251y0122 10/23/01
ECO 251 QUANTITATIVE BUSINESS ANALYSIS I
SECOND EXAM
OCTOBER 23, 2001
NAME: ___KEY____________
SECTION ENROLLED: MWF TR 10 11 12:30
(Circle both days and time separately or just write down days and time)
Part I. Multiple Choice (20 Points) Remember that 'At least one' means 'one or two or three or four or……'.
It does not mean 'exactly one.'
Mothers Against Drunken Driving has compiled the following data about drinking and
driving:.
Alcohol Involved?
Yes
No
Total
Number of Vehicles Involved
1
2
3
Total
50
100
20
170
25
175
30
230
75
275
50
400
Define the following events: A  Involved Alchohol, M  Involved MultipleVehicles ,
S  Involved SingleVehicle , All probabilities in questions 1-4 are given to only 4 places.
Solution: Make a joint probability table. P A  S   50  .1250 , P A  M   10020  .3000, etc.
400
Use the joint probability table:
S
M
A .1250 .3000  .4250


A .0625 .5125  .5750
.1875 .8125 1.000
1.
400
PS  A  P A  S   .1250
P A  S  .1250
P A S  

 .6667
PM 
.8125
P A  M   P A  PM   P A  M 
 .4250  .8125  .3000  .9375
From the table above find the proportion of single vehicle accidents that involved alcohol.
a.
P S A  .1250
b.
c.*
 
PA S   .1250
PA S   .6667
d.
P A  S   .6667
e.
None of the above. Write in both parts of the answer.
Solution: See above or note that P A S  50 75  .6667
 
2.
From the same table find the probability that, if we pick an accident at random, it will involve
alcohol or multiple vehicles (or both)?.
P A  M   .3000
b.
P A  M   .3000
c.
P A  M   .9375
d.
P A  M   .4250
e.*
None of the above. Write in both parts of the answer
Solution: See above or note that P A  M   501002017530 400 or 2755017010020 400  .9375
a.
251y0122 10/23/01
3.
If we pick an accident at random, what is the probability that it involves both a single vehicle and
alcohol?
a.
P A  S   .6667
b.*
P A  S   .1250
c.
P A  S   .4875
PS A  .2941
e.
None of the above. Write in both parts of the answer.
Solution: See above or note that P A  S   50 400  .6667
d.
4.
A friend of yours claims to have a psychic gift. You are skeptical of her claim. To test her gift you
take five cards from a deck, the ace of spades, the ace of hearts, the ace of clubs, the king of hearts
and the ace of diamonds. You randomly choose one of these cards and ask your friend to tell you
what it is without looking at it. You replace the card and repeat the experiment three times. If your
friend is really not a psychic, what is her chance of guessing all three cards correctly?
a.*
.0080
b.
.2000
c.
.5000
d.
.6000
e.
.8000
Solution: The probability of getting the right answer on any one guess is 15  .20. So the probability of
three successive right guesses, assuming independence is .20 3  .0080 .
5.
(MBS 9) A dice game called Chuck-a-Luck involves throwing three dice and betting on one of the
six numbers, one through six. The game costs $2 to play, and your winnings are determined by the
number of dice that come up with your number. Four example if you get no dice with your number,
you win nothing, so your profit is -$2.00(It's negative!). If one die comes up with you number, you
win $2, but when your cost is deducted, your profit is $0.00. If your number comes up twice, your
profit is $1.00 and if it comes up on all three dice, your profit is $2.00. The following table gives
the possible profits and the chances of winning.
Profit Chance of Winning
-$2
125/216
$0
75/216
$1
15/216
$2
1/216
What is your expected (mean) profit from a round of Chuck-a-Luck? (to 4 decimal places)
a.
$0.2500
b.
-$0.5000
c.*
-$1.0787
d.
-$1.1574
e.
-$1.2778
Solution:
Using decimals:
Profit
Probability
x P x 
x 2 Px
-2
125/216
-1.1574
2.3148
0
75/216
0
0
1
15/216
0.0694
0.0694
2
1/216
0.0093
0.0185
1.00000
-1.0787
2.4027
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251y0122 10/23/01
Using fractions:
Profit
x P x 
-250/216
0
15/216
2/216
-233/216
Probability
-2
0
1
2
125/216
75/216
15/216
1/216
216/216
x 2 Px
500/216
0
15/216
4/216
519/216
We have a valid distribution since all the probabilities are between one and zero and add to 1. The formula
xPx 
for a population variance for a distribution is  x2  E x 2   x2 , where   E x  

233
216 
   x
1.0787 and E x
2
 
2
Px  
519
216 
2.4027 . So

 x2
 2.4027   1.0787 2  1.2391
and  x  1.2391  1.1132 .
6
In problem 5, what is the standard deviation of the profit from a round of Chuck-a-Luck?
a.
1.2454
b.
1.2392
c.*
1.1132
d.
0.0062
e.
None of the above - Supply answer and show your work.
7.
The number of ways that 5 items can be taken from 15 if order is important and replacement is not
allowed is
a.
1
b.
3
c.
3003
d.
60360
e.
759375
f.*
None of the above - Supply answer and show your work.
n!
15!
 15 14 13 12 11  360360 . Sorry! A digit got dropped in d.
Solution: Prn 
so P515 
n  r !
5!
8.
Suppose that there is a 20% chance that a risky stock investment will end in a total loss of your
investment. Because the rewards are so high, you decide to invest in four independent risky stocks.
What is the probability that at least one of your investments is a total loss? (Hint: Complement?)
a.
.8000
b.
.0016
c.*
.5904
d.
.2000
e. None of the above.
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251y0122 10/23/01
Solution: According to Problem H1, :
P A  B  C  D  P A  PB   PC   PD   P A  B   P A  C   P A  D  PB  C   PB  D  PC  D 
 P A  B  C   P A  B  D  P A  C  D  PB  C  D   P A  B  C  D 
But note that it might be easier to compute P A  B  C  D   P A  B  C  D   1  PA  B  C  D  .


 
So if the event L1 is a total loss on investment 1 and we take this suggestion, PL1   .20, P L1  .80 etc.




So PL1  L2  L3  L4   P L1  L2  L3  L4   1  P L1  L2  L3  L4  1  .80 4  1  .4096  .5904 . If you

try to do it the other way, you get PL1  L2  L3  L4   4.20   6.20 2  4.20 3  .20 4  .5904 . Why is any
answer below .20 unreasonable?
9.
As part of a give away promotion, a local cellular phone company gave away 200 cellular phones.
Unfortunately, someone has bugged 20 of the phones. Both you and your roommate have received
a free phone. What is the chance that both are bugged?
a.
.01000
b.
.20000
c.*
.00955
d.
.00400
e.
.10000
Solution: To do this 'right,' we can say that we are taking a sample of two from a population of 200
2019
1 20 19
C 20 C 180
21
containing 20 bugged phones and 180 normal phones. P2  2 2000  200


 00955 .
199
200 199
C2
21
(Remember C rn 
phone is
result.
20
200
n!
n  r ! r!
.) But, if we are in a hurry, we can say that the probability that you get a bugged
and the probability that your roommate gets a bugged phone is
19
199 ,
and get the same
10.
In the above problem, what is the probability that at least one of you received a bugged phone?
a.
.36000
b.*
.19045
c.
.19000
d.
.20000
e.
.20055
Solution: To do this 'right,' we can say that we are taking a sample of two from a population of 200
containing 20 bugged phones and 180 normal phones and that what we want is the complement of the
180179
C 20 C 180 1 1
180 179
probability that no one gets a bugged phone . P0  0 2002  2002199


 .80955 . But, if we
200 199
C2
21
are in a hurry, we can say that the probability that you get a normal phone is
your roommate gets a normal phone is
1  P0  1  .80945  .19045 .
179
199 ,
180
200
and the probability that
and get the same result. The probability of the complement is
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251y0122 10/23/01
Part II. Show your work!
The number of cars you sold on five successive Saturdays is represented by the results below. (This is a
sample not a probability distribution!) compute the sample standard deviation (5 Points - 2 Point Penalty
for not trying.)
Number of
Cars Sold
0
5
13
3
6
Solution:
x
0
5
13
3
6
27
x2
0
25
169
9
36
239
x
 x  27  5.4
s2 
n
x
5
2
 nx 2
n 1

239  55.42
 23 .3
4
s  23.3  4.8270 .
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251y0122 10/23/01
Part III. Do at least 2 1/2 (25 Points) of the four problems on the next four pages. Show your work! You
receive extra credit for extra work!
1. You are dealt six cards from a deck. Remember that there are 4 cards of each denomination (Aces. 2's,
etc.) but 13 cards of each suit (Hearts, Clubs, Spades, Diamonds)
a. How many possible hands are there? (2) Work out answers to a) and b).
b. What is the probability of getting three kings and three queens? (2)
You may leave the answers to the remaining sections of this question in factorial form.
c. What is the probability of getting all hearts? (2)
d. What is the probability of getting at least one king? (2)
e. What is the probability of getting two hearts ? (2)
f. What is the probability of two hearts, two clubs, and two diamonds? (2)
52! 52  51  50  49  48  47 52  51 10  49  2  47
n!


 20358520
Solution: a) C rn 
, C 652 
46!6!
6  5  4  3  2 1
6
n  r ! r!
b) P3K ,3Q  
C 34 C 34
C 652
4!  4! 


3!1!  3!1! 
42
16



 .000000786
52!
52  51  50  49  48  47 20358520
46! 6!
6  5  4  3  2 1
13!  39! 
13 12 11 10  9  8


1
7! 6!  39! 0! 
1716
6  5  4  3  2 1
c) P6 H  



 8.4289  10 5  .000084289
52!
52  51  50  49  48  47 20358520
46! 6!
6  5  4  3  2 1
d) As usual, most people gave me the probability of exactly one king.
4!  48! 


1 48  47  46  45  44  43 1
4 48
4!0!  42!6! 
C0 C6
12271512
6  5  4  3  2 1
1  P0 K   1 
 1
 1
 1
 1  .60277  .39722
52!
52  51  50  49  48  47
20358520
C652
46!6!
6  5  4  3  2 1
C 613C 039
C 652
or 1  P0 K   1 
48  47  46  45  44  43
 1  .60277  .39722
52  51  50  49  48  47
e)
P2 H  

C 213C 439
C 652
13!  39!  13 12 39  38  37  36 13 12



 39  38  37  36
11! 2!  35! 4! 
2

1
4

3

2

1
2

1



52!
52  51  50  49  48  47 52  51
50  49  48  47
46! 6!
6  5  4  3  2 1
65
13 12  39  38  37  36 6  5
 .315136
52  51  50  49  48  47 2 1
f) P2 H ,2C ,2 D  
C 213C 213C 213
C 652
3
13!  13!  13! 
 13 12 





11! 2! 11! 2!  11! 2! 
78 3
 2 1 



 .023310
52!
52  51  50  49  48  47 20358520
46! 6!
6  5  4  3  2 1
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251y0122 10/23/01
2. The In-Your-Face Custard Pie Company bakes 200 pies at a cost of $2.00 each every day. It sells 100
pies on 50% of days, 150 pies on 30% of days and 200 pies on 20% of days. Each pie sells for $4.00.
Unsold pies are thrown out at the end of the day. Figure out what the profit will be in each of the 3
situations (Sales of 200, 150 or 100 pies).
Profit answer: 100 pies 4100   2200   0 , 150 pies 4150   2200   200 , 200 pies
4200   2200   400 . The equation for profit is y  4x  400 .
a. Find the probability distribution of the daily profit and fill in the following table. (3)
Profit  y 
Probability P y 
0
200
400
.50
.30
.20
1.00
b. Find the expected value, variance and standard deviation of profit  y  . (3)
Solution:
Profit  y 
Probability P y 
y P y 
0
.50
0
200
.30
60
400
.20
80
1.00
140
 y  E y 
yP y   140 and E y 2 
y 2 P y   44000 .
So  y2

 E y   
2
y 2 P y 
0
12000
32000
44000
  
2
y
 44000  140 2  24400 and  y  24400  156.2049
c. Make a similar table for the number of pies sold and find the expected value, variance and
standard deviation for the number of pies sold. (3)
Solution:
Pies x 
Probability Px 
x P x 
x 2 Px
100
.50
50.0
5000
150
.30
45.0
6750
200
.20
40.0
8000
1.00
135.0
19750
x 2 Px   19750 .
  E x  
xPx   135 .0 and E x 2 

 
  
So  x2  E x 2   x2  19750  135 .02  1525 .00 and  x  1525 .00  39 .0512
d. Find an equation relating the number of pies sold to the profit and show that the equations
relating variances and expected values of linear functions of random variables to the variances and
expected values of the random variables would have enabled you to answer b) using the answer to
c) without making the table in a). (4)
Solution: From the outline Eax  b  aEx   b and Varax  b  a 2Varx  . Since y  4x  400 ,
a  4 and b  400 . So  y  E y   Eax  b  4135 .0  400  140 and  y2  Var  y 
 Var ax  b  42 Var x   161525 .00   24400 . Finally,  y  24400  156.205
or
 y  a  x  439 .0512   156 .205 .
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251y0122 10/23/01
3. All probabilities you get in this problem should be stated to four decimal places (i.e. .1575 not .157 or
.158).
a. Assume that P A  .75 and PB   .55 Make a joint probability table showing the events A ,
 
A (not A ), B and B on the assumption that A and B are independent. What is P B A ? (4)
b. Assume that P A  .75 and PB   .55 Make a joint probability table showing the events A ,
A (not A ), B and B on the assumption that A and B are collectively exhaustive. What is
PB A now? (4)
c. Assume that P A  .75 and PB   .55 Try to make a joint probability table showing the
events A , A , B and B on the assumption that A and B are mutually exclusive. Why is this
impossible? What must P B A be for the two events to be mutually exclusive? Use the addition
 
rule to show that two events cannot have probabilities that sum to more than one and be mutually
exclusive. (5)
A
A
P B 
B P A  B  P A  B
Solution: a) The joint probability table shows
and A and B
PB
B P A B P A  B

P  A

 
 


 
PA
1
B
A
.4125
are independent, which means that P A  B   P APB  . If P A  .75 and PB   .55 ,
P A  B   .75 .55   .4125 . If we start to fill in the table, we have
fill in the blanks to make it add up, we get
B
B
A
.4125
A
.1375
.3375
.75
.1125
.25
 
.55
.45
B
A
___
___ ___
.75
.25
.55
.45
. If we just
1.00
.
1.00
-------Because A and B are independent, P B A  PB  .55.
b) If A and B are collectively exhausted, there is nothing that is not in A or B . This means
A
A
.55
B
__ __
. If we just fill in the blanks
P A  B  0 . If we start to fill in the table, we have
.45
B
__ 0
.75
.25 1.00


to make it add up, we get
B
B
A
A
.30 .25
.55
.45
0 .45
.75 .25 1.00
 
------- By the Multiplication Rule, P B A 
.
PB  A .30

 .40
P A
.75
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251y0122 10/23/01
c) If A and B are mutually exclusive, P A  B   0 . If we start to fill in the table, we have
B
A
A
0 __
.55
. To make column 1 and row 1 add up we fill in
B
A
A
0 .55
.45
__ __
B .75 __
.75 .25 1.00
.75 .25
impossible, because the numbers inside the table now add up to more than 1.
B
.55
.45
. But this is
1.00
By the addition rule, P A  B   P A  PB   P A  B  , but if A and B are mutually exclusive,
P A  B   0 , so P A  B   P A  PB   0. If P A  PB  add up to more than 1, P A  B  must be
more than 1, which is impossible since all probabilities are between zero and one.
9
251y0122 10/23/01
4. (Bowerman, O'Connell and Hand, simplified) You are drilling for oil.
On the basis of the geology of the site, you have estimated the probability of oil as none N , some S or
much M as follows: PN   .6, PS   .3, PM   .1 . You now perform a seismic experiment that can yield
either affirmative A or negative A results. On the basis of your experience, you know that you will get an
affirmative reading 4% of the time on sites with no oil, 2% of the time on sites with some oil and 96% of the
time on sites with much oil. You do the seismic experiment and get affirmative results.
a. Using these numbers construct a joint probability table with columns representing the
affirmative and negative events and rows representing 'none', 'some' or 'much'. (3)
b. Find the joint probability that you will get an affirmative reading and find no oil. (1)
c. Now, using your table or Bayes' rule, revise your probabilities. You need 3 probabilities, the
probability that you will find much oil when you have gotten an affirmative reading, the probability
that you will find some oil after getting an affirmative reading and the probability that you will find
no oil after getting an affirmative reading. (6)
Solution: We have defined the following events: N  No Oil , S  Some Oil , M  Much Oil  ,
A  Affirmativ e Seismic Experiment and A  Negative Seismic Experiment.


 
a) Then the problem tells us that PN   .6, PS   .3, PM   .1 P A N  .04 , P A S  .02 and
PA M   .96 .
Then, Using the Multiplication Rule,
P A  N   PA N PN   .04.6  .024 , PA  S   PA S PS   .02.3  .006 and
PA  M   PA M PM   .96.1  .096 . If we put these numbers into a table, we find
A A
N
.024
S
.006
M
.096
.6
N
.3 . If we fill in the blanks to make it add up we get S
.1
M
1.0
A
.024
.006
.096
A
.576
.294
.004
.126 .874
.6
.3 .
.1
1.0
b) We have already found on the table P A  N   .024 .
c) If we use the table and the Multiplication Rule,
P A N PN  .04 .6
PN  A .024

 .19048 .

 .19048 or by Bayes' Rule PN A 
P  A
.126
P A
.126
(For Bayes' Rule we need P A  PN  A  PS  A  PN  A
PN A 
 PA N PN   PA S PS   PA M PM   04.6  02.3  96.1  .024  .006  .096  .126 )
PS A 
PA S PS  .02 .3
PS  A .006

 .04762 .

 .04762 or PS A 
P  A
.126
P A
.126
P A M PM  .96 .1
PM  A .096

 .76190 . Note that these

 .76190 or P M A 
P  A
.126
P A
.126
three probabilities must add to one.
PM A 
10
251y0122 10/23/01
e. You have a standardized random variable, z . Find the mean and standard deviation of the following (1.5
each):
a. y  6 z
b. y  z  1
c. y  6z  1
Solution: To summarize the rules in J3 and J4 in the outline
If y 
 y  E y  
 y2  Var  y  
b
ax
b
aEx 
xb
E x   b
aEx   b
ax  b
0
a 2Varx
Var x 
a 2Varx
If we replace x with z , and z is a standardized variable so that  z  Ez   0 , and Varz    z2  1
(  z  1 .), we find:
a) y  6z  az , where a  6. Then  y  E y   aEz   60  0 and
 y2  Var  y   a 2Var z    62  1  36 . So  y  36  6.
b) y  z  1  z  b , where b  1 . Then  y  E y   Ez   b  0  1  1 and
 y2  Var  y   Varz   1 So  y  1  1.
c) y  6z  1  ax  b , where a  6 and b  1 . Then  y  E y   aEz   b  60  1  1 and
 y2  Var  y   a 2Var z    62  1  36 . So  y  36  6.
11
251y0122 10/23/01
5. You roll a pair of dice. Let x represent the number of dots on the top of the first die, y represent the
number on the second die and w  x  y the total number of dots on both dice.
a. Find the probability that w is less than 6. (1)
b. Find the probability that x is less than 4. (1)
c. Find the probability of the intersection of the events in a and b. (1)
d. Find the probability of the union of the events in a and b. (1)
e. If you have not used the addition rule already, show that the probabilities in a-c obey the
addition rule. (If you used the addition rule correctly in d, you already have credit for this.) (1)
f. Make a table of the distribution of w , with values of w in one column and their probabilities in
the next. Use it to find the mean (expected value) and variance of w. (5)
g. Do the same for x . Note that the distribution of x is the same as the distribution of y. (3)
h. The random variables x and y are said to be independent of one another because the
occurrence of any one value of one does not affect the probabilities of values of the other. You
already know that the sum of the expected values of x and y will give the expected value of
w  x  y . What can you say about the relationship between the variances? (Warning! This only
works if they are independent.) (3)
Solution: The diagram below was presented in 251solnH3 ( x and y are reversed) and the probabilities of
getting various sums were discussed. Let event A be w  6 and event B be x  4 .
x
1
2
y 3
4
5
6
1
2
3
4
5
6
7
2
3
4
5
6
7
8
3
4
5
6
4 5 6 7
5 6 7  8
6 7 8 9
7  8  9  10 
8  9  10  11 
9  10  11  12 
a) If we count points. A consists of 10 points ,
x  1, y  1, x  1, y  2, x  1, y  3,
x  1, y  4, x  2, y  1, x  2, y  2,
x  2, y  3, x  3, y  1, x  3, y  2
x  4, y  1 .
and
Of course, it would be easier to use the distribution for w in part f) and say that P A 
Pw  2  Pw  3  Pw  4  Pw  5  136  2 36  3 36  4 36  10 36  .2778
You should shade the points in A in the diagram.
b) It would be a good idea to shade the 18 points in the diagram that are in B , they are x  1, y  1,
x  1, y  2, x  1, y  3, x  1, y  4, x  1, y  5, x  1, y  6, x  2, y  1, x  2, y  2,
x  2, y  3, x  2, y  4, x  2, y  5, x  2, y  6, x  3, y  1, x  3, y  2,
x  3, y  3, x  3, y  4, x  3, y  5 and x  3, y  6 . So
Px  1  Px  2  Px  3  18 36  .5000 . Of course it would be easier to use the distribution for x in
part g) and to say that PB   Px  1  Px  2  Px  3  16  16  16  1 2  .5000
c) If you shaded the events in a) and b), you would now notice that A B consists of the 9 points,
x  1, y  1, x  1, y  2, x  1, y  3, x  1, y  4, x  2, y  1, x  2, y  2, x  2, y  3,
x  3, y  1 and x  3, y  2 . Its probability must be P A  B  9 36  1 4  .2500 .
d) A B consists of all the points mentioned above. These 19 points are x  1, y  1, x  1, y  2,
x  1, y  3, x  1, y  4, x  1, y  5, x  1, y  6, x  2, y  1, x  2, y  2, x  2, y  3,
x  2, y  4, x  2, y  5, x  2, y  6, x  3, y  1, x  3, y  2,
x  3, y  3, x  3, y  4, x  3, y  5, x  3, y  3, x  3, y  6 and x  4, y  1 . Its probability
must be P A  B   19 36  .5278 .
12
251y0122 10/23/01
e) The addition rule says P A  B   P A  PB   P A  B  . Substituting from above
19  10  18  9 .
36
36
36
36
f) Using
fractions:
Note that
Pw
w Pw
w 2 P w 
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
36/36
2/36
6/36
12/36
20/36
30/36
42/36
40/36
36/36
30/36
22/36
12/36
252/36
4/36
18/36
48/36
100/36
180/36
294/36
320/36
324/36
300/36
242/36
144/36
1974/36
w
2
3
4
5
6
7
8
9
10
11
12
 Pw  1 , 
 
w
 E w 
Varw   w2  E w 2   w2 
g) Using
fractions:
Note that
Px 
x
1
2
3
4
5
6
 w Pw  
2
x Px 
1/36
1/36
1/36
1/36
1/36
1/36
6/36
1/6
2/6
3/6
4/6
5/6
6/6
21/6
 Px   1 , 
 E x  
 
x
 wPw 
Varx   x2  E x 2   x2 
x
2
2
w
252
36
  w
 7. E w 2 
2
Pw  197436  54 .8333 .
 197436  25236  21036  5.8333.
2
x 2 Px 
1/6
4/6
9/6
16/6
25/6
36/6
91/6
 xPx 
21
6
  x
 3.5 . E x 2 
2
Px   916  15 .1667 .
Px   x2  916  216   10536  2.91667 .
2
h) Since E x   E y   3.5 , it should be no surprise that E x   E x  y)  E x   E  y   3.5  3.5  7.0 .
But note that Varx   Varx  y)  Varx   Var y   10536  10536  21036  5.8333 . Too bad that this only
works when x and y are independent.
13