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ECO 251 QUANTITATIVE BUSINESS ANALYSIS I
SECOND EXAM
OCTOBER 29, 2002
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 (40 Points) Remember that 'At least one' means 'one or two or three or four or……'.
It does not mean 'exactly one.'
1. For the joint probability table below, the probability of the union of A (not A ) and B is:
B
.20
.10
A
A
a.
b.*
c.
d.
e.
B
.20
.50
.42
.80
.52
Above 1
None of the above. For credit, write in an answer, showing your work
Solution: By the addition rule


PA  B   PA   PB   PA  B   .6  .7  .5  .8 . You


could also say that P A  B  P  A  B   1  P A  B   1  .20  .8 .
2.
In problem 1, A and B are
a. Complements
b. Mutually exclusive
c. Collectively exhaustive
d. Independent
e.* None of the above.
3.
The following table shows the probabilities of the Whizbang Corporation’s stock movement:
Economic
Stock Price
Stock Price
Conditions
Rises
Falls
Good
.30
.03
__
Fair
.20
__
__
Poor
__
.22
__
Total
__
.45
Complete the table and then find the probability that the stock price goes up given that economic
conditions are fair.
a. .550
b.* .500
c. .200
d. .220
e. .364
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Solution: Here is the completed table.
Economic
Conditions
Good
Fair
Poor
Total
Stock Price
Rises
.30
.20
.05
.55
Using the notation in the next problem,
4.
PU F  
Stock Price
Falls
.03
.20
.22
.45
PU  F  .20

 .500
P F 
.40
Total
.33
.40
.27
1.00
Let us name the events in the previous problem as follows: G Economic Conditions Good; F
Economic Conditions Fair; Po Economic Conditions Poor; U Stock price goes up and D Stock
price goes down. Then the probability that the stock price is going down if conditions are poor is
a. P D  Po

D Po
PPo D

b. * P
c.


d. P D  Po
e. None of the above.
5.
In the two problems above, the joint probability of a falling stock price and fair economic
conditions is:
a.* P D  F  .20
b.
c.
d.
e.


PD  F   .20
PD  F   .50
PD  F   .65
PD  F   .65
6.
If the probability that a child is a boy or a girl is equally likely and I have 3 children, what is the
probability that at least one is a girl?
a. .750
b. .375
c.* .875
d. .625
e. .667
Solution: The probability of at least one girl could be the sum of the probabilities of 1, 2 and 3 girls.
To see how to do it this way look at the examples where we found the probabilities of 1, 2 or three


heads. To do it right, find P BBB  1  PBBB   1  .5  .5  .5  1  .125  .875 .
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7.
A factory has three veeblefetzers. Veeblefetzer A produces defective items 3 percent of the time
and thus is only used for 10% of the output. Veeblefetzer B produces 20% of the output, and 2% of
its output is defective. Because only 1% of the output of Veeblefetzer C is defective, 70% of the
output is produced on Veeblefetzer C. If a defective item is found, the probability that it comes
from Veeblefetzer C is closest to
a.* 50%
b. 40%
c. 30%
d. 20%
e. 10%
Solution: Here is the completed table.
Total
A
B
C
0.3
0.4
0.7
1.4
D
9.7
19.6
69.3
98.6
D
Total
10
20
70
100
To get the ‘total’ row, take 100 and multiply it by the percents given for production by each machine.
To get the D row, multiply the total by the percent defective. For example, 3% of 10 is 0.3. Get the
total defective by adding the row. Then the fraction of defective items that come from C is 0.7 out
P A  .10, PB  .20, PC   .70,
PD A  .03, PD B   .02 and PD C   .01 . You were asked for PC D . By Bayes’ rule,
of 1.4 or 50%. Formally, you were given the following:
PC D  
PD C PC 
.01.70  .007  0.50
, where we found PD using
PD 
.014
.014
PD  PD  A  PD  B  PD  C   PD AP A  PD BPB  PD C PC 

 .03 .10  .02 .20  .01 .70  .003  .004  .007  .014
8.
In problem 7, the proportion of output that comes from Veeblefetzer C and is not defective is
a. Between 90% and 100%
b. Between 80% and 90%
c. Between 70% and 80%
d.* Between 60% and 70%
e. Between 50% and 60%
Solution: From the table, 69.3 out of 100 units. Formally, since
PD C   .01, PD C   .99 . So
PD  C   PD C  PC   .99 .70  .693
9. The event A has a probability of p ; the event B also has a probability of p . If the two events
are independent
P A  B   2 p  p 2
b. P A  B  2 p
2
c.* P A  B   2 p  p
d. P A  B  0
2
e. P A  B   2  p  p
Solution: From the addition rule P A  B  P A  PB  P A  B . P A  p ,
PB  p and, since A and B are independent, P A  B   P APB   p 2 . So
a.
P A  B   p  p  p 2  2 p  p 2 .
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10. If the events A and B each have probabilities above zero and below 1, A and B are not
independent,
PA B   .6 and PB A  .5 , then the following must be true:.
P A  B is below .5 and above zero.
b. P A  B is between .5 and .6.
a.
c.* Either a. or b. could be true.
d. P A  B is 1
e. There is not enough information to answer this question.


Solution: From the multiplication rule,
P A  B  PA B PB  and
P A  B  PB  A  PB A P A . Since P A and PB  are both less than 1,
P A  B must be less than both PA B  and PB A.
11. The following table shows the number of days absent among your employees in a month and the
probability.
Number of days absent
Probability
0
.60
1
.20
2
.12
3
.04
4
.04
5
0
What is the expected value (mean) of the number of days absent?
a. 0.20.
b.* 0.72
c. 2.50
d. 2.00
e. 1.00
Solution: Here is the completed table for both the definitional and the computational formula as
generated by Minitab. There is no reason to bother with the definitional method.



 
row
xP x x Px  x   x   P x x    Px 
x
Px
1
0
0.60
0.00
0.00 -0.72
-0.4320
0.311040
2
1
0.20
0.20
0.20
0.28
0.0560
0.015680
3
2
0.12
0.24
0.48
1.28
0.1536
0.196608
4
3
0.04
0.12
0.36
2.28
0.0912
0.207936
5
4
0.04
0.16
0.64
3.28
0.1312
0.430336
sum
1.00
0.72
1.68
0.0000
1.1610
We have a valid distribution since all the probabilities are between one and zero and add to 1. The formula
xPx   0.72 and
for a population variance for a distribution is  x2  E x 2   x2 , where   E x  
   x
Ex
2
2
 
2
Px   1.68 . So
  1.68  0.72  1.1616
2
x
2
2

and
 x  1.1616  1.0778 .
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12. What is the value closest to the standard deviation of the number of days absent in problem 11?
a.* 1.08
b. 1.87.
c. 2.50
d. 1.16
e. 1.68
13. What is F 2 using the table in problem 10?
a. 1.00
b. .08.
c. .20
d.* .92
e. None of the above. For credit, write in an answer, showing your work.
Solution: You have been asked for the probability that x is less than or equal to 2.

F 2  Px  2
 Px  0  Px  1  Px  2  .6  .2  .12  .92
14. If I know that eight workers are available to fill four openings, in how many ways can I fill them?
a. 4098
b.* 70
c. 32
d. 1680
e. 65536
Solution: Since order is not important and C r 
n

n!
8!
8!
8

, C4 
8  4!4! 4!4!
n  r !r!
8  7  6  5 1680

 70 .
4  3  2 1
24
15. The number of ways that 4 items can be taken from 13 if order is important and replacement is not
allowed is
a. 0
b. 1
c. 15444
d. 715
e.* None of the above. For credit, write in an answer, showing your work.
Solution: Since order is important and Pr 
n
n!
13!
13!
13
, P4 

n  r !
13  4 9!
 13 12 1110  17160 .
16. If x is a standardized variable (z-score) and y  2  8 x , the expected value of y is
a.* 2
b. 8
c. -6
d. -8
e. None of the above. For credit, write in an answer, showing your work
Solution: Since if y  ax  b,
 y  E  y   aE x   b
and
 y2  Var y   a 2Varx and a
a  8 and b  2 and
2
compute  y  E  y   8 E  x   2  80   2  2 and   Var  y    8 1  64
standardized variable has a mean of zero and a standard deviation of 1, let
2
y
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17. If x is a standardized variable (z-score) and y  2  8 x , the standard deviation of y is
a. 36
b. -8
c. -6
d.* 8
e. None of the above. For credit, write in an answer, showing your work
Solution: From the above
 y2  Var  y    82 1  64 , so  y  64  8 .
18. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that as
long as one of the two components is working, the jorcillator works. If the probability of the
phillinx failing in the first month of service is .5, and the probability of the phillinx failing in the
second month is .4, while the probability of the flubberall failing in the first month is .4 and
probability of the flubberall failing in the second month is .4, what is the probability that both
components fail in the first month? Assume that the failure of one component is independent of
failure of the other.
a. 0
b.* .20
c. .90
d. .70
e. None of the above. For credit, write in an answer, showing your work.
Solution: The problem asks nothing about the jorcillator. Let A1 be the probability that the phillinx
A2 be the probability that the phillinx fails in the second period, and let A3
be the probability that the phillinx fails in the third period (after the first and second period). Let B1 be
the probability that the flubberall fails in the first period, let B2 be the probability that the flubberall
fails in the second period, and let B3 be the probability that the flubberall fails in the third period. The
events are independent. Then we have been asked for P A1  B1   P A1 PB1 
 .5 .4  .20 .
fails in the first period, let
19. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that as
long as one of the two components is working, the jorcillator works. If the probability of the
phillinx failing in the first month of service is .5, and the probability of the phillinx failing in the
second month is .4, while the probability of the flubberall failing in the first month is .4 and
probability of the flubberall failing in the second month is .4, what is the probability that the
jorcillator will last beyond 2 months? Assume that the failure of one component is independent of
failure of the other.
a. .02
b. .30
c.* .28
d. .32
e. None of the above. For credit, write in an answer, showing your work
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Solution: The problem asks for the probability that the jorcillator lasts beyond 2 months. We already
know that P A1  B1  P A1 P B1  .5 .4  .20 . Make the following table.
Total
A
A
A


   
B1
B2
B3
The event
A1
 
3
1
2
.20
.16
.04
0.40
.20
.16
.04
0.40
.10
.08
.02
0.20
Total
.50
.40
.10
1.00
 B3 , for example will down the jorcillator in the third period because as long as one of
the two components is working, the jorcillator works. If we add together all the events that down the
jorcillator in the third period, we find
P A1  B3   P A2  B3   P A3  B3   P A3  B2   P A3  B1 
 .10  .08  .02  .04  .04  .28 . This turns out to be the same as
P A3  B3   P A3   PB3   P A3  B3   .1  .2  .1.2  .28
20. If A and B are complementary events:
a. P A  B  0


P A  B  1
P A  1  PB
b.
c.
d.* All of the above.
e. None of the above
Part II. Show your work!
The number of days of absence last month for a sample of 6 employees appears below. (This is a sample not
a probability distribution!) Compute the sample standard deviation (5 Points - 2 Point Penalty for not
trying.)
Number of
x2
Days x
0
0
4
16
0
0
5
25
16
256
1
1
26
298
 
So
 x 26
 x  26,  x  298 and n  6 . x  n  6  4.333
 x  nx  298  64.333  37.06667 s  37.06667  6.08824

2
2
s2
n 1
2
2
5
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Part III. Do at least 1 (5+ Points) of the two following problems: Show your work! You receive extra
credit for extra work!
1. You are dealt four 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 etc). Let A be the event that you get three diamonds and
B the event that you get one heart.
a. How many possible hands are there? (2) Work out answer to a) . You may leave the answers
to the remaining sections of this question in factorial form.
n!
,
n  r !r!
52  51 50  49 6497400
52!
52!




 270725
52  4!4! 48!4!
4  3  2 1
24
Solution: C r 
n
C 452
b.
Solution:
c.
Solution:
P A (2)
P A 
C313C139
C 452
 13!  39! 



10!3!  38!1! 



 52! 


 48!4! 
 13 12 11   39 

 

 3  2 1   1   11154  .0412
270725
270725
P A  B (2)
P A  B 
13
3
13
1
52
4
C C
C
 13!  13! 



10!3!  12!1! 



 52! 


 48!4! 
 13 12  11   13 

  
 3  2 1   1   3718  .01373
270725
270725
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2. We are trying to sell residential real estate. Let A be the event that the selling price of the property is at
least 98% of the asking price. P A  .25 . . Let B be the event that the property takes more than 3
months to sell. If the selling price is at least 98% of the asking price, there is a 30% probability that the
property takes more than 3 months to sell. If the selling price is less than 98% of the asking price, there is a
40% probability that the property takes more than 3 months to sell.
 
a. Find
PB 
 A  .25 , PB A  .30 and PB A   .40
Solution: Since P
, it must be true that
PA   .75 and
for 100 properties, we can make the following table
B
A
A
B
7.5
30
37.5
17.5
45
25
75
62.5
100
 
P A  B  PB A P A  .30 .25  .075 , PA  B   P B A PA   .40 .75  .30 , so
PB   P A  B   PA  B   .075  .30  .375 .
b. If the property is on the market for more than 3 months, what is the probability that it will sell
for more than 98% of the asking price? (5)
 
7.5
 .20 , or by Bayes’ rule,
37.5
PB A P A .30 .25
P A B  

 .20
P B 
.375
Solution: From the table, P A B 
9