MA 114 Introduction to Finite Mathematics
Midterm Exam 1 Practice Run Solution
May 31, 2010
Instructions: Show all work and justify all your steps relevant to the solution of each problem. No texts,
notes, or other aids are permitted. Calculators are not permitted. Please turn off your cellphones. You have
90 minutes to complete this exam.
1. A college has 400 students. A survey has determined that
• 42 read French
• 32 read German
• 56 read Spanish
• 14 read French and German
• 24 read French and Spanish
• 10 read German and Spanish
• 7 read all three languages
Answer each of the following.
a. Create a Venn Diagram of the above information
(*) The information necessary is
• 11 read only French
• 15 read only German
• 29 read only Spanish
• 7 read only French and German
• 17 read only French and Spanish
• 3 read only German and Spanish
• 7 read all three languages
b. How many students read at least 2 languages
(*) 34
1
c. How many students read Spanish, but not French?
(*) 32
d. How many students read Spanish, but not French or German?
(*) 29
2. An appliance manufacturer has 6 models of autmatic washers and 3 models of automatic dryers. How
many washer-dryer combinations can be bought?
(*) 6 × 3 = 18, or C(6, 1) × C(3, 1)
3. A student is required to answer 3 questions out of 6 on a math test. In how many ways can the student
choose the 3 questions?
(*) C(6, 3) = 20
4. Ann is shopping and has found 5 coats and 7 hats that she likes.
a. If Ann decides to buy one coat and one hat, how many different options does she have for the items
she will buy?
(*) 5 × 7 = 35, or C(5, 1) × C(7, 1)
b. If Ann decides to buy 2 of the 5 coats and 2 of the 7 hats, how many options does she have?
(*) C(5, 2) × C(7, 2) = 210
c. If Ann decides to buy 3 items altogether, including at least one coat and at least one hat, how many
options does she have?
(*) She can buy 2 coats or 1 hat, or 1 coat and 2 hats. This is represented
C(5, 2) × C(7, 1) + C(5, 1) × C(7, 2) = 175, where the + separates the two cases.
5. A group of 4 boys and 3 girls are going to sit together in a row of 7 theater seats.
a. In how many ways can they seat themselves?
(*) 7!
b. In how many ways can they seat themselves if all the boys and all the girls sit together?
(*) There are two distinct cases. Either the group of boys sit to the left of the girls, or
vice-versa.
(4! × 3!) + (3! × 4!) = 288
c. What is the probability they will be seated together in alphabetical order from left to right?
(*) There is 1 way this can happen, and from part A, 7! = 5040 possibilities total. Hence,
1
5040 .
2
d. What is the probability the boys will be seated together and the girls will be seated together?
(*) From part B there are 288 ways this can happen. Hence,
288
5040 .
6. Suppose P(A) = .5, P(B) = .3, and P(A ∩ B) = .2. Find:
a. P(A ∩ Bc )
(*) .3
b. P(Ac ∪ B)
(*) .7
c. P(A|B)
(*) =
P(A∩B)
P(B)
=
2
3
d. P(Bc |A)
(*) =
P(Bc ∩A)
P(A)
= .6
7. Mac is one of 3 men attending a party. There are also 6 women attending.
a. If one person is randomly selected the job of washing dishes, what is the probability it will be a
man?
(*) There are 9 ways any one person can be selected, and 3 ways a man can be selected.
Hence, P(E) = 39
b. If it is known a man will be selected the job of washing dishes, what is the probability it will be
Mac?
(*) Let E be the event Mac does the dishes, and A the event a man does the dishes. P(E|A) =
P(E∩A)
1
P(A) = 3
3