose a pit bull. 5 ∗ 200 200 = 5 Answer: 5 students Homework #9 – 22: Write your answer as a reduced fraction. 9) A TV remote control has keys for channels 0 through 9. If you select a key at random, a) What is the probability you press channel 6? There are 10 possible numbers that can be pushed. This gives the 10 for the denominator. There is only one 6 so numerator is 1. Answer: 1/10 b) What is the probability you press a key for an odd number? There are 10 possible numbers that can be pushed. This gives the 10 for the denominator. There are 5 odd numbers (1,3,5,7,9), so the numerator is 5. 5/10 Answer: 1/2 c) What is the probability you press a key for a number less than 7? There are 10 possible numbers that can be pushed. This gives the 10 for the denominator. There are 7 numbers less than 7 (0,1,2,3,4,5,6), so the numerator is 7. Answer: 7/10 Problems #11 - 30: one card is selected from a deck of cards. Find the probability the card selected is: Write your answer as a reduced fraction. (a diagram of a deck of cars is located at the end of this section) The denominator for each of these problems is 52 as there are 52 cards in a deck of cards. 11) a seven The denominator is 52 as there are 52 cards in a deck of cards. There are 4 sevens in a deck of cards, so the numerator is 4. 4/52 Answer: 1/13 13) a heart The denominator is 52 as there are 52 cards in a deck of cards. There are 13 hearts in a deck of cards, so the numerator is 13. 13/52 Answer: 1/4 15) a black card The denominator is 52 as there are 52 cards in a deck of cards. There are 26 black cards (13 clubs and 13 spades) in a deck of cards, so the numerator is 26. 26/52 Answer: 1/2 17) a seven and a red card The denominator is 52 as there are 52 cards in a deck of cards. There are 2 red sevens in a deck of cards (7 of diamonds and 7 of hearts), so the numerator is 2. 2/52 Answer: 1/26 19) a red card and a black card The denominator is 52 as there are 52 cards in a deck of cards. There is no card that is both red and black. This is not possible. The numerator will be 0. 0/52 Answer: 0 21) a black card and a seven The denominator is 52 as there are 52 cards in a deck of cards. There are 2 black sevens in a deck of cards (7 of clubs and 7 of spades), so the numerator is 2. 2/52 Answer: 1/26 23) a diamond and a four The denominator is 52 as there are 52 cards in a deck of cards. The numerator is the 1 card that is both a diamond and a 4 (the 4 of diamonds) Answer: 1/52 25) a queen or a king The denominator is 52 as there are 52 cards in a deck of cards. The word or allows me to count all 4 kings and all 4 queens for the numerator. The numerator must be 8. 8/52 Answer: 2/13 27) a queen and a king The denominator is 52 as there are 52 cards in a deck of cards. There is no card that is both a queen and a king. This is not possible. The numerator will be 0. 0/52 Answer: 0 29) a club or a spade The denominator is 52 as there are 52 cards in a deck of cards. The word or allows me to count all 13 spades and all 13 clubs for the numerator. The numerator must be 26. 26/52 Answer: 1/2 Homework #31 – 34: A dart is thrown at this target and the color it lands on is noted.. Find the requested probabilities. Write your answer as a reduced fraction. 31) Probability that the dart lands in the blue region. Think of the purple region being cut into 2 pieces. I need visualize the graph as if all the regions were the same size. The denominator will be 4, as this graph needs to be divided up into 4 equal size regions. The denominator will be 4. There is one blue region, the numerator will be 1. Answer: 1/4 33) Probability the dart does not land in blue region. The denominator is still the 4 equal regions that the graph must be broken in to. The numerator is the three regions that are not blue. Answer: 3/4 Homework #35-38: The following spinner is spun one time. It is spun again if it lands on a line. Find the requested probabilities. Write your answer as a reduced fraction. 35) Probability that spinner lands in the blue region. I need to break this circle up in to 6 regions. 1 yellow, 1 red and 4 blue. 4/6 Answer: 2/3 37) Probability the spinner does not land in blue region. 2 of the 6 regions are not blue. 2/6 Answer: 1/3 #39 – 42: Each individual letter of the word Mississippi is placed on a piece of paper, and all 11 pieces of paper are placed in a hat. If one letter is selected at random from the hat, find the probability that: Write your answer as a reduced fraction. 39) the letter s is not selected The denominator will be 11 as there are 11 letters in the word Mississippi. The numerator will be 7 (M i i i p p i) Answer: 7/11 41) the letter w is selected There are no w’s so the numerator is 0. 0/11 Answer: 0 #43 – 46: refer to the table which shows the results of a survey regarding the manufacturer of cars driven by the people who were interviewed. men Women total American 260 273 533 Foreign 85 97 182 Total 345 370 715 If one person who completed the survey is selected at random, determine the probability the person selected is: Write your answer as a reduced fraction. 43) A man. The denominator will be 715, which is the total number of people interviewed. The numerator will be the 345 men. 345/715 Answer: 69/143 45) A woman that drives a foreign car. The denominator will be 715, which is the total number of people interviewed. The numerator will be the 97 women that drive foreign cars. Answer: 97/715 #47-50: refer to the table which shows the results of a survey regarding the type of cell phone owned by people who were interviewed. Write your answer as a reduced fraction. men Women total IPhone 150 200 350 Galaxy 100 50 150 Other 125 175 300 Total 375 425 800 If one person who completed the survey is selected at random, determine the probability the person selected is 47) A man The denominator will be 800, which is the total number of people surveyed The numerator will be the 375 men. 375/800 Answer: 15/32 49) Is a woman that doesn’t own an IPhone? The denominator will be 800, which is the total number of people surveyed The numerator will be the 50 women own a Galaxy plus 175 women that own other (50+175 = 225). 225/800 Answer: 9/32