Math 300 Exam 4 Review (Chapter 11) Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Give the probability that the spinner shown would land on the indicated color. 1) black 1) A) 1 3 B) 2 3 C) 1 6 D) 1 2 Comment: 2/6 = 1/3 Find the probability. 2) Two 6-sided dice are rolled. What is the probability that the two numbers obtained differ by more than 2? A) 11 36 B) 1 3 C) 1 4 D) 13 36 Answer: B Comment: We know there are 36 possible outcomes when rolling a pair of dice. The outcomes in which the two numbers obtained differ by more than 2 are: (1,4), (1,5), (1,6), (2,5), (2,6), (3,6), (4,1), (5,1),(5,2),(6,1),(6,2), (6,3). Thus 12 of the 36 outcomes are in the desired event, so the probability is 12/36 = 1/3 2) Solve the problem. 3) 3) What are the odds in favor of drawing an even number from these cards? A) 5:2 B) 2:5 C)2:3 D) 3:2 Comment: 4) The table shows the number of college students who prefer a given pizza topping. 4) toppings freshman sophomore junior senior cheese 16 16 21 28 meat 24 28 16 16 veggie 16 16 24 28 Find the empirical probability that a randomly selected student prefers cheese toppings. A) 0.325 B) 0.337 C) 0.112 D) 0.346 Comment: There are 81 students who prefer cheese. There are 249 total students. Therefore, the probability that a randomly selected student prefers cheese is 81/249 0.325
5) Two distinct even numbers are selected at random from the first ten even numbers greater than zero. What is the probability that the sum is 30? 5) A) 1 45 B) 1 10 C) 2 45 D) 1 15 Answer: D Comment: Using combinations, the number of pairs of even numbers (selected from the first 10 positive even numbers) is 10 C 2 = 45. The pairs whose sum is 30 are: (20, 10), (18, 12), and (16, 14). Thus the probability is 3/45 = 1/15. We could also use permutations (if we wanted order to matter). Then the number of ordered pairs possible would be 10 P 2 = 90, and the ordered pairs summing 30 would be (10, 20), (12, 18), (14, 16), (16, 14), (18, 12), and (20, 10). Thus the probability is 6/90 = 1/15 6) Mr. Larsen's third grade class has 22 students, 12 girls and 10 boys. Two students must be selected at random to be in the spring play. What is the probability that one boy and one girl will be chosen? Order is not important. A) 40 B) 1 C) 5 D) 2 77 11 6 11 Comment: The number of combinations with one boy and one girl is 12 C 1 10 C 1 = 120. The number of possible combinations of two students chosen from the 22-student class is 22 C 2 = 231. Thus the probability is 120/231 = 40/77. 6) Find the probability. 7) When two balanced dice are rolled, there are 36 possible outcomes. Find the probability that the sum is a multiple of 3 or greater than 4. A) 8 9 B) 11 12 C) 13 36 Comment: Multiples of 3: (1,2), (2,1), (1,5), (2,4), (3,3), (4,2), (5,1), (3,6), (4,5), (5,4), (6,3), (6,6) Not greater than 4: (1,1), (1,2), (1,3), (2,1), (2,2), (3,1). Thus there are 30 greater than 4. Notice that 10 of these 30 are also multiples of 3. P(3k or >4) = P(3k) + P(>4) - P(3k and >4) = 12/36 + 30/36-10/36 = 32/36 = 8/9 D) 5 6 7)
Find the indicated probability. 8) The table shows the distribution of family size in a certain U.S. city 8) Family Size Probability 2 0.451 3 0.243 4 0.190 5 0.071 6 0.027 7+ 0.018 A family is selected at random from the city. Find the probability that the size of the family is less than 5. Round approximations to three decimal places. A) 0.884 B) 0.433 C) 0.504 D) 0.071 Comment: P(<5) = P(2 or 3 or 4) = P(2) + P(3) + P(4) = 0.451 + 0.243 + 0.190 = 0.884 9) The distribution of B.A. degrees conferred by a local college is listed below, by major. 9) Major Frequency English 2073 Mathematics 2164 Chemistry 318 Physics 856 Liberal Arts 1358 Business 1676 Engineering 868 9313 What is the probability that a randomly selected degree is not in Mathematics? A) 0.768 B) 0.682 C) 0.303 D) 0.232 Comment: P(not Math) = 1-P(Math) = 1-2164 9313 0.768 10) The table below describes the smoking habits of a group of asthma sufferers. 10) Light Heavy Nonsmoker smoker smoker Total Men 366 80 60 506 Women 360 73 78 511 Total 726 153 138 1017 If one of the 1017 subjects is randomly selected, find the probability that the person chosen is a nonsmoker given that the person is a woman. A) 0.496 B) 0.705 C) 0.354 D) 0.502 Answer: B Comment: P(N W) = P(N and W) P(W) = n(n W) n(w) = 360 511 0.705
Find the probability. 11) In one town, 70% of adults have health insurance. What is the probability that 8 adults selected at random from the town all have health insurance? A) 0.114 B) 0.7 C) 0.058 D) 5.6 Comment: (0.7)8 0.058 Use the general multiplication rule to find the indicated probability. 12) An IRS auditor randomly selects 3 tax returns from 55 returns of which 10 contain errors. What is the probability that she selects none of those containing errors? A) 0.5477 B) 0.006 C) 0.5409 D) 0.0046 Comment: There are 45 C 3 ways to select 3 returns containing no errors., and there are 55 C 3 ways to 11) 12) select 3 returns. Thus the probablility of selecting 3 with no errors is 45C 3 55 C 3 0.5409 Find the conditional probability. 13) If two cards are drawn at random without replacement from a standard deck, find the probability that the second card is a spade, given that the first card was a spade. A) 1 4 B) 3 13 C) 4 17 D) 11 51 Comment: If the first card was a spade, then there are 12 spades left in the remaining 51 cards. Thus the probability that the second card is a spade is 12 51 = 4 17 13) Solve the problem. 14) A basket contains 6 oranges and 4 tangerines. A sample of 3 is drawn. Find the probability that they are all oranges. 14) A) 1 5 B) 1 3 C) 1 6 D) 4 9 Comment: 6 10 5 9 4 8 = 3 5 5 9 1 2 = 1 6 Use counting rules to determine the probability. 15) Determine the probability that in a class of 8 students, at least two students have the same birthday. Assume that there are always 365 days in a year and that birth rates are constant throughout the year. (Hint: First determine the probability that no two students have the same birthday and then apply the complementation rule.) A) 0.926 B) 0.154 C) 0.074 D) 0.114 Comment: The probability that no students have the same birthday is 365 364 365 363 365 362 365 361 365 360 365 359 365 358 0.926 Thus the probability that at least 2 students have the same birthday is 1-0.926 = 0.074 15)
Find the probability. 16) Find the probability that when a gardener plants 20 seeds, she harvests 16 radishes given the probability that a radish seed will germinate is 0.7. A) 0.075 B) 0.130 C) 0.571 D) 0.068 Answer: B Comment: This is the probability of 16 "successes" out of 20 "trials." We use the binomial probability formula with n = 20, p = 0.7, q = 0.3. P(16) = 20 C 16 (0.7)16(0.3)4 = 0.130 17) A test consists of 10 true/false questions. To pass the test a student must answer at least 7 questions correctly. If a student guesses on each question, what is the probability that the student will pass the test? A) 0.117 B) 0.055 C) 0.172 D) 0.945 Comment: P(at least 7) = P(7 or 8 or 9 or 10) = P(7) + P(8) + P(9) + P(10) = 10 C 7 (0.5)7(0.5)3 + 10 C 8 (0.5)8(0.5)2 + 10 C 9 (0.5)9(0.5)1 + 10 C 10 (0.5)10 0.172 16) 17) Solve the problem. 18) An ice cream store has 5 flavors. If we pick flavors successively at random, what is the probability that the flavor strawberry will be selected for the first time on pick 3? [the same flavor can be picked more than once] A) 0.032000 B) 0.640000 C) 0.128000 D) None of the above is correct. Comment: This is the probability that the first and second picks are not strawberry, and the third is strawberry. This is 4 5 4 5 1 5 = 16 125 = 0.128 19) If 5 apples in a barrel of 25 apples are rotten, what is the expected number of rotten apples in a random sample of 2 apples? 18) 19) A) 3 5 B) 1 C) 2 5 D) 4 5 Comment: Let x be the number of rotten apples picked. See the table of the probability distribution below: x P(x) x P(x) 0 20 24 0 1 25 20 5 24 +20 25 5 24 2 5 25 4 24 1 3 1 15 Adding the entries in the product column gives 2/5.
20) A certain game involves tossing 3 fair coins. It pays 28 cents for 3 heads, 14 cents for 2 heads, and 6 cents for 1 head. What is a fair price to pay to play this game? A) 16 cents B) 9 cents C)10 cents D) 11 cents Answer: D Comment: The "fair" price is equal to the expected winnings. Let x = number of heads 20) x P(x) Winnings Product 0 1/8 $0 0 1 3/8 $0.06 0.0225 2 3/8 $0.14 0.0525 3 1/8 $0.28 0.035 Adding the product column gives the fair price, which is $0.11.