Document 6530673

Hackernotes, Pima Community College
© Wayne Hacker 2007. All rights reserved.
Business Mathematics I: Math 173
Sample Exam 2-Solutions
Name:
Print your name neatly. If you forget to write your name, or write so sloppy that I can’t
read it, you can lose all of the points! Answer all the questions that you can. Circle your
answer.
All problems are worth 1 point each.
Do any 25 of the 26 problems. You may do all 26 if you like, but I will stop grading
after you have 25 problems correct. In other words, you cannot score more than 100%
on this exam. You must show how you arrived at the answer in order to receive any
credit for the problem. Problems for which only an answer is given with no
supporting work, will receive a score of zero for that problem.
DO NOT WRITE ON THIS PAGE IN THE SPACE BELOW
1. _______
6. _______
11. _______
16. _______
21. _______
2. _______
7. _______
12. _______
17. _______
22. _______
3. _______
8. _______
13. _______
18. _______
23. _______
4. _______
9. _______
14. _______
19. _______
24. _______
5. _______
10. _______
15. _______
20. _______
25. _______
26. _______
STUDY GUIDE FOR EXAM 2:
Topics covered on this exam: Basic Probability, Conditional Probability and
Independence, The Law of Total Probability and Bayes’ Theorem.
What homework sets should you review for this exam: Homework sets 5, 6, 7.
Warning: The actual exam will need not be the same questions, but the number of
problems, types of problems, and level of difficulty will be similar.
- Business Mathematics I, Sample Exam2: page 2 -
Problems 1-15 are designed to test your skills at using formulas. These are all
considered elementary problems.
1. (Basic Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∩ F ) = .2. Find P ( E ∪ F ) .
Solution: P ( E ∪ F ) = P ( E ) + P ( F ) − P ( E ∩ F ) = .4 + .5 − .2 = .7
2. (Basic Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∪ F ) = .7. Find P ( E ∩ F ) .
Solution: P ( E ∩ F ) = P ( E ) + P ( F ) − P ( E ∪ F ) = .4 + .5 − .7 = .4
3. (Basic Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∪ F ) = .7.
Find P(( E ∪ F ) C ) .
Solution: P(( E ∪ F ) C ) = 1 − P( E ∪ F ) = 1 − .7 = .3
4. (Basic Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∪ F ) = .7.
Find P ( E C ∩ F C ) .
Solution: P( E C ∩ F C ) = P(( E ∪ F ) C ) = 1 − P( E ∪ F ) = 1 − .7 = .3
The first equality follows from De Morgan’s Law.
5. (Basic Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∩ F ) = .2. Find P ( F − E ) .
Solution: Let F be the blue circle in the Venn diagram. Then we want the mass of the “right
moon” = mass of the blue disk − mass of the “football” (the intersection)
P ( F − E ) = P ( F ) − P ( E ∩ F ) = .5 − .2 = .3.
E
6. (Basic Probability) Suppose that P (E ) = .6, P (F ) = .5, and P ( E ∩ F ) = .2.
Find P (( F − E ) ∪ ( E − F )) = P(exactly one of the events E or F, but not both).
Solution: P (( F − E ) ∪ ( E − F )) = P ( E ∪ F ) − P ( E ∩ F ) = P ( E ) + P ( F ) − 2 ⋅ P ( E ∩ F )
= .6 + .5 −2(.2) = .7 (the mass of the left and right moons)
7. (Conditional Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∩ F ) = .2.
Compute P ( E | F ) .
Solution: Applying the definition of conditional probability yields
P( E | F ) =
P( E ∩ F ) .2 2
= = = .4 .
P( F )
.5 5
F
- Business Mathematics I, Sample Exam2: page 3 -
8. (Conditional Probability) Suppose that P (E ) = .4, P (F ) = .5, and P ( E ∩ F ) = .2.
Compute P ( F | E ) .
Solution: Applying the definition of conditional probability yields
P( F | E ) =
P( E ∩ F ) .2 1
= = = .5 .
P( E )
.4 2
Notice that P ( E | F ) ≠ P ( F | E ) . This is generally the case, given two events E and F. That
means that when we write conditional probabilities, the order of the events is important. We
can’t interchangeably write P ( E | F ) when we mean P ( F | E ) , and vice versa. This makes realworld sense. Suppose E is the event that a person is a woman, and F is the event that a person is
Condoleezza Rice. Then P ( E | F ) = 1 (if someone’s Condoleezza Rice, then she’s certainly a
woman), but P ( F | E ) is a very small number—out of the 3 billion or so women in the world,
only one is Condoleezza Rice. Be careful: order matters!
9. (Independence) Suppose that P (E ) = .4, P (F ) = .5 and that E and F are independent events.
Compute P ( E ∩ F ) .
Solution: Applying the definition of independent events
P ( F ∩ E ) = P ( E ) P ( F ) = (.4)(.5) = .2.
10. (Independence) Suppose that P (E ) = .5, P (F ) = .8 and that E and F are independent
events.
Compute P ( E | F ) .
Solution: Applying the definition of conditional probability (first equality) together with the
definition of independence (the second equality) yields
P( E ∩ F ) P( E ) P( F )
=
= P( E ) = .5
P( F )
P( F )
Recall: A second definition of independence is two events E and F are independent if
P ( E | F ) = P ( E ) . Thus a second way to solve this problem is just to write P ( E | F ) = P ( E ) =.5.
P( E | F ) =
11. (Independence) Suppose that P (E ) = .5, P (F ) = .8 and that E and F are independent
events.
Compute P ( F | E ) .
- Business Mathematics I, Sample Exam2: page 4 -
Solution: Applying the definition of independent events
P ( F | E ) = P ( F ) = .8.
Since F is independent of E, the probability of F does not change if the event F is conditioned on
the event E.
Problems 12-15. (Law of Total Probability and Bayes’ Theorem) Let B1 and B2 partition
the sample space Ω.
Suppose we are given the following information:
2
1
2
1
P ( B1 ) = , P ( B2 ) = , P ( A | B1 ) = , and P ( A | B2 ) = .
3
3
3
2
12. Find P ( A) .
Solution: P ( A) = P ( B1 ) P ( A | B1 ) + P ( B2 ) P ( A | B2 ) =
2 2 1 1 4 1 11
⋅ + ⋅ = + =
3 3 3 2 9 6 18
13. Find P( B1 | A) .
4
P ( B1 ) P( A | B1 )
8
= 9 =
Solution: P( B1 | A) =
11
P ( B1 ) P ( A | B1 ) + P ( B2 ) P ( A | B2 )
11
18
Suppose we are given the following information:
1
4
1
1
P( B1 ) = , P ( B2 ) = , P( A | B1 ) = , and P ( A | B2 ) = .
5
5
3
2
14. Find P ( A) .
Solution: P ( A) = P ( B1 ) P ( A | B1 ) + P ( B2 ) P ( A | B2 ) =
1 1 4 1
2
4⋅3
14 7
⋅ + ⋅ =
+
=
=
5 3 5 2 5 ⋅ 3 ⋅ 2 5 ⋅ 3 ⋅ 2 30 15
15. Find P( B1 | A) .
1
P( B1 ) P( A | B1 )
1
= 15 =
Solution: P( B1 | A) =
7
P( B1 ) P( A | B1 ) + P( B2 ) P( A | B2 )
7
15
- Business Mathematics I, Sample Exam2: page 5 -
16. (Basic Probability) Red Hat and SUSE are companies that package the GNU-Linux
operating system for distribution. The history of the stock prices of these two companies has
been analyzed using the frequency definition of probability. It was found that the probability that
Red Hat goes up tomorrow is 40% and that the probability that SUSE goes up tomorrow is 50%,
and that the probability that both Red Hat and SUSE go up tomorrow is 20%. What is the
probability that either Red Hat or SUSE go up tomorrow?
(a) 0
(c) .2
(e) .4
(g) .6
(i) .8
(k) 1
(b) .1
(d) .3
(f) .5
(h) .7
(j) .9
(l) none of the these
Solution: Answer: (h)
Step 1: Write down the events. Let
R = event Red Hat’s stock price goes up
S = event SUSE’s stock price goes up
Step 2: Write down the given information.
P ( R ) = .4, P(S) = .5, P(R and S) = P ( R ∩ S ) = .2
Step 3: Write down what you want.
P(R or S) = P ( E ∪ F ) = P ( E ) + P ( F ) − P ( E ∩ F ) = .4 + .5 − .2 = .7
17. (Basic Probability) A health inspector will inspect your restaurant on one of the weekdays
{M, T, W, R, F} during the coming two weeks, and is equally likely to arrive on any of the days.
What is the probability that the health inspector will show up on a Monday or a Wednesday?
(a) 0
(c) .2
(e) .4
(g) .6
(i) .8
(k) 1
(b) .1
(d) .3
(f) .5
(h) .7
(j) .9
(l) none of the these
Solution: Answer: (e) Since the events M and W are disjoint, we have
1 1 2 4
P(Monday or Wednesday) = P(M) + P(W) = + = =
= 0 .4
5 5 5 10
- Business Mathematics I, Sample Exam2: page 6 -
18. (Basic Probability) You are promoting an outdoor concert for the coming Saturday (2 days
from now). It is a fact that weather records are fairly accurate over 3 day periods. Records
indicate that the probability of rain is 0.4, the probability of gale-force winds is 0.5, and the
probability of rain or gale-force winds is 0.8. What is the probability that it neither rains nor has
high winds?
(a) 0
(c) .2
(e) .4
(g) .6
(i) .8
(k) 1
(b) .1
(d) .3
(f) .5
(h) .7
(j) .9
(l) none of the these
Solution: Answer: (c)
Step 1: Write down the events. Let
R = event of rain
W = event of gale-force winds
Step 2: Write down the given information.
P(R) = .4, P(W) = .5, P(R or W) = P ( R ∪ W ) = .8
Step 3: Write down what you want.
P(neither R nor W) = P(not R and not W)
= P( R C ∩ W C ) (convert words into mathematical symbols)
= P(( R ∪ W ) C ) (by De Morgan’s Law)
= 1 − P( R ∪ W )
= 1− .8
= .2
- Business Mathematics I, Sample Exam2: page 7 -
19. (Basic Probability) You have just opened a restaurant and have applied for a liquor license
and a smoking permit (to allow smoking in a closed-off part of the restaurant). You are friends
with the city inspector and he has told you that based on past experience you have a 60% chance
of getting a liquor license, a 40% chance of getting a smoking permit, and a 20% chance of
getting both permits approved. What is the probability that you will be given exactly one of the
permits?
(a) 0
(c) .2
(e) .4
(g) .6
(i) .8
(k) 1
(b) .1
(d) .3
(f) .5
(h) .7
(j) .9
(l) none of the these
Solution: Answer: (g)
Step 1: Write down the events. Let
L = event of obtaining a Liquor license
S = event of obtaining a Smoking license
Step 2: Write down the given information.
P(L) = .6, P(S) = .4, P(L and S) = P ( L ∩ S ) = .2
Step 3: Write down what you want.
P(( L − S ) ∪ ( S − L)) = P( L − S ) + P( S − L) (disjoint sets)
= ( P( L) − P( L ∩ S )) + ( P( S ) − P( L ∩ S ))
= P ( L) + P ( S ) − 2 P ( L ∩ S )
= .6 + .4 − 2(.2)
= .6
The idea behind the solution:
P(Exactly one permit) = mass of the left moon + mass of the right moon =
P (( A − B ) ∪ ( B − A)) . The Venn diagrams below show that these sets are disjoint
=
---
Notice that we must subtract A ∩ B out from the union, because we want to exclude the
possibility of being in both sets A and B. We can also write the probability of “exactly one” as
P (( A − B ) ∪ ( B − A)) = P ( A ∪ B ) − P ( A ∩ B ) .
- Business Mathematics I, Sample Exam2: page 8 -
20-22. (Independence) Consider three identical boxes that we shall refer to as box1, box2, and
box3. Each box contains 100 parts, one of which is defective. You randomly choose one part
from each box. Start by writing down the events and their associated probabilities.
Solution:
Step 1: Write down the events.
⎧ Di = the event of choosing a defective part from box i
Let ⎨ C
⎩ Di = the event of choosing a non - defective part from box i
Step 2: Write down what you know.
P( Di ) = .01 and P ( DiC ) = 1 − P ( Di ) = .99
NOTE: The first two steps are the same for problems 16-18. The key fact to be aware of in this
problem is that all of our selections are independent. Our selection from box1 has no bearing on
our selection in box2, or box3.
20. What is the probability of selecting 3 non-defective parts?
Solution: To choose three non-defective parts you must choose a non-defective part from box1
and a non-defective part from box2 and a non-defective part from box3. This can be expressed
more compactly by the set: D1C ∩ D2C ∩ D3C . We want to compute P ( D1C ∩ D2C ∩ D3C ) . Using the
independence assumption:
P ( D1C ∩ D2C ∩ D3C ) = P ( D1C ) P ( D2C ) P ( D3C ) = (.99)(.99)(.99) = (.99) 3 ≈ .9703
21. What is the probability of selecting a non-defective part from box1, a defective part from
box2, and a non-defective part from box3?
Solution: We seek the event D1C ∩ D2 ∩ D3C . We want to compute P ( D1C ∩ D2 ∩ D3C ) . Using
the independence assumption:
P ( D1C ∩ D2 ∩ D3C ) = P ( D1C ) P ( D2 ) P ( D3C ) = (.99)(.01)(.99) = (.99) 2 (.01) ≈ .0098
- Business Mathematics I, Sample Exam2: page 9 -
22. What is the probability of selecting 1 defective part and 2 non-defective parts?
Solution: We can select one defective part and 2 non-defective parts in 3 different ways:
D1 ∩ D2C ∩ D3C
( the defective part comes from box 1)
D1C ∩ D2 ∩ D3C
( the defective part comes from box 2)
D1C ∩ D2C ∩ D3
( the defective part comes from box 3)
We want the probability of selecting D1 ∩ D2C ∩ D3C , or D1C ∩ D2 ∩ D3C , or D1C ∩ D2C ∩ D3 :
P (( D1 ∩ D2C ∩ D3C ) ∪ ( D1C ∩ D2 ∩ D3C ) ∪ ( D1C ∩ D2C ∩ D3 ))
= P ( D1 ∩ D2C ∩ D3C ) + P ( D1C ∩ D2 ∩ D3C ) + P ( D1C ∩ D2C ∩ D3 )
= 3P( D )( P ( D C )) 2
≈ 3(.0098) = .0294,
where the event D is short-hand for D1 , D2 , or D3 .
- Business Mathematics I, Sample Exam2: page 10 -
23. (Conditional Probability) In a certain region of the country, atmospheric conditions are
classified into a finite number of categories, and weather records involving the events of rain and
gale-force winds are kept for each of these categories. Weather records indicate that under
category 1 atmospheric condition there is a 40% chance of rain, a 30% chance of gale-force
winds, and a 50% chance that it either rains or has gale-force winds. Suppose you wake up on a
day with category 1 conditions and find that it is raining. What is the probability of gale-force
winds? (Assume that you know the all of the categories and can classify them correctly.)
Solution: Step 1: Write down the events.
⎧ R = the event of rain
Let ⎨
⎩W = the event of gale - force winds
Step 2: Write down what you know.
P(R) = .4,
P(W) = .3, and P ( R ∪ W ) = .5 .
Step 3: Write down what you want.
We are told that it is raining. Thus the event R has occurred. This means that we want to know
the probability of gale-force winds given that it is raining. In symbols, we want P(W | R) .
Step 4: Solve
Using the definition of conditional probability and P(R) = .4 leads to the following equation:
P(W | R) =
P( R ∩ W ) P( R ∩ W )
=
.
P( R)
.4
(Equation 1)
We cannot evaluate the expression on the right-hand side because we don’t know P ( R ∩ W ) .
However, we can use the formula P ( R ∪ W ) = P ( R ) + P (W ) − P ( R ∩ W ) to solve for P ( R ∩ W ) .
Substituting the values found in step 2 into the formula yields
.5 = .4 + .3 − P ( R ∩ W ) .
Solving for P ( R ∩ W ) in the equation gives us P ( R ∩ W ) = .2 . For the last step, we must
substitute this value in to equation 1 to find the desired result.
P (W | R ) =
P ( R ∩ W ) .2 1
= =
.4
.4 2
- Business Mathematics I, Sample Exam2: page 11 -
24. (Law of Total Probability) You are selling a product in an area where 30% of the people
live in the city; the rest live in the suburbs. 20% of the city dwellers (urbanites) use your product;
and 10% of the suburbanites use your product. What fraction of the people in the area use your
product?
Solution:
Step 1: Define the events: let
Ω = all of the potential customers in your area
U = the person is an urbanite
S = the person is a suburbanite
C = the person is a customer
C C = the person is not a customer
Step 2: Write down the given information:
First, notice that the events U and S partition the sample space Ω. The events C and CC
also partition the set Ω; however, as shown below, the given information suggests that we take U
and S as our partition for this problem.
Given:
P (U ) = 0.30,
P ( S ) = 1 − P (U ) = 1 − 0.30 = 0.70 (since U and S partition Ω),
P (C | U ) = 0.20 , (since the events U and S have occurred, they should be the partitions of Ω)
P (C | S ) = 0.10 .
Step 3: Write down what you are trying to solve for:
Want P (C ). We can compute this term using the law of total probability with U and S as our
partition.
P (C ) = P (U ) P (C | U ) + P ( S ) P (C | S ) = (.3)(.2) + (.7)(.1) = .06 + .07 = .13
- Business Mathematics I, Sample Exam2: page 12 -
25. (Law of Total Probability) A high school conducts random drug tests on its students. Of
the student body, it is known that 8% use marijuana regularly; 17% use it occasionally; and 75%
never use it. The testing regime is not perfect: regular marijuana users falsely test negative 5%
of the time; occasional users falsely test negative 13% of the time; and non-users falsely test
positive 11% of the time. What percentage of the student body will test positive for marijuana
use?
Solution:
Step 1: Define the events: let
Ω = all of the students in the high school (the entire student body)
R = the person uses marijuana regularly
O = the person uses marijuana occasionally
N = the person never uses marijuana
T + = the person tests positive for marijuana use
T − = the person tests negative for marijuana use
Notice that (T + ) C = T − . If we assume that there were no inclusive tests, and if all of the students
were tested, then T + and T − would partition Ω. However, the data suggests that we use a
different partition. The sets R, O, and N, also partition the sample space, since R ∪ O ∪ N = Ω ,
R ∩ O = Ø, R ∩ N = Ø, and O ∩ N = Ø. We will use these sets as our partition.
Step 2: Write down the given information:
Given:
P ( R ) = 0.08,
P (O ) = 0.17 ,
P ( N ) = 0.75 ,
From the data we are given the probability of testing positive or negative given that the events
R, O, and N have occurred. Thus based on the data, we should use R, O, and N to partition Ω.
From the problem description we can infer the following:
P(T − | R) = 0.05 ⇒ P(T + | R) = 1 − P(T − | R) = 1 − 0.05 = .95 ,
P(T − | O) = 0.13 ⇒ P(T + | O ) = 1 − P(T − | O) = 1 − 0.17 = .87 ,
P (T + | N ) = 0.11 ⇒ P(T − | N ) = 1 − P(T + | N ) = 1 − 0.11 = .89 .
Step 3: Write down what you are trying to solve for:
Want P(T + ) . We can compute this term using the law of total probability with R, O, and N as our
partition.
- Business Mathematics I, Sample Exam2: page 13 -
P(T + ) = P( R) P(T + | R) + P (O) P(T + | O) + P( N ) P (T + | N )
= (.08)(.95) + (.17)(.87) + (.75)(.11) = .3064
26. (Bayes’ Theorem) Suppose that we have two identical boxes: box 1 and box 2. Box 1
contains 5 red balls and 3 blue balls. Box 2 contains 2 red balls and 4 blue balls. A box is
selected at random and exactly one ball is drawn from the box. Given that the selected ball is
blue, what’s the probability that it came from box 2?
Solution: Use the law of total probability.
Step 1: Define the events: let
⎧Ω = all balls
⎪ B = event you select box 1
⎪⎪ 1
⎨ B2 = event you select box 2
⎪ R = event you select a red ball
⎪
⎪⎩ B = event you select a blue ball
Step 2: Write down the given information:
First, notice that the events B1 and B2 partition the sample space Ω.
Given:
P ( B1 ) =
3
4 2
1
1
, P ( B2 ) = , P ( B | B1 ) = , P ( B | B2 ) = =
8
6 3
2
2
.
Step 3: Write down what you are trying to solve for:
Want: P( B2 | B) . We will need Bayes’ theorem to compute this probability. We first apply the
law of total probability to compute P ( B ) :
P ( B ) = P ( B1 ) P ( B | B1 ) + P ( B2 ) P ( B | B2 ) =
1 3 1 2 3 1 9 + 16 25 1
⋅ + ⋅ =
+ =
=
≈
2 8 2 3 16 3
48
48 2
Next, we apply Bayes’ theorem to arrive at the desired solution:
1
P ( B2 ) P ( B | B2 )
16
P ( B2 | B ) =
= 3 =
.
P( B1 ) P( B | B1 ) + P( B2 ) P( B | B2 ) 25 25
48