AP Statistics
Unit 5 review
1. The TV weatherman says, “There’s a 30% chance of rain tomorrow.” Explain what this statement means.
When the weather conditions are like those seen today, it has rained on the following day about 30% of
the time.
2. From police records, it has been determined that 15% of drivers stopped for routine license checks are not
wearing seatbelts? If a police officer stops 10 vehicles, how likely is it that two consecutive drivers won’t
be wearing their seat belts? (This problem makes more sense when all sentences are typed)
A. Describe the design of a simulation to estimate this probability. Explain clearly how you will use the
partial table of random digits below to carry out your simulation.
Let the numbers 01 – 15 represent drivers who do not have their seat belts on and 16 – 99 and 00
represent drivers who do have their seat belts on. Using the table of random digits, read 10 sets of 2digits. We do not need to discard repeated numbers because they are not referring to specific
individuals. For each set of 10 2-digit numbers, record whether there are two consecutive numbers
between 01 – 15 or not.
B. Carry out three repetitions of the simulation. Copy the random digits below onto your paper. Then
mark on or directly above the table to show your results.
The first sample is 29/07/7 1/48/63/ 61/68/3 4/70/52 In this sample, there was only one not
wearing a seat belt, so there were not two consecutive people not wearing their seat belts.
The second sample is 62/22/4 5/10/25/95/05/2 9/09/08 In this sample, there were two
consecutive drivers not wearing their seatbelts.
The third sample is 73/59/2 7/51/86/ 87/13/6 9/57/61 In this sample, there were not two
consecutive drivers not wearing their seat belts.
Based on these three simulations, we’d suggest that the probability of finding two consecutive people
out of a sample of 10 not wearing their seat belts to be about 0.33.
27102 56027 55892 33063 41842 81868
43367 49497 72719 96758 27611 91596
3. Eight teams of the world’s best sand sculptors gathered for an “extreme” competition. At five points
during the competition, a randomly selected team’s sculpture was blown up. That team was then forced
to build a new sculpture from scratch. No team could be selected for destruction more than once. Last
year’s winning team did not have a sculpture blown up either last year or this year. Other teams were
suspicious. Should they be? Design and carry out a simulation to answer this question.
Assign each of the eight teams a single-digit number. Now use technology or random digit table to pick 5
different random numbers between 1 and 8. These are the teams that had their sculpture blown up last
year. Pick 5 more numbers between 1 and 8 for teams to have their sculpture blown up this year. Record
whether a number was not chosen in two consecutive years. Repeat many times.
I used the rest of the random digit table from above.
First sample: 2/7/1/0/2/ 5/6 for the first year and 0/2/7/ 5/5/8/9/2/ 3 for the second year. Team 4
was not chosen either year.
Second sample: 3/0/6/3/ 4/1/8 for the first year and 4/2/ 8/1/8/6 for the second year. Team 5 and 7
were not chosen either year.
You should repeat this simulation many times, but with just these 2 simulations, it seems likely that a
particular team might not be chosen two consecutive years.
4. Nonstandard dice can produce interesting distributions of outcomes. Suppose you have two balanced, sixsided dice. Die A has faces with 2, 2, 2, 2, 6, and 6 spots. Die B has three faces with 5 spots and three
faces with 1 spot. Imagine that you roll both dice at the same time.
A. Find a probability model for the difference (Die A – Die B) in the total number of spots on the up-faces.
Die A →
Die B ↓
5
5
5
1
1
1
2
2
2
2
6
6
-3
-3
-3
1
1
1
-3
-3
-3
1
1
1
-3
-3
-3
1
1
1
-3
-3
-3
1
1
1
1
1
1
5
5
5
1
1
1
5
5
5
The table above gives all possible differences. Therefore the probability model is:
X = Die A – Die B
Probability
-3
12/36 = 1/3
1
18/36 = 1/2
5
6/36 = 1/6
B. Which die is more likely to roll a higher number? Justify your answer.
Die A is more likely to be higher. If Die A – Die B is positive, Die A is higher than Die B. The probability
that Die A is higher is ½ + 1/6 = 2/3
5. The Census Bureau allows each person to choose from a long list of races. That is, in the eyes of the
Census Bureau, you belong to whatever race you say you belong to. Hispanic (also called Latino) is a
separate category. Hispanics may be of any race. If we choose a resident of the United States at random,
the Census Bureau gives these probabilities.
Not
Hispanic
Hispanic
Asian
0.001
0.044
Black
0.006
0.124
White
0.139
0.674
Other
0.003
0.009
A. Verify that this is a legitimate assignment of probabilities.
It is legitimate because every person must fall into exactly one category, the probabilities are all
between 0 and1, and they add up to 1.
B. What is the probability that a randomly chosen American is Hispanic?
The probability that a randomly chosen person is Hispanic is 0.0001 + 0.0006 + 0.139 + 0.003 = 0.149.
C. Non-Hispanic whites are the historical majority in the United States. What is the probability that a
randomly chosen American is not a member of this group?
The probability that a randomly chosen person is not a Non-Hispanic white is 1 – 0.674 = 0.326
D. Explain why P(white or Hispanic) ≠ P(white) + P(Hispanic). Then find P(white or Hispanic).
Being white and being Hispanic are not mutually exclusive events. So
P(white or Hispanic) = P(white) + P(Hispanic) – P(White and Hispanic) = 0.813 + 0.149 – 0.139 = 0.823
6. Ramon has applied to both Princeton and Stanford. According to his counselor, the probability that
Princeton will admit him is 0.4, the probability that Stanford will admit him is 0.5, and the probability that
both will admit him is 0.2.
A. Make a Venn diagram to model this chance process.
Princeton
0.2
Stanford
0.2
0.3
0.3
B. What is the probability that neither university admits Ramon?
The probability that neither admits him is 0.3.
C. What’s the probability that he gets into at least one of the two schools? Use the general addition rule
to confirm that your answer is correct.
The probability that at least one admits him is 0.7. We can see this from the general addition rule .
P  P  S   P  P   P  S   P  P  S   0.4  0.5  0.2  0.7
7. A company has developed a drug test to detect steroid use by athletes. The test is accurate 95% of the
time when an athlete has taken steroids. It is 97% accurate when an athlete hasn’t taken steroids.
Suppose that the drug test will be used in a population of athletes in which 10% have actually taken
steroids. Let’s choose an athlete at random and administer the drug test.
A. Make a tree diagram showing the sample space of this chance process.
Test positive
Steroids
Test negative
Athlete
Test positive
No Steroids
Test negative
B. What’s the probability that the randomly selected athlete tests positive? Show your work.
Add the branches that end in a positive test. P  positive   0.095  0.027  0.122.
C. Suppose that the chosen athlete tests positive. What’s the probability that he or she actually used
steroids? Show your work.
P  actually used steroids | positive  
P  actually used steroids and positive  0.095

 0.7787.
P  positive 
0.122
8. You work at Mike’s pizza shop. You have the following information about the 7 pizzas in the oven: 3 of the
7 have thick crust, and of these 1 has only sausage and 2 have only mushrooms. The remaining 4 pizzas
have regular crust, and of these 2 have only sausage and 2 have only mushrooms. Choose a pizza at
random from the oven.
A. Are the events {getting a thick-crust pizza} and {getting a pizza with mushrooms} independent?
Explain.
4
These events are not independent, because P  pizza with mushrooms   , but
7
2
P  pizza with mushrooms | thick crust   . If the events had been independent, these probabilities
3
would have been equal.
B. You add an eighth pizza to the oven. This pizza has thick crust with only cheese. Now are the events
{getting a thick-crust pizza} and {getting a pizza with mushrooms} independent? Explain.
4 1
With the eighth pizza P  pizza with mushrooms    , and
8 2
2 1
P  pizza with mushrooms | thick crust    , so these events are independent.
4 2
9. As suburban gardeners know, deer will eat almost anything green. In a study of pine seedlings at an
environmental center in Ohio, researchers noted how deer damage varied with how much of the seedling
was covered by thorny undergrowth.
Deer Damage
Thorny Cover
Yes
No
None
60
151
< 1/3
76
158
1/3 to 2/3
44
177
> 2/3
29
176
A. What is the probability that a randomly selected seedling was damaged by deer?
There are a total of 871 pine seedlings, of which 209 had damage from deer, so the probability of a
209
 0.24.
randomly selected pine seedling having damage by deer is
871
B. What are the conditional probabilities that a randomly selected seedling was damaged, given each
level of cover?
P  damage | no cover  
60
1  76

 0.2844, P  damage | <  
 0.3248,
211
3  234

1
2  44
2  29


P  damage | to  
 0.1991, and P  damage | >  
 0.1415.
3
3  221
3  205


C. Does knowing about the amount of thorny cover on a seedling change the probability of deer damage?
Justify your answer.
Yes, knowing the amount of cover does change the probability of deer damage. It appears that deer
do much more damage when there is no cover or less than 1/3 cover and much less damage when
there is more cover than that.
10. The “random walk” theory of stock prices holds that price movements in disjoint time periods are
independent of each other. Suppose that we record only whether the price is up or down each year, and
that the probability that our portfolio rises in price in any one year is 0.65. This probability is
approximately correct for a portfolio containing equal dollar amounts of all common stocks listed on the
New York Stock Exchange.)
A. What is the probability that our portfolio goes up for three consecutive years?
P  up three consecutive years    P  up one year    0.65  0.274625.
3
3
B. What is the probability that the portfolio’s value moves in the same direction (either up or down) for
three consecutive years?
P  same diretion for 3 years   P  up 3 years   P  down 3 years    0.65   0.65
3
3
 0.2746  0.0429  0.3175.
11. Each of us has an ABO blood type, which describes whether two characteristics called A and B are present.
Every human being has two blood type alleles (gene forms), one inherited from our mother and one from
our father. Each of these alleles can be A, B, or O. Which two we inherit determines our blood type. The
table shows what our blood type is for each combination of two alleles. We inherit each of a parent’s two
alleles with probability 0.5. We inherit independently from our mother and father.
Alleles inherited
Blood type
A and B
A
A and B
AB
A and O
A
B and B
B
B and O
B
O and O
O
A. Hannah and Jacob both have alleles A and B. Diagram the sample space that shows the alleles that
their next child could receive. Then give the possible blood types that this child could have, along with
the probability for each blood type.
A from father
A from mother
B from father
Child
A from father
B from mother
B from father
The possible blood types and their probabilities are:
Blood type
A
AB
B
Probability
0.25
0.5
0.25
B. Jennifer has alleles A and O. Jose has alleles A and B. They have two children. What is the probability
that at least one of the two children has blood type B? Show your method
A from father
A from mother
B from father
Child
A from father
O from mother
B from father
The possible blood types and their probabilities are:
Blood type
A
AB
B
Probability
0.5
0.25
0.25
So the probability that at least one of two children are type B is 1   0.75  0.4375
2