Questions 20 – 26 pertain to the table below which describes the smoking habits
ID: 3172988 • Letter: Q
Question
Questions 20 – 26 pertain to the table below which describes the smoking habits of a group of students.
Non Smoker
Occasional Smoker
Regular Smoker
Heavy Smoker
Total
Men
382
37
60
34
513
Women
403
31
74
37
545
Total
785
68
134
71
1058
20. If one person is randomly selected, find the probability that the person would be a heavy smoker.
(a) 0.067 (b) 0.478 (c) 0.484 (d) 0.671
21. If one person is randomly selected, find the probability that the person would be a non-smoker OR a regular smoker?
(a) 0.0868 (b) 0.869 (c) 0.742 (d) 1.868
22. If one person is randomly selected, find the probability that the person would be a non-smoker OR a woman?
(a) 0.382 (b) 0.742 (c) 0.876 (d) 0.515
23. If one person is randomly selected, what is the probability that the person would be a woman, given that a regular smoker is selected?
(a) 0.552 (b) 0.515 (c) 0.127 (d) 0.136
24. If two people are randomly selected, what is the probability that they are both men who are heavy smokers? Assume the selections are made with replacement.
(a) 0.235 (b) 0.005 (c) 0.229 (d) 0.001
25. If two people are randomly selected, what is the probability of getting a woman who is a regular smoker and a man who is a non-smoker? Assume the selections are made with replacement.
(a) 0.025 (b) 0.101 (c) 0.269 (d) 0.614
26. If three people are randomly selected, what is the probability of getting at least one heavy smoker?
(a) .811 (b) .933 (c) .188 (d) .067 (e) .158
27. A student experiences difficulties with malfunctioning alarm clocks. Instead of using 1 alarm clock, he decides to use 3. What is the probability that at least 1 alarm clock works correctly if each individual alarm clock has a 90% chance of working correctly?
(a) 0.729 (b) 0.271 (c) 0.001 (d) 0.999
28. The sample space for tossing 3 coins consists of how many outcomes?
(a) 2 (b) 4 (c) 6 (d) 8 (e) 10
29. Which method of probability is used to estimate the likelihood that interest rates will rise in the next 6 months?
(a) Empirical (b) Subjective (c) Classical (d) relative Frequency
30. In a scientific study there are 8 guinea pigs, 5 of which are pregnant. If 3 are selected at random without replacement, find the probability that all are pregnant.
(a) 3/28 (b) 4/28 (c) 5/28 (d)7/28 (e) .159
31. In a statistics class of 30 students, there were 13 men and 17 women. Two of the men and three of the women received an A in the course. If a student is chosen at random from the class, find the probability that the student is a woman.
(a) .176 (b) .567 (c) .167 (d) .433 (e) .742
32. The probability P(E) for any event E is always between 0 and 1, inclusive.
(a) True (b) False
33. Whi4ch method of probability should be used to determine the likelihood of you retiring at age 50?
(a) Classical (b) Relative Frequency (c) Subjective
34. On an ACT or SAT test, a typical multiple-choice question has 5 possible answers. If you make a random guess on one such question what is the probability that your response is wrong?
(a) .20 (b) 0.25 (c) 0.80 (d) 0.75 (e) 0.50
35. Claudia has gotten A’s in 10 of her 30 college courses. Find the probability that Claudia gets a grade other than an A in a randomly chosen college course.
(a) .3000 (b) .3333 (c) .667 (d) unable to determine
Explanation / Answer
20)probability that the person would be a heavy smoker = number of heavy smokers / total number of persons
= (71)/1058 = 0.0671 OPTION D
21) probability that the person would be a non- smoker or regular smoker =
probability of non-smoker + probabaility of regular smoker (since they both are mutually exclusive categories) = (number of non-smokers / total number of persons) + (number of regular smokers/total number of persons)
= (785)/1058 + (134)/1058 = 0.8686 ~ 0.869 OPTION B
22) probability of woman or non-smoker = probability of women + probability of non - smoker - probability of women and non smoker
As we know that P(A U B) = P(A) +P(B) - P(A B).
probability of woman or non-smoker = (number of women/total number of persons) +(number of non-smokers /total number of persons) - (number of women who are non smokers/total number of persons)
probability of woman or non-smoker = (545)/1058 + (785)/1058 - (403)/1058
= 0.8761 OPTION C
23)probability that the person would be a woman, given that a regular smoker is selected = probability of woman who is a regular smoker/ probability of regular smoker
As we know that P(A|B) = P(AB)/P(B)
probability that the person would be a woman, given that a regular smoker is selected = (74/1058) / (134/1058)
= 0.552 OPTION A
Men Women Non Smoker 382 403 785 Occasional Smoker 37 31 68 Regular Smoker 60 74 134 Heavy Smoker 34 37 71 Total. 513 545 1058Related Questions
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