Download Department of Mathematics Math 132 Spring 2007 Test 3 Solution

Math 132
Spring 2007
Test 3
Department of Mathematics
Math 132 Spring 2007
Test 3
Solution
Instructor: Eric Agyekum
Question 1.
Time: 90min.
[7 marks]
At the Milex tune-up and brake shop, the manager has found out that a car will require
tune-up with a probability of 0.6, a brake job with a probability of 0.3, and both a tune-up
and a brake job with a probability of 0.1. What is the probability that a randomly chosen
car at the shop
[Hint: Draw a Venn Diagram]
a) requires a tune-up or a break job (or both)?
P (B or T) = P(B) + P(T) – P(B and T) = 0.6 + 0.3 – 0.1 = 0.8
b) requires a tune-up but not a brake job?
P (T and B ) = P (T) – P (B and T) = 0.6 – 0.1 = 0.5
c) requires neither type of repair?
(
)
P T and B = 1- P ( T or B) = 1- 0.8 = 0.2
d) Are the events “requiring a tune-up” and “needing a break job” mutually exclusive?
Explain.
No, since P (T and B) ≠ 0 , it means the two events can occur at the same time.
[3 marks]
Question 2:
a) If a man owns 5 shirts and 6 pairs of slacks, how many shirt-slacks combinations
does he have?
5 × 6 = 30
5 ways to choose a shirt AND 6 ways to choose a slack.
b) A coin is tossed and a die is rolled. In how many ways can the outcome consist of a
head and an even number?
1× 3 = 3 A head can occur in one way and an even number can occur in 3 ways.
Please use the reverse side of the sheet if needed.
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Math 132
Spring 2007
Test 3
Question 3:
[7 marks]
The “WE LOVE MATH CLUB” has 10 members.
a) In how many ways can all 10 members line up for a picture?
10! = 3, 628,800
b) In how many ways can all 10 members line up for a picture if PI and ALPHA are
members and would not stand next to each other?
10!− 2 × 9! = 2,903, 040 (If we combine PI and ALPHA as one, then we have 9
distinct objects to arrange. We can arrange ALPHA and PI in 2 ways. Thus,
there are 2 × 9! arrangements in which PI and ALPHA stand next to each
other).
c) In how many ways can the club choose a president, vice-president and a secretary?
10
P3 = 720
d) In how many ways can the club choose 3 members to attend the regional MATH
COMPETITON?
10
C3 = 120
Question 4:
[4 marks]
A committee has 4 male and 4 female members. In how many ways can a sub-committee of
three members be selected if it must contain at least one male and at least one female?
Need 1M, 2F OR 2M, 1F
n( E ) = 4 C1 × 4 C2 + 4 C2 × 4 C1 = 4 × 6 + 6 × 4 = 48
OR Total number of ways we could form a committee of 3 people is 8 C3 = 56 . We
exclude cases where all 3 are males or females.
Thus, n( E ) = 8 C3 − [ 4 C3 + 4 C3 ] = 56 − 8 = 48
Please use the reverse side of the sheet if needed.
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Math 132
Spring 2007
Test 3
Question 5:
[4 marks]
a) How many four-digit natural numbers can be named using each of the digits 1, 2, 3,
4, 5, and 6 at most once? (That is repetition is not allowed)
6
P4 = 6 × 5 × 4 × 3 = 360
b) What is the probability that the number in (a) begins with an odd digit?
The first digit can be chosen in 3 ways.
n( E ) = 3 × 5 × 4 × 3 = 180; ∴ P( E ) =
180 1
=
360 2
c) What is the probability that the number in (a) is odd?
The last digit can be chosen in 3 ways.
180 1
n( E ) = 5 × 4 × 3 × 3 = 180; ∴ P( E ) =
=
360 2
d) How many of the numbers in part (a) are greater than 4000?
The first digit can be chosen in 3 ways. n( E ) = 3 × 5 × 4 × 3 = 180
Question 6:
[4 marks]
Your drawer contains 2 green socks (that is, one pair), 4 red socks and 6 blue socks. It is
dark and the power is off so you cannot see which is which. So you select one at random
and pull it onto one foot. Then you select another sock at random from the drawer and put
it onto the other foot.
a) Draw a tree diagram to show the possible outcomes.
2/12
G
4/12
R
6/12
B
1
4 6 2 3 6 2 4 5
11 11 11 11 11 11 11 11 11
G
R B G R B
G R B
b) What is the probability that you are wearing two socks of the same colour?
P( E ) =
2 1 4 3 6 5
44 1
× + × + × =
=
12 11 12 11 12 11 132 3
Please use the reverse side of the sheet if needed.
OR P ( E ) =
2
C2 + 4 C2 + 6 C 2 1
=
3
12 C2
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Math 132
Spring 2007
Test 3
Question 7:
[5 marks]
A company considers two business ventures. The executives believe that venture A has a
0.60 probability of a $50,000 profit, a 0.30 probability of breaking even, and a 0.10
probability of a $70,000 loss. Venture B has a 0.40 probability of $100,000 profit and 0.60
probability of a $60,000 loss. Determine the expected value of each venture. Which venture
appears to be more profitable?
Venture A
Venture B
e = 50, 000 × 0.6 + 0 × 0.3 − 70, 000 × 0.1
= $23, 000
e = 100, 000 × 0.4 − 60, 000 × 0.6
= $4, 000
Venture A appears to be more profitable.
Question 8:
[7 marks]
a) How many different arrangements are there of the letters in
i)
MATH?
4! = 4 × 3 × 2 × 1 = 24
ii)
MATHYUMMY?
9!
= 30, 240
3!× 2!
b) How many different arrangements are there of the letters HHHTT?
5!
= 10
3!× 2!
c) A coin is tossed 5 times, what is the probability of obtaining three heads and two
tails?
C2 10 5
=
=
25
32 16
5
Please use the reverse side of the sheet if needed.
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Math 132
Spring 2007
Test 3
Question 9:
[5 marks]
Acme Auto Rental has three red Fords, four white Fords, and two black Fords. Acme also
has six red Hondas, two white Hondas, and five black Hondas. If a car is randomly selected
for a customer for rental, what is the probability that
a) it is a white Ford?
Ford
3
4
2
9
Red
White
Black
Total
4
2
=
22 11
Honda
6
2
5
13
Total
9
6
7
22
b) it is a Ford given that the customer demands a white car?
P( F | W ) =
4 2
=
6 3
c) it is white given that the customer demands a Ford?
P (W | F ) =
4
9
Question 10:
[4 marks]
a) If P ( A) = 0.12, P( B) = 0.34, and A and B are mutually exclusive, compute the odds
in favour of A or B.
[Hint: Need to find P(A or B) first]
P ( Aor B) = P( A) + P( B) = 0.12 + 0.34 = 0.46 Since A and B are mutually exclusive.
Odds in favour of A or B are 0.46 : 0.54 = 23 : 27
( )
b) If the odds in favour of A are 3:5, find P A .
( )
P A =
5
8
Please use the reverse side of the sheet if needed.
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