MATH 2280 PRACTICE TEST 2 ANSWERS
Name:__________________________
1. Maggie watched 100 cars drive by her window and compiled the following data:
Model Number
Ford
23
Toyota
25
GM
18
Chrysler
17
Honda
17
What is the empirical probability that the next car is
(a) A Ford?
‘Empirical probability’ is what you would expect the probability to be based on past
events.
23
100
(b) Not a Toyota?
75 cars were not Toyotas, so P(not a Toyota) =
75 3

100 4
You could also say P(not a Toyota) = 1 – P(is a Toyota) = 1 
25
75 3


100 100 4
(c) A Chrysler or a GM?
18  17 35
7


100
100 20
(d) A Rolls Royce?
Based on past observations it is not possible for the next car to be a Rolls Royce.
0
2. In a class of 100 students, 75 take statistics, 21 take calculus and 13 take both subjects.
What is the probability that a randomly selected student takes statistics or calculus?
Use P( S or C )  P( S )  P(C )  P( S and C )
P( S or C )  P( S )  P(C )  P( S and C )
75 21 13



100 100 100
83

 .83
100
3. In an English class there 12 juniors and 15 seniors; 7 of the juniors are males and 5 of
the seniors are females. If a student is selected at random, find the probability of
selecting the following:
(a) A senior or a female
It can be helpful to put the information into a chart.
Male Female
Junior
7
5
12
Senior
10
17
5
10
15
27
P(senior or female) = P(senior) + P(female) – P(senior and female)
15 10 5 20




27 27 27 27
(b) A junior or a senior
The events ‘being a junior’ and ‘being a senior’ are mutually exclusive so,
12 15

1
P(junior or senior) = P(junior) + P(senior) 
27 27
Note that you may say that the answer is one because it is certain that a student will be
either a junior or a senior.
4. In an election 47% of eligible voters did not vote. If three eligible voters are selected
at random find:
(a) The probability that none of them voted in the election.
In this type of problem you assume that the events are independent, in which case you
multiply probabilities.
P(none voted) = (.47)(.47)(.47) =.103823
(b) The probability that at least one of the three voted in the election.
At least one is the complement of none
P(at least one voted) = 1 – P(none voted) =1 – .103823 = .896177
5. If two cards are selected from a 52 card deck without replacement, find these
probabilities:
(a) Both are kings
P(both are kings) = P(first is a king)P(second is a king, given that the first is a king)

4 3
1
 
52 51 221
(b) The cards are different suits
The first card can be any card. Whatever suit it is there will be 51 cards left of which 39
will be in the other three suits.
P(different suits) =
52 39 13
 
52 51 17
6. At a small college, the probability that a student takes physics and sociology is 0.18.
The probability that a student takes sociology is 0.72. Find the probability that a student
is taking physics, given that he is taking sociology. To get credit you must show the
formula you use to get your answer.
P( A and B)
You use the formula P( A | B) 
P( B)
P(physics | sociology) 

P(physics and sociology)
P(sociology)
.18 1
  .25
.72 4
7. In a board of directors composed of eight people, now many ways can one chief
executive officer, one director, and one treasurer be selected?
By the counting principle it is 8  7  6  336
You may also figure it as 8 P3 . You use P rather than C because picking the same three
people in a different order constitutes a different selection.
8. In how many ways can a committee of four people be selected from a pool of ten
people?
10 C4  210
You use C rather than P because picking the same four people in a different order
constitutes the same committee.
9. (a) In how many ways can a jury of six women and six men be selected from nine
women and twelve men?
The six women can be chosen in 9 C 6 ways. The six men can be chosen in 12 C6 ways.
By the counting principle the number of ways six women and six men can be chosen is
9 C6  12 C6  84  924  77616 .
10. A bag contains five red balls and seven white balls. If you select four balls at
random without replacement, find the probability that you get two red balls and two
white balls.
The two red balls can be chosen in 5 C 2 ways. The two white balls can be chosen in 7 C2
ways. By the counting principle the two red balls and two white balls can be chosen in
5 C2  7 C2 ways.
There are 12 balls so the 4 balls can be chosen in 12 C4 ways.
C  C 10  21
 0.424
P(2 red balls and 2 white balls) = 5 2 7 2 
495
12 C 4
11. An English teacher wants to select 2 plays and 3 novels from 6 plays and 9 novels,
and then decide on a reading order. How many possibilities are there?
The two plays can be chosen in 6 C2  9 C3  15  84  1260 ways.
Then each of these selections can be ordered in 5! Ways.
Total number of ways = 120 1260  151, 200 ways.