Practice Quiz with Solutions: Chapter 5 / MATH200 (TC3, Brown)

TC3 → Stan Brown → Statistics → Chapter 5 Practice Quiz

revised 4 Oct 2012 (What’s New?)

Practice Quiz with Solutions: Chapter 5 (50 min)

Disclaimer: This quiz is representative of the level of difficulty you should expect, but it doesn’t include every single topic from the week’s work. The real quiz may include some other topics and may skip some that are in this practice quiz. (The real quiz also may not word questions in the same way as the practice quiz. You should focus on the concepts, not a particular form of words.)

These solutions show about the same level of work I expect from you, though I often add some extra commentary. Please see Show Your Work for the what, why, and how.

1(points: 2) A seed packet is labeled “probability of germination guaranteed 95%”. Give two interpretations of this probability.

Remark: Remember your two interpretations: probability of one = proportion of all.

2(points: 3) You toss three coins.
(a) Construct the sample space of equally likely events.
(b) Find P(2H), the probability of getting exactly two heads.

S = {	`HHH HTH THH TTH`	}
`HHT HTT THT TTT`

Common mistake: Sometimes students write the sample space correctly but miss one of the combinations of 2 heads. I wish I could offer some “magic bullet” for counting correctly, but the only advice I have is just to be really careful.

3(points: 3) P(A), the probability of event A, is 0.7. A and B are complementary events.
(a) Find P(not A).
(b) Find P(B).
(c) Find P(A and B).
If any of the above cannot be determined from the information given, say so.

Solution: (a) P(not A) = 1−P(A) = 1−0.7 → P(not A) = 0.3
(b) That A and B are complementary means that one or the other must happen, but not both. Therefore P(B) = P(not A) → P(B) = 0.3
(c) Since the events are complementary, they can’t both happen: P(A and B) = 0

Common mistake: Many students get (c) wrong, giving an answer of 1. If events are complementary, they can’t both happen at the same time. That means P(A and B) must be 0, the probability of something impossible.

Maybe those students were thinking of P(A or B). It’s true that if A and B are complementary then P(A or B) = P(A) + P(B) = 1. But part (c) was about probability and, not probability or.

4(points: 2) You draw a card at random from a standard 52-card deck. What’s the probability that it’s an ace or a spade?

Solution: “Ace” and “spade” are non-disjoint events, since a given card can be both an ace and a spade. Therefore, you don’t have a formula for this problem. But you can go back to the sample space and compute the probability by counting.

There are thirteen spades in the deck, and three aces in addition to the ace of spades, which you already counted as one of the spades. 13+3 = 16, out of 52 cards. P(Ace or Spade) = 16/52

5(points: 2) In Monopoly, if you roll doubles you get an extra roll, but if you roll doubles three times in a row you go to jail.
(a) On any given turn, what’s the probability you’ll roll doubles? (The picture in the handout may help.)
(b) On any given turn, what’s the probability you’ll go to jail for rolling doubles three times in a row?

Solution: (a) There are 6 ways to roll doubles, and the sample space is 36 equally likely events.

(b) Successive rolls are independent events, so you use the simple multiplication rule:

6(points: 2) At a job fair, 200 students comprised 62 nursing majors, 45 business majors, 12 hospitality, 8 construction, 10 science, and 63 liberal arts. If you randomly select a student, what’s the probability that he or she is majoring in nursing or liberal arts?

Solution: 62+45+12+8+10+63 = 200, so you know that nobody has two majors and therefore all the majors are disjoint events. That lets you use the simple addition rule:

To find those probabilities, you might find it easier to think of them as proportions of the group instead of probabilities for one student.

	Male	Female	Total
	Registrations (thousands)
Democrat	21	28	49
Republican	24	21	45
Green	2	3	5
other	0	1	1
Total	47	53	100

Voter registrations in a fictional New York county are summarized in the table. Use the table to answer these questions about a randomly selected registered voter:

7(points: 2) P(Republican or Female) =

Solution: The events “Republican” and “Female” are not disjoint, so you add 53 thousand females plus the 24 thousand male Republicans, to get 77 thousand who are Republican or female. P(Republican or female) = 77/100 or 77%.

Solution: The denominator again is 100, the total registrations. The numerator is 21, the number of female Republicans.

Common mistake: You don’t need for any kind of formula here. The “and”condition is already covered by the fact that the cell containing 21 is the intersection of the Republican row and the Female column.

Common mistake: Some students try to multiply P(Republican) times P(Female). This is wrong, because the simple multiplication rule applies only when the events are independent. Here the events “Republican” and “Female” are not independent, so the short form of the rule for “and” does not apply.

Solution: There are 53 thousand females, and 21 thousand of them are Republicans, so the probability is 21/53 or about 39.62%

10(points: 4) Tom Turkey invested in two stocks, A and W. There is a 0.90 probability that company A will go bankrupt, and a 0.80 probability that company W will go bankrupt. Assuming the two companies have no connection, find the probabilities that (a) both will go bankrupt; (b) one of them, but not both, will go bankrupt; (c) neither will go bankrupt.

Solution: You’re being asked about all three possibilities: two fail, one fails, none fail. Therefore the three probabilities must add up to 1, and you need to compute only two of them. It’s also important to note that the companies are independent: whether one fails has nothing to do with whether the other fails.

(a) Since the companies are independent, you can use the simple multiplication rule:

At this point you could compute (b), but it’s hard: you need the probability that A fails and W is okay, plus the probability that A is okay and W fails. (c) looks easier, so do that first.

(c) “Neither bankrupt” means both are okay. Again, the events are independent so you can use the simple multiplication rule.

Solution: If you have time, it’s always good to check your work using a different method. You can work out (b) the long way. You have only independent events (whether A is okay or fails, whether W is okay or fails) and disjoint events (A fails and W okay, A okay and W fails). The “okay” probabilities were computed in part (c).

Common mistake: A very common mistake is to solve only half the problem when working this out the long way. When you have probability of one not the other, you have to consider both A-and-not-W and W-and-not-A.

Remark: If you computed all three probabilities the long way, pause a moment to check your work by adding them to make sure you get 1. Whenever possible, check your work with a second type of computation.

11(points: 2) A poll found that 45% of baseball fans had attended a game in person within the past year. Of five randomly selected baseball fans, find the probability that at least one had not attended a game within the past year.

Solution: In “at least” and “no more than” probability problems, the complement is often your friend. The complement of “at least one had not attended” is “all had attended”. If the fans are randomly selected, their opinions are independent and you can use the simple multiplication rule.

12(points: 2) A second poll found that 86% of baseball fans favored drug testing for baseball players. What’s the probability that a given baseball fan had attended a game within the past year and favored drug testing?

Solution: This is a probability “and”, so you need independent events. But you don’t have any basis to think that fans who attend games in person have the same opinions as fans who do not. In other words, you don’t know that “attended a game within the past year” and “favors drug testing” are independent events. Answer: events may not be independent; can’t solve

13(points: 3) In 2003 a federal government survey estimated that 58.2% of US households had both a cell phone and a landline, 2.8% had only cell service, and 1.6% had no phone service at all.
(a) Construct a probability model for type of phone service to US households.
(b) Polling agencies try (in theory) not to call cell phones, because consumers object to paying for the calls. What proportion of US households could be reached by a landline in 2003?

Service type	Prob.
Landline only	37.4%
Landline and cell	58.2%
Cell only	2.8%
No phone	1.6%
Total	100.0%

Solution: (a) In a probability model, the probabilities must add to 1 (= 100%). The given probabilities add to 62.6%. What is the missing 37.4%? They’ve accounted for cell and landline, cell only, and nothing; the remaining possibility is landline only. The model is shown at right.

Remark: “Landline” and “cell” are not disjoint events, because a given household could have both. But “landline only” and “landline and cell” are disjoint, because a given house can’t both have a cell phone with landline and have no cell phone with landline.

What’s New

21 Jun 2012: Rule out double majors here and here; thanks to Lotte Hammond for pointing out the ambiguity. Also, simplify the explanation of the recording devices (and the entire problem would be removed four months later).

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