Practice Quiz with Solutions: Chapter 5

Before looking at these solutions, please work the practice quiz.

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.)

See also:  How to Take a Math Test

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

Solution: There are 26 letters in the alphabet, but you are told not to use I, O, or Q, so 23 letters are available. Three of those letters, without repetition, give 23×22×21 = 10,626 combinations. There are ten ways to choose the first digit, nine ways to choose the second, and so on:

ways to choose three letters = 23×22×21 = 10,626

ways to choose three letters and one digit = 10,626×10 = 106,260

ways to choose three letters and two digits = 106,260×9 = 956,340

ways to choose three letters and three digits = 956,340×8 = 7,650,720

ways to choose three letters and four digits = 7,650,720×7 = 53,555,040

Therefore three digits aren’t enough, and four digits are required.

Solution:  (a)

(b) Three events out of eight equally likely events: P(2H) = 3/8

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.

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.

Solution:  (a) When you have duplicates, to construct the sample space you need subscripts. The sample space for drawing just one cookie would not be {M, C} because chocolate is more likely than molasses. Instead, it’s {M, C₁, C₂, C₃}. From that, you can easily write the sample space for drawing two cookies without replacement:

(b) Since there’s only one molasses, “two of the same type” must mean two chocolate. Counting, there are six possibilities out of the 12-event sample space; P(2 same type) = 6/12 or ½

Common mistake: Students often have trouble writing the sample space correctly. The most common mistake is forgetting that order matters, so you need both MC₁ and C₁M.

Answer: Some possibilities are rolling two dice and getting doubles or not; rolling two dice and getting a 7 or not; rolling one die and getting a 3 or not; drawing one card and getting a spade or not; drawing one card and getting a 2–9 or a 10-J-K-Q-A

Common mistake: Did you list only one event, or two events that weren’t complementary, or more than two events? A tip for test taking: always reread the question when you think you’ve answered it, and make sure that you actually have. For more test-taking tips, see How to Take a Math Test.

Solution: A (slightly) clever shortcut is to realize that the opposite of “the program is recorded” is “both recordings fail”. Remember, the complement is your friend when you suspect that a probability computation is more complicated than it needs to be.

The success or failure of the Tivo and VCR are independent, so you can use the simple multiplication rule for probability “and”:

P(both fail) = P(Tivo fails) × P(VCR fails) = 0.2 × 0.3 = 0.06

P(one or both records) = 1 − P(both fail) = 1−0.06 = 0.94 or 94%

Alternative solution: To get your program, one or both of the devices must record it. The events are not mutually exclusive, so you must use the long form of the addition rule for “or”:

P(Tivo records or VCR records) = P(Tivo records) + P(VCR records) − P(Tivo records and VCR records)

To find the first two probabilities, use the complement rule:

P(Tivo fails) = 0.2; therefore P(Tivo records) = 1−0.2 = 0.8

P(VCR fails) = 0.3; therefore P(VCR records) = 1−0.3 = 0.7

To find the third probability, P(Tivo records and VCR records), you must realize that the two events are independent, so you can use the short rule for probability “and”:

P(Tivo records and VCR records) = P(Tivo records) × P(VCR records) = 0.8×0.7 = 0.56

Now you can put together the full computation:

P(Tivo records or VCR records) = P(Tivo records) + P(VCR records) − P(Tivo records and VCR records)

P(Tivo records or VCR records) = 0.8 + 0.7 − 0.56 = 0.94

Solution:  The denominator is 100, the total of all people. The numerator is 45 (Republicans) plus 28+3+1 (Females of other parties).

p = (45+28+3+1)/100 = 77/100 = 0.77 or 77%

Alternative solution: You could also compute the numerator as 53 (Females) plus 24 (Male Republicans). p = (53+24)/100 = 0.77 or 77%

Alternative solution: You could even use the general formula for “or” probabilities:

P(Republican or Female) = P(Republican) + P(Female) − P(Republican and Female)

p = 45/100 + 53/100 − 21/100 = 77/100 = 0.77 or 77%

Common mistake: You can’t simply add the Females (53) and Republicans (45) to get 98/100. That would count the 21 (21,000) Republican Females twice.

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

P(Republican and Female) = 21/100 = 0.21 or 21%

Common mistake: There is no 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:  P(Republican | Female), “probability of Republican given Female”, is concerned only with Republicans in the Female column, not with Republicans in the whole table. The denominator is 53 (all Females, the “given that” condition that follows the vertical bar “|”), and the numerator is 21 (Republican Females).

p = 21/53 ≈ .396 or 39.6%

Common mistake: The order matters with probability “given that”. If you computed 21/45, you were finding P(Female | Republican), which is different from what was asked.

Solution: Set up your sequences:

(A bankrupt) and (W okay) or (A okay) and (W bankrupt)

.9 × (1 – 0.8) + (1 – 0.9) × 0.8

.9 × (0.2) + (0.1) × 0.8

0.26

Solution: There are two ways to get one of each flavor: cherry then orange, and orange then cherry. Those two are mutually exclusive, so you can add their probabilities once you compute them:

P(one of each) = P(cherry first and orange second) + P(orange first and cherry second)

Now, how are those two probabilities to be computed? Well, Grace is choosing the sourballs without replacement (one would hope!), so the probabilities are not independent and you must use the long form of the rule for probability “and”:

P(cherry first and orange second) = P(cherry first) × P(orange second | cherry first)

There are 20 sourballs to begin with, so P(cherry first) = 11/20. If Grace’s first sourball is a cherry, then there are 9 orange left out of 19 (not 20), so P(orange second | cherry first) = 9/19. Therefore

P(cherry first and orange second) = P(cherry first) × P(orange second | cherry first)

P(cherry first and orange second) = 11/20 × 9/19 = 99/380

In a similar way, compute the other probability that is needed:

P(orange first and cherry second) = P(orange first) × P(cherry second | orange first)

P(orange first and cherry second) = 9/20 × 11/19 = 99/380

(In this particular case the two probabilities are the same, but in more complex problems that may not be true.) Putting all the probabilities together,

P(one of each) = P(cherry first and orange second) + P(orange first and cherry second)

P(one of each) = 99/380 + 99/380 = 198/380 ≈ 0.521 or 52.1%

Alternative solution: Again, you can use the complement as a possibly shorter solution, or to check your solution. The complement of “one of each” is “two the same”:

P(one of each) = 1 − P(both the same)

There are two ways to get both the same: two cherry and two orange. Since those possibilities are mutually exclusive, the probabilities simply add:

P(both the same) = P(two cherry) + P(two orange)

P(both the same) = (11/20)×(10/19) + (9/20)×(8/19)

P(both the same) = 110/380 + 72/380 = 182/380

Remember, that’s the complement of the actual event you’re interested in:

P(one of each) = 1 − P(both the same)

P(one of each) = 1 − 182/380 = 198/380 as before.