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MDM 4U Data Management and Statistics Probability Test 17 /40 Multiple Choice [12 K/U] Identify the choice that best completes the statement or answers the question. 1. A coin is tossed three times. What is the probability of tossing three heads in succession? 3 1 1 7 a. b. c. d. 8 2 8 8 Answer: c. 2. A coin is tossed three times. What is the probability of tossing exactly two heads? 5 1 1 3 a. b. c. d. 8 2 4 8 Answer: d. 3. If the probability of rain tomorrow is 65%, what is the probability that it will not rain tomorrow? a. 0.65 b. 0.55 c. 0.45 d. 0.35 Answer: d. 4. Two identical spinners each have five equal sectors that are numbered 1 to 5. What is the probability of a total of 7 when you spin both three spinners? 4 7 1 1 a. b. c. d. 25 25 4 5 Answer: a. 5. If the odds in favour of snow tomorrow are 4:7, what is the probability of snow tomorrow? 4 7 4 7 a. b. c. d. 7 11 11 4 Answer: c. 6. If you were to toss three coins, what are the odds against them landing heads up? a. 1:8 b. 8:1 c. 7:8 d. 7:1 Answer: d. 7. What are the odds against randomly picking a green apple from a fruit bowl containing seven green apples and four red ones? a. 7:4 b. 4:11 c. 7:11 d. 4:7 Answer: d. 8. Suppose you simultaneously roll a standard die and spin a spinner that is divied into 10 equal sectors, numbered 1 to 10. What is the probability of getting 4 on both the die and the spinner? 1 1 1 1 a. b. c. d. 4 16 15 60 Answer: d. 9. Suppose you simultaneously roll a standard die and spin a spinner with 8 equal sectors, numbered 1 to 8. What is the probability of both rolling an even number and spinning an odd number? 1 1 1 1 a. b. c. d. 8 16 4 48 Answer: c. 10. If Mike does his mathematics homework today, the probability that he will do it tomorrow is 0.8. The probability that he will do his homework today is 7.0. What are the odds that he will do it both today and tomorrow? a. 4:1 b. 8:7 Answer: d. c. 3:2 d. 14:11 11. When two cards are drawn from a standard deck, what is the probability of drawing a face card or an ace? 4 2 3 5 a. b. c. d. 13 13 13 13 Answer: a. 12. What is the probability of randomly selecting either a club or a nonface card from a standard deck of cards? 1 1 43 10 a. b. c. d. 5 4 52 13 Answer: c. Show your work [12 A] 13. A standard die is rolled. What is the probability of rolling a prime number? [2 A] Answer: 1 2 14. If the odds against the Blue Jays winning this year’s World Series are 20:1, what is the probability that the Blue Jays will win this series? [2 A] Answer: 1 21 15. What are the odds in favour of a total greater than 9 in a given roll of two standard dice? [2A] Answer: 1 : 5 16. If a satellite launch has a 97% chance of success, what is the probability of three consecutive successful launches? [2A] Answer: (0.7)3 ≈ 91.3% 17. The probability that Sergey will play golf today is 60%, the probability that he will play golf tomorrow is 75%, and the probability that he will play golf on both days is 50%. What is the probability that he does not play golf on either day? [2A] Solution Event A = “Sergey will play golf today”, P (A) = 0.6. Event B = “Sergey will play golf tomorrow”, P (B) = 0.75. Event A∩B = “Sergey will play golf on both days”, P (A∩B) = 0.5. P (A∪B) = P (A) + P (B) – P (A∩B) = 0.6 + 0.75 – 0.5 = 0.85. ̅̅̅̅̅̅̅ P (𝐴̅ ∩ 𝐵̅) = P (𝐴 ∪ 𝐵 ) = 1 – P (A∪B) =1 – 0.85 = 0.15. 18. If 28% of the population of Statsville wears contact lenses, 37% have blue eyes, and 9% are blue-eyed people who wear contact lenses, what is the probability that a randomly selected resident has neither blue eyes nor contact lenses? [2A] Solution Event A = “a randomly selected resident wears contact lenses” Event B = “a randomly selected resident has blue eyes” A∩B = “a randomly selected resident wears contact lenses and has blue eyes” P (A) = 0.28, P (B) = 0.37, P (A∩B) = 0.09, P (A∪B) = P (A) + P (B) – P (A∩B) = 0.28 + 0.37 – 0.09 = 0.56. ̅̅̅̅̅̅̅ P (𝐴̅ ∩ 𝐵̅) = P (𝐴 ∪ 𝐵 ) = 1 – P (A∪B) =1 – 0.56 = 0.44. Communication [6C] 19. Explain the difference between the empirical probability, theoretical probability, and subjective probability. [2C] a) Empirical probability Answer The empirical probability, also known as relative frequency, or experimental probability, is the ratio of the number of outcomes in which a specified event occurs to the total number of trials, not in a theoretical sample space but in an actual experiment. For example, Olga tossed a coin 100 times and got 53 tails and 47 heads. The empirical probability of getting a tail equals 0.53. b) Theoretical probability Answer Theoretical Probability of an event is the ratio of the number of favorable outcomes to the total number of equally likely outcomes. For example, probability of getting an even number when rolling a die equals 3 1 P (even) = = 6 2 since the total number of equally likely outcomes equals 6 and the number of favorable outcomes equals 3. c) Subjective probability Answer A probability derived from an individual's personal judgment about whether a specific outcome is likely to occur. Subjective probabilities contain no formal calculations and only reflect the subject's opinions and past experience. For example, Abdul states that the probability that he will pass this Probability Test equals 60%. 20. Explain the meaning of the terms in the probability formula, P (A) 𝑛 (𝐴) = . [2C] 𝑛 (𝑆) Answer P (A) – the probability of an event A 𝑛 (𝐴) 𝑛 (𝑆) – the ratio of numbers of favourable for the event A outcomes and outcomes in the sample space S. 21. The probability that Jacqueline will be elected to the student’s council is 0.6, and the probability that she will be selected to represent her school in a public-speaking contest is 0.75. The probability of Jacqueline achieving both of these goals is 0.5. [2C] a) Are these two goals mutually exclusive? Explain your answer. Answer No, not mutually exclusive. She can be elected to both of these goals. b) What is the probability that Jacqueline is either elected to the student’s council or picked for the public-speaking contest? Note: "either p or q" means "either p or q, but not both" Answer: P (either council or contest) = P (council) + P (contest) – 2 P (council and contest) = = 0.6 + 0.75 – 2(0.5) = 0.35. Thinking [10T] 22. The Royals’ coach stated that “the odds in favour of us winning the next game are 5:7, the odds of tying the next game are 1:3, and the odds of losing the next game are 2:3”. Can the coach’s predictions be correct? Justify your answer. [4T] Answer: The odds in favour of us winning the next game are 5:7, therefore, 5 5 P (winning) = = 5 + 7 12 The odds of tying the next game are 1:3, therefore, 1 1 P (tying) = = 1+3 4 The odds of losing the next game are 2:3, therefore, 2 2 P (losing) = = 2+3 5 P (winning) + P (tying) + P (losing) = = 5 1 + + 2 12 4 5 25+15+24 64 60 16 = = ≠ 1. 60 15 Therefore, the coach’s predictions are not correct. 23. A test for the presence of E. coli in water detects the bacteria 97% of the time when the bacteria is present, but also gives a false positive 2% of the time, wrongly indicating the presence of E. coli in uninfected water. If 10% of the water samples tested contain E. coli, what is the probability that a test result indicating the presence of the bacteria is accurate? [4 T] Solution Event E = “the bacteria E. coli is present in water”. Event D = “The test detects the bacteria”. Given: P (D|E) = 0.97, P (D|𝐸̅ ) = 0.02, P (E) = 0.1 Find: P (E|D) P (𝐸̅ ) = 1 – P (E) = 1 – 0.1 = 0.9 P (D) = P (DE) + P (D𝐸̅ ) = P (D|E) P (E) + P (D|𝐸̅ ) P (𝐸̅ ) = (0.97) (0.1) + (0.02) (0.9) = 0.115 (0.1)(0.97) 𝑃(𝐷𝐸) 𝑃(𝐸)𝑃(𝐷|𝐸) 𝑃(𝐸 |𝐷) = = = ≈ 0.84. 𝑃(𝐷) 𝑃(𝐷) 0.115 24. A study on the effects that listening to loud music through headphones had on teenager’s hearing found that 12% of those teenagers in the sample who did listen to music in this way showed signs of hearing problems. If 60% of the sample reported that they listened to loud music on headphones regularly, and 85% of the sample were found not to have hearing problems, are the events {having hearing problems} and {listening to loud music on headphones} independent? Explain your reasoning. [3T] Answer Event A = {having hearing problems}, Event B = {listening to loud music on headphones} P (A) = 100% – 85% = 15%, P (B) = 60% P (A|B) = 12% P (A|B) ≠ P (A) ⟹ the events A and B are dependent.