Practice 3-5/3-7 Answers
Cholesterol
1) The probabilities that an adult American man has high blood pressure and/or high cholesterol are
shown in the table.
Blood Pressure
High
High
OK
OK
0.11
0.21
0.16
0.52
a. What’s the probability that a man has both conditions?
P(High Cholesterol and High Blood Pressure) = 0.11
b. What’s the probability that he has high blood pressure?
P(High Blood Pressure) = 0.11 + 0.16 = 0.27
c. What’s the probability that a man with high blood pressure has cholesterol?
P( HC and HP) 0.11

 0.4074
P(High Cholesterol | High Blood Pressure) =
P( HP)
0.27
d. What’s the probability that a man has high blood pressure if it’s known that he has high
cholesterol?
P( HC and HP) 0.11

 0.344
P(High Blood Pressure | High Cholesterol) =
P( HC )
0.32
2) USA Today (June 6, 2000) gave information on seat belt usage by gender. Assume that these
proportions are representative of adults in the United States and that a US adult is selected at
random. The proportions in the following table are based on a survey of large number of adult men
and women in the United States:
Uses seat belts regularly
Does not use seat belts regularly
Male
0.10
0.40
Female
0.175
0.324
a. What is the probability that the selected adult regularly uses a seat belt?
P(Adult regularly uses seat belt) = P(Male) + P(Female) = 0.10 + 0.175 = 0.275
b. What is the probability that the selected adult regularly uses a seat belt given that the
individual selected is male?
P( S and M ) 0.10

 0.20
P(Adult regularly uses seat belt | male) =
P( M )
0.50
1
c. What is the probability that the selected adult does not use a seat belt regularly gives that the
selected individual is female?
P( NS and F ) 0.324

 0.65
P(Adult does not use seat belt | female) =
P( F )
0.499
d. What is the probability that the selected individual is female given that the selected individual
does not use a seat belt regularly?
P( F and NB) 0.324

 0.448
P(Female | adult does not use seat belt) =
P( NB)
0.724
3) Jeanie is a bit forgetful, and if she doesn’t make a “to do” list, the probability that she forgets
something she is supposed to do is 0.1. Tomorrow she intends to run three errands, and she fails
to write them on her list.
a. What is the probability that Jeanie forgets all three errands?
P(Forgets all 3) =  0.1  0.001
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b. What is the probability that Jeanie remembers at least one of the three errands?
P(Remembers at least 1) = 1 – P(Forgets all 3) = 1 – 0.001 = 0.99
c. What is the probability that Jeanie remembers the first errand but not the second and third?
P(Remembers 1st, Forgets 2nd and Forgets 3rd) = P( R)  P( F )  P( F )  (0.99)(0.1)(0.1)  0.0099
Gender
4) Two psychologists surveyed 478 children in grades 4, 5, and 6 in elementary schools in Michigan.
They stratified their sample, drawing roughly 1/3 from rural 1/3 from suburban and 1/3 from urban
schools. Among other questions, they asked the students whether their primary goal was to get
good grades, to be popular or to be good at sports. One question of interest was whether boys and
girls at this age had similar goals. Below is table giving counts of the students by their goals and
gender.
Boy
Grades
117
Girl
130
Total
247
Goals
Popular
50
Sports
60
Total
227
91
30
251
141
90
478
a. What is the probability of selecting a girl?
251
P(girl) =
 0.525
478
b. What is the probability of selecting a girl whose goal is to be popular?
91
P(girl and popular) =
 0.190
478
2
c. What is the probability of selecting a student whose goal is to excel at sports?
90
P(sports) =
 0.188
478
d. What is the probability of selecting a student whose goal is to excel at sports given that we
have selected a girl?
30
P( S and G) 478 30
P(sports | girl) =


 0.120
251 251
P(G)
478
e. What is the probability of selecting a girl given that the selected student’s goal is popularity?
91
P(G and P) 478 91
P(girl | popularity) =


 0.645
141 141
P( P)
478
5) You bought a new set of four tires from a manufacturer who just announced a recall because 2% of
those are defective. What is the probability that at least one of yours is defective?
P(At Least One is Defective) = 1 – P(Not Defective)
 1  (0.98) 4  0.078
6) Suppose that E and F are two events and that P(E and F) = 0.24, P(E) = 0.4 and P(F) = 0.6. Are E and
F independent? Why? Hint: Is P(E|F) = P(E)
P( E and F) 0.24
P( E | F ) 

 0.40
P( F )
0.60
P( E | F )  P( E )
0.40  0.40
Yes they are independent
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7) For men, binge drinking is defined as having five or more drinks in a row, and for women as
having four or more drinks in row. (The difference is due to the average difference in weight.)
According to a study by the Harvard School of Public Health (H. Wechsler, G.W. Dowdall, A.
Davenport and W. Dejong, “Binge Drinking on Campus: Results of a National Study”), 44% of
college students engage in binge drinking, 37% drink moderately, and 19% abstain entirely.
Another study, published in the American Journal of Health Behavior, finds that among binge
drinkers ages 21 to 34, 17% have been involved in an alcohol-related automobile accident, while
among non-bingers of the same age, only 9% have been involved in such accidents.
a. Make a tree diagram of students drinking habits. Add in the probabilities too. (Hint: Binge,
Moderate and Abstain)
b. Extend the tree diagram by adding in the car accidents.
SKIP
8) Suppose 40 cars start at the Indianapolis 500. In how many ways can the top three cars finish the
race?
40P3 = 59,280
9) How many different simple random of size 5 can be obtained from a population whose size is 50?
50C5 =
2118760
SKIP
10) How many different 10-letter words (real or imaginary) can be formed from the letters in the word
STATISTICS?
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