251y0341 05/05/03
ECO 251 QBA1
FINAL EXAM
MAY 7, 2003
Name KEY
Class ________________
Part I. Do all the Following (14 Points) Make Diagrams! Show your work!
x ~ N 3, 4 . How many of you really believe that a probability can be negative?
1.85  3 
  2.30  3
z
 P 1.32  z  0.29 
1. P2.30  x  1.85   P 
4
4 

 P1.32  z  0  P0.29  z  0  .4066  .1141  .2925
or  P1.33  z  0  P0.29  z  0  ..4082  .1141  .2941
or , better,  P1.325  z  0  P0.29  z  0  .4074  .1141  .2933
17  3 
1  3
z
 P 0.50  z  3.50 
2. P1  x  17   P 
4
4 

 P0.50  z  0  P0  z  3.50   .1915  .4998  .6913
1.85  3 

 Pz  0.29   P 0.29  z  0  Pz  0
3. Px  1.85   P  z 
4 

 .1141  .5  .6141
4  3

4. F 4.00  (Cumulative Probability)  Px  4  P  z 
4 

 Pz  0.25   Pz  0  P0  z  0.25   .5  .0987  .5987
4  3
 4 3
z
 P 1.75  z  0.25 
5. P4.00  x  4.00   P 
4 
 4
 P1.75  z  0  P0  z  0.25   .4599  .0987  .5586
6. x.015 (Find z .015 first) Make a diagram for z . Show a Normal curve with a mean of zero in its center.
Remember that z .015 is a point with 1.5% above it and 98.5% below it. Since 50% of the distribution is
bellow zero P0  z  z.015   .9850  .5  .4850 . According to the Normal table P0  z  2.17   .4850 .
So z .015  2.17 and x    z.015  3  2.17 4  3  8.68  11.68 .
11 .68  3 

Check: Px  11 .68   P  z 
  Pz  2.17   Pz  0  P0  z  2.17   .5  .4850  .0150
4


7. A symmetrical region around the mean with a probability of 25%. Make a diagram for z . Show a
Normal curve with a mean of zero in its center. If we split 25% in two, we get two areas, one on either side
of the mean with probabilities of 12.5%. We can call the point we want z .375 , because, since the area above
zero is 50%, the area above z .375 must be 50% - 12.5% = 37.5%. But we have already decided that the
probability between z .375 and zero is 12.5%. The closest we can come on the Normal table is
P0  z  0.32   .1255 , so z .375  0.32 and x    z.375  3  0.32 4  3  1.28 or 1.72 to 4.28,
4.28  3 
1.72  3
z
 P 0.32  z  0.32   2 P0  z  0.32   2.1255 
Check: P1.72  x  4.28   P 
4 
 4
 .2510
Exam is normed on 75 points. There are actually 128 possible points.
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II. (10 points+-2 point penalty for not trying part a .) Show your work!
The following numbers apply to 9 developed countries and give deaths per 100 million miles and speed
limits. (xy was not supplied and is calculated in red.)
Row
deaths
SpLim
y
x
1
2
3
4
5
6
7
8
9
3.1
3.4
3.5
3.6
4.2
4.4
4.8
5.0
6.2
xy
55
55
55
70
55
60
55
60
75
170.5
187.0
192.5
252.0
231.0
264.0
264.0
300.0
465.0
2326.0
These sums have been calculated for you.
y
2
 x  38.2 ,  x
2
 169 .86 ,
 y  540 and
 32850 . Please calculate the following:
a. The sample standard deviation of x (4) Note that y  60.00 and s y  7.50 .
Many of you wasted time and energy computing the x squared and y squared columns and then decided that
you could get the xy sum by multiplying the sums of x and y. Where have you been?
b. The sample covariance between x and y . (3)
c. The sample correlation between x and y . (2)
d. Given the size and sign of the correlation, what conclusion might you draw on the relation
between speed and safety if this were the only evidence available? (1)
e. Assume that the death rate in all 9 countries fell by .1. What would be the new values of
x , s x , s xy and rxy . Use only the values you computed in a-c and rules for functions of x and y to get
your results. If you state the results without explaining why, or change x and recompute the results, you
will receive no credit. (4). How many of you recomputed the results anyway?
Solution: a) x 
 x  38.2  4.24444
n
9
sx  0.9653  0.9825 .
y
 y  540  60.00
n
9
b) We found above that
c) rxy 
s xy
sx sy

s x2 

s 2y 
y
2
 ny 2
n 1
x

xy  2326 , so s xy 
4.25
 .5768
0.9825 7.50 
2
 nx 2
n 1

169 .86  94.24444 2 7.72256

 0.9653
8
8
32850  960 .00 2 450

 56 .25
8
8
 xy  nxy
n 1
2

s y  56.25  7.50
2326  94.24444 60 .00  34

 4.25
8
8
This must be between -1 and 1.
d) If we square the correlation we get 0. 333, which in a zero to one scale is not impressive. I would not
want to conclude that speed limits and safety are closely related, though speed limits may be a factor.
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e) From the syllabus supplement article:
“Let us introduce two new variables, w and v , so that w  ax  b , and v  cy  d , where
a, b, c, and d are constants. From the earlier part of this section we know the following:
 w2  Varw  a 2Varx  a 2 x2
 w  Ew  Eax  b  aEx   b
 v2  Var v   c 2Var  y   c 2 y2
 v  E v   E cy  d   cE y   d
To this we now add a new rule: Covw, v   wv  acCovx, y   ac xy
To find the correlation between w and v , recall that  wv 
 w2  a 2 x2
 wv 
 wv
. But since
 w v
and  v2  c 2 y2 , then
ac xy
a 2 x2 c 2 y2

ac xy
ac  x y

ac  xy
 signac xy .”
ac  x y
Since these rules work for sample statistics too, then w  x  0.1 , v  y , so a  1, b  0.1,
c  1 and d  0. u  x  .1  4.24444  .1  4.24444
s wv  11s xy  4.25 . rwv 
acs xy
a 2 s x2 c 2 s 2y

acs xy
ac s x s y
s w2  12 s x2  s x2 , so s w  56 .25  7.50 .

ac s xy
 signacrxy  .5768
ac s x s y
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III. Do at least 4 of the following 6 Problems (at least 12 each) (or do sections adding to at least 48 points Anything extra you do helps, and grades wrap around) . Show your work! Please indicate clearly what
sections of the problem you are answering! If you are following a rule like E ax  aEx  please state it! If
you are using a formula, state it! If you answer a 'yes' or 'no' question, explain why! If you are using the
Poisson or Binomial table, state things like n , p or the mean. Avoid crossing out answers that you think
are inappropriate - you might get partial credit. Choose the problems that you do carefully – most of us are
unlikely to be able to do more than half of the entire possible credit in this section!)
1. Assume that the amount of paid time (in days) lost by a blue-collar worker during a 3-month period is
N 1.4, 1.3 . I take a random sample of 10 workers and record the time they lost in the last 3 months..
a. What is the probability that a randomly picked worker lost paid time exceeding 1.5 days in the
3-month period? (2)
b. What is the probability that all 10 workers in the sample lost paid time exceeding 1.5 days in
the 3-month period? (2)
c. What is the probability that at least one of the workers in the sample lost paid time exceeding
1.5 days in the 3-month period? (2)
d. What is the probability that the average amount of paid time lost time exceeded 1.5 days in the
three month period? (2)
e. What is the probability that the total amount of time lost by the sample of 10 workers exceeded
15 days in the three month period. (2)
f. Looking at the distribution of the sample mean in this problem, give a value of the sample
mean that will be above the mean we actually observe 95% of the time (the 95th percentile) (2)
Solution: a) x ~ N 1.4,1.3 Make a diagram.
1.5  1.4 

Px  1.5  P  z 
 Pz  0.08   Pz  0  P0  z  0.08   .5  .0319  .4681
1.3 

b) Binomial n  10, p  .4681 . P10   .4681 10  .0005051 .
c) Binomial n  10, p  .4681 . Px  1  1  P0  1  1  .4681 10  1  .0018126  .99819 .

1.3 
  N 1.4, 0.411 
d) x ~ N  ,  x   N 1.4,

10 

1.5  1.4 

Px  1.5  P  z 
 Pz  0.24   Pz  0   P0  z  0.24   .5  .0948  .4052
0.411 

e) Answer is the same as for d because if the total time for 10 workers was 15, the mean was 15 divided by
10.
f) According to the t table, the 95th percentile for z was z .05  1.645 . x ~ N  ,  x   N 1.4, 0.411  , so
x    z.05 x  1.4  1.645 0.411   2.076 .
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2. (Bowerman and O’Connell) A retailer that sells home entertainment systems accumulated 10,451 sales
invoices during the last year.
a. An auditor takes a sample of 16 invoices and computes mean sales of x  $532 . If the
population standard deviation was known to be $168, find a 99% confidence interval for the
mean sales per invoice. (4)
b. I lied. Though the sample mean was $532, $168 was a sample standard deviation. Do the
99% confidence interval again. (4)
c. I lied. Though it is true that the sample mean was $532 and the sample standard deviation was
$168, the actual sample was 650 invoices out of the 10451 invoices that were collected. Do the
99% confidence interval again. (4)
d. (Extra credit) Assume that the confidence interval in c is correct, and that the 10451 invoices
were all that were generated, using these two facts, create a confidence interval for total sales in
the last year. (3)
e. (Extra credit) The firm claims that its total sales were above $5.75 million last year. In view of
your results in d, does that seem likely? Would you change your mind if I insisted on a
confidence level of 99.8%? (3)
f. Using the data in a) create a 97% Confidence interval for the mean sales per invoice. (You
might want to look at page 1.) (2)
Solution: The solution to problem O1 gives the following formulas:
i)   x  z  x and  x 
2
x
when  is known and the sample is small relative to the
n
population. ii)   x  z  x and  x 
2
x
n
N n
when  is known and the sample is large relative to
N 1
s
the population. iii)   x  tn1 s x and s x  x
2
n
when  is unknown and the sample is small relative
s
N n
to the population and iv)   x  tn1 s x and s x  x
when  is unknown and the sample is
2
n N 1
large relative to the population.
So a) x  $532,   168 , n  16 and N is more than 20 times 16.

168
x  x 
 42 . z  z.005  2.576   532  2.576 42   532  108 or P424    640   .99 .
2
n
16
b) x  $532, s  168 , n  16 and N is more than 20 times 16.
s
168
sx  x 
 42 . t n 1  t .15
005  2.947 .   532  2.947 42   532  124 or P408    656   .99 .

n
16
2
c) x  $532, s  168 , n  650 and N  10451 , which is less than 20 times the sample size.
168 2 9801 
N n
168 10451  650


 40 .7248  6.38 . t n 1  t .649
005  2.576 .

650 10450 
10451  1
2
n N 1
650
  532  2.576 6.38   532  16.43 or P515 .57    548 .43  .99.
The only place to find z is the very bottom of the t table, unless you have to figure it out as in part f. On the
other hand, if you need t, you need the rest of the t table. If you don’t know the difference between sigma
and s, please find out!
d) Let’s say the average sales invoice was $548.43, then 10451 customs would generate $10451(548.43)
= $5,731,642. Likewise if average sales were $515.57, 10451 customers would generate $5,388,222. We
can say P5388222  Total sales  5731642   .99.
sx 
sx
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e) Obviously the high estimate in the last interval was below $5.75 million. If we wanted to use a
confidence level of 99.8%, we would use t n 1  t .649
001  3.092 .   532  3.092 6.38   532  19 .77 . The

2
upper limit is now $551.77. If we multiply this by 10451, we get $5,756,097, which might make us change
our minds.
f) On page 1 we found z .015  2.17 , which is the value of z  that we would use if   .03 . So we have
2
x  $532,   168 , n  16 and N is unspecified and assumed much larger than 16.  x 
z 2  z.015  2.17   532  2.17 42   532  91.
x
n

168
 42 .
16
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3. a. Assume that the entire amount of a product made by a supplier is a population of 100 units and that
you buy the whole batch. Assume that 15% of the batch is defective. Take a sample of 5 items and give me
the probability that at least one is defective. (3)
b. Assume that the batch you buy is much larger, well over 200 and that you still take a sample of 5. What is
the chance that at least one is defective? (You should not need to use the number 200 or any larger number
in your calculations.) (2)
c. Assume that you have bought at least a million units, and that 15% still represents the proportion of the
product that is defective. This time you take a sample of 80. Find the probability that at least 10 are
defective using the Poisson distribution. First, show that it is legitimate to use the Poisson distribution in
this case. (3)
d. Do part c using the Normal distribution. That is assume that you have a large population that is 15%
defective and that you take a sample of 80. Show that the Normal distribution can be used here and find the
probability of at least 10 defective items in the sample. (3 points without continuity correction, 3.5 with)
e. (Extra credit) In section 7.3 of the text (on the CD) the author tells us we should use a finite population
correction with the variance if it is justified. Assume that in part d, the population is 200 and we take a
sample of 80, what is the probability of at least 10 defective items in the sample now? (2)
f. If we are taking a sample from a large population, find the probability that the first defective item is
between the 6th to the 10th item we test. (2)
Solution: At least half of you ignored the fact that I changed the sample size from 10 to 5; the only penalty
was that it made the problem easier if you made the change. Much of this problem was a repeat of the last
take-home exam.
Most of you seem to have completely forgotten the following:
(i) If you are looking for numbers of successes when the number of tries is given and
the probability of success is constant, you want the Binomial distribution.
(ii) If you are looking for the try on which the first success occurs out of many possible
tries when the probability of success is constant, you want the Geometric distribution.
(iii) If you are looking for numbers of successes when the number of tries is given and
the average number of successes per unit time or space is given, you want the Poisson
distribution.
(iv) If you are looking for numbers of successes when the number of tries is given and
the probability of success is not constant because the total number of successes in the
population is limited, you want the Hypergeometric distribution.
a) Hypergeometric n  5, N  100 , p  .15 and M  p100   15 are defective.
Px  1  1  P0 Px  
1  P0   1 
C xM C nNxM
C 010 C 585
C 5100
Crn 
C nN
 1
n!
n  r !r!
 85! 

 80! 5! 
1
 100! 


 95! 5! 
85  84  83  82  81
5  4  3  21
 1
100  99  98  97  96
5  4  3  21
85  84  83  82  81
 1  .43568  .5643
100  99  98  97  96
b) Binomial n  5 and p  .15
Px  1  1  P0  1  .44371  .55625 .
 1
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251y0341 05/05/03
c) We can use the Poisson distribution if
n
n 80

 533 .33  500
 500 . In this case
p .15
p
Px  10   1  Px  9  1  .24239  .7576 From table.
m  np  80.15   12
d) We can use the Normal distribution if expected successes np and expected failures nq are both above
five. In this case n  80, p  .15 q  1  p  1  .15  .85  , so   np  80.15   12  5 and
nq  80.85   80  12  68  5 and the Normal distribution is valid.   npq  12 .85   10 .2
10  12 

 3.1937 . So without the continuity correction Px  10   P z 
  Pz  0.63 
3.1937 

 P0.63  z  0  Pz  0  .5  .2357  .7357 , and with the continuity correction
9.5  12 

Px  9.5  P z 
  Pz  0.78   P 0.78  z  0  Pz  0  .5  .2823  .7823 .
3.1937 

N n
200  80
120
npq 
12 .85  
10 .2  6.15075  2.4801
N 1
200  1
199
and we are doing a normal approximation to the Hypergeometric distribution. So without the continuity
10  12 

correction Px  10   P z 
  Pz  0.81  P 0.81  z  0  Pz  0  .5  .2910  .7910 and
2.4801 

e) n  80, N  200 , p  .15 .  
9.5  12 

with the continuity correction Px  9.5  P z 
  Pz  1.01  P 1.01  z  0  Pz  0
2.4801 

 .5  .3438  .8438 .
f) This has the geometric distribution with p  .15 q  1  p  1  .15  .85  . F x  1  q x .


P6  x  10   Px  10   Px  5  F 10   F 5  1  .85 10  1  .85 5  .85 5  .85 10
 .44371  .19687  .24987 .
8
251y0341 05/05/03
4. As everyone knows, a jorcillator has two components, a Phillinx and a Flubberall. It seems that the
jorcillator only works as long as both components work.
a. The distribution of failure times for the Phillinx is Normal with a mean of 8 months and a standard
deviation of 2 months.
(i) What is the probability that the Phillinx dies in the first six months Px  6 ?
(ii) What is the probability that the Phillinx dies in months 6 – 12 ?
(iii) What is the probability that the Phillinx lasts beyond 12 months? (4 total)
b. The distribution of the failure times for the Flubberall is described by the continuous uniform
distribution between 0 and 11.
(i) What is the probability that the Flubberall dies in the first six months?
(ii) What is the probability that the Flubberall dies in months 6-12?
(iii) What is the probability that the Flubberall lasts beyond 12 months?
(iv) What is the mean and standard deviation of the Fluberall’s life? (4.5 total)
c. Now let’s see if you learned anything about combining these probabilities.
(i) What is the probability that the jorcillator will fail in the first six months?
(ii) What is the probability that the jorcillator will fail in the second six months?
(iii) What is the probability that the jorcillator will last beyond 12 months? (5 total)
d. Rework (c) assuming that the jorcillator works as long as one component works. (6)
Solution: a) x ~ N 8, 2 Make a diagram. Draw a Normal curve with a mean at 8. Represent Event B1
by the area below 6, Event A2 by the area between 6 and 12 and Event A3 by the area above 12.
6 8

 Pz  1.00 
(i) Event A1 : The Phillinx dies in the first six months. Px  6  P  z 
2 

 Pz  0  P1.00  z  0  .5  .3413  .1587 .
12  8 
6 8
z
(ii) Event A2 : The Phillinx dies in months 6 – 12 . P6  x  12   P 
2 
 2
 P1.00  z  2.00   P1.00  z  0  P0  z  2.00   .3413  .4772  .8185 .
12  8 

 Pz  2.00 
(iii) Event A3 : The Phillinx lasts beyond 12 months. Px  12   P  z 
2 

 Pz  0  P0  z  2.00   .5  .4772  .0228 .
b) y is continuous uniform c  0 and d  11 . Make a diagram. Draw a box between 0 and 11 with a
1
1

. Represent Event B1 by the area below 6 and Event B 2 by the area above 6.
d  c 11  0
Event B 3 is not in the box.
height of
 60 
(i) Event B1 : The probability that the Flubberall dies in the first six months. P y  6   

11  0 
 .54545
(ii) Event B 2 : What is the probability that the Flubberall dies in months 6-12? P6  y  12 
 11  8 

  .45455 .
11  0 
(iii) Event B3 : What is the probability that the Flubberall lasts beyond 12 months? Px  12   0.
(iv)  
c  d 0  11

 5.5.  
2
2
d  c 2
12

11  02
12
 10 .0833  3.175
9
251y0341 05/05/03
c, d) Now we put this all together in a humongous table. From here on, c is just a rerun of a take-home exam
problem.
Period of Failure
Joint Event
If both components If only one
Probability
are needed to keep component is
it working
needed to keep it
working.
0 – 6
0 – 6
.1587(.54545) = .0866
A1 B1
A1 B2
A1 B3
A2 B1
A2 B 2
A2 B3
A3 B1
A3 B2
A3 B3
0
0
0
6
6
0
6
>
– 6
– 6
– 6
– 12
– 12
– 6
- 12
12
6
>
6
6
>
>
>
>
– 12
12
– 12
– 12
12
12
12
12
.1587(.45455)
.1587(0) = 0
.8185(.54545)
.8185(.45455)
.8185(0) = 0
.0228(.54545)
.0228(.45455)
.0228(0) = 0
= .0721
= .4465
= .3720
= .0124
= .0104
c) So we get the following if both components are required:
(i) Probability that the jorcillator will fail in the first six months = .0866 + .0721 + 0 + .4465 + .0124 =
.6176
(ii) Probability that the jorcillator will fail in the second six months = .4465 + 0 + .0104 = .3824
(iii) Probability that the jorcillator will last beyond 12 months = 0
Note that these add to one.
d) So we get the following if only one component is required:
(i) Probability that the jorcillator will fail in the first six months = .0866
(ii) Probability that the jorcillator will fail in the second six months = .0721 + .4465 + .3720 = .8906
(iii) Probability that the jorcillator will last beyond 12 months = 0 + 0 + .0124 + .0104 + 0 = .0228
Note that these add to one.
10
251y0341 05/05/03
5. I am a sales representative and I have two visits scheduled for this morning. Let A represent the event
that I make a sale on visit 1 and B represent the event that I make a sale on visit 2.
A A
a. The joint probability table for the two events is
B
.4
B
and the distribution table
.7
w Pw
0
for w , the total number of sales is 1
2
.
Fill in these tables on the assumption that A and B are independent (How many of you ignored
independence?) and find the expected value and standard deviation for w . (5)
Solution: Because A and B are independent, P A  B   P A PB   .4.7  .28 . We also know that
 
P A  P A  1 , so we have
B
A
.28
A
.4
B
.6
. If we just fill in the rest, we get
B
B
A
.28
A
.12
.4
.42
.7
.18
.3
.6
1 .0
.7
.3 1.0
Now if we use this information to fill in the distribution, we note that, first, probabilities must be between
zero and one, second, that the probabilities in a valid distribution must add to one, and third,
Pw  0  P A  B , Pw  1  P A  B  P A  B and Pw  3  P A  B .


 


w Pw wPw w Pw
0 .18
0
0
2
2
1 .54
.54
.54 . We get   1.10 and   1.66  1.10   0.45 .
2 .28
.56
1.12
1.00 1.10
1.66
2
Finally   0.45  0.6708.
b. Fill in the two tables again on the assumption that A and B are collectively exhaustive.
w Pw
A A
0
B
.4
and compute the expected value and variance for
1
B
2
.7
w . (4)


Solution: Because A and B are collectively exhaustive, P A  B   1, and its complement P A  B  0 .
A
If we use this information we get
A
B
.4
0
B
.7
.6
. Now just fill in the blanks.
.3 1.0
11
251y0341 05/05/03
B
B
A
.10
A
.30
.60
.7
0
.3
w Pw wPw w 2 Pw
0
0
0
0
.4
and do the distribution table. 1 .90
.90
.90 . We get   1.10 and
.6
2 .10
.20
.40
1.0
1.00 1.10
1.30
 2  1.30  1.10 2  0.09 . Finally   0.09  0.3 .
c. Let x represent the number of sales I make on visit 1 ( x can only be 0 or 1) and y represent
the number of sales I make on visit 2. What relation must exist between the variances of x and
y and the variance of w in the case where A and B are independent that cannot exist when
they are mutually exclusive? Why? (2).
Solution: In general w  x  y and Varw  Varx   Var y   2Covx, y  . But if x and y are
independent Covx, y   0. So Varw  Varx   Var y  . I was afraid that I’d get questioned on this, so I
actually did a general proof of this. Don’t read it unless you really like to think about this stuff. Assume that
P A  Px  1  a  1 , that PB   P y  1  b  1 , that a  b  1 , and that A and B are mutually
exclusive. The joint probability table and computation of the means etc follow.
x
0
0 1  a  b 
1 
b
1  a 
Px 
xPx 
0
y
x 2 Px 
1 P y  yP y  y 2 P y 
a 1  b
0
0
0
b
b
b . Thus  x  E x   a, E x 2  a,  y  E y   b,
a
1.0
b
b
a
a
 
0 a
a
   b, and we can see that Exy  1  a  b00  a10  b01  011  0 . So we can say
Varx  Ex    a  a , Var  y   E y     b  b and
Ey
2
2
2
x
2
2
2
y
2
Covx, y   E xy    x  y  0  ab  ab . So if w  x  y ,
Varw  Varx   Var y   2Covx, y   a  a 2  b  b 2  2ab  a  b  a  b2 .
d. Assume that PB   .4 . Use the addition rule to show that A and B cannot be both
collectively exhaustive and independent if A has a probability below 1. (3)
Solution: If A and B are collectively exhaustive P A  B   1 , PB   .4 and if A and B are
independent, P A  B  P A PB . So we have
P A  B  P A  PB  P A  B   P A  PB   P APB   1 . This means that
P A  B   P A  PB   P A  B   P A  .4  P A.4  1 . But this can only be true if P A  1 .
More generally, 1  PB  P A  P APB . If we factor out P A , we get 1  PB   P A 1  PB ,
which can only be true if P A  1. If we do this with PB   .4 , we get 1  .4  P A  P A.4 or
.6  .6 P A . If we divide through by 0.6, we get 1  P A .
12
251y0341 05/05/03
e. Let us define the following events S1  No Oil , S 2  Some Oil and S 3  Much Oil . Assume
PS1   .7 , PS 2   .2 and PS 3   .1 . We run a seismic experiment that has three possible
readings H (high), M (medium) and L (low). All you have to know for this problem is
P H S1  .04 , P H S 2  .02 and P H S 3  .96








(i) Explain the difference between P H S 3 and PH  S 3  and show how we get the
second of these from the first. (3)
(ii) Find PH  (2)



(iii) Find P S1 H (4)

Solution: (i) P H S 3 is a conditional probability which means that it is the probability of getting a ‘high’
reading when it is true that there is much oil. PH  S 3  is the joint probability of both getting a ‘high’
reading and much oil out of all 9 possible combinations of readings and oil. The multiplication rule says
that PH  S 3   P H S 3 PS 3   .96.1  .096 .


(ii) If we get a ‘high’ reading, it must be true that there is either no oil, some oil or much oil. Thus it must be
true that PH   PH  S1   PH  S 2   PH  S 3 
 PH S1 PS1   PH S 2 PS 2   PH S 3 PS 3   .04.7  .02.2  .96.1  .028  .004  .096  .128 .
(iii) We had to bring in Bayes’ rule sometime! PS1 H  
P H S1 PS1 
P H 

.04 .07  .028

 .21875 .
.128
.128
Another way of looking at this is to assume that we check 1000 locations. Then 70% or 700 will have no
oil, 20% or 200 will have some oil and 10% or 100 will have much oil. P H S1  .04 means that of the




700 locations with no oil, 4% or 28 will give a ‘high’ reading. P H S 2  .02 means that of the 200


locations with some oil, 2% or 4 will give a ‘high’ reading. P H S 3  .96 means that of the 100 locations
with much oil, 96% or 96 will give a ‘high’ reading . We thus get a total of 28 + 4 + 96 = 128 ‘high’
28
 .21875 .
readings. 28 of these locations have no oil, so P S1 H  
128
13
251y0341 05/05/03
6. The Phillies are in the 2003 World Series. I estimate that they have a constant .6 chance of winning a
game. There are seven games in a series and the series stops if one team wins four games.
a. If there are seven games played what is the mean and variance of the number of games the
Phillies win? (2)
b. What is the chance that they will win at least 4 of the seven games (You can assume all seven
games are played)? (2)
c. What is the chance that they will win the series by winning the first four games? (1)
d. What is the chance that the first game that they win is the third game? (2)
e. What is the chance that they win the series on the fifth game (but not the 4th)? (Ask yourself
what has to happen in the first 4 games so that they do not win their 4th game until the 5th try.)
(3)
f. What is the chance that they lose the first three games and win the series? (2)
g. Let x represent the number of the first game they win (so that Px  3 is the probability that
the first game that they win is the third game). What is the mean and standard deviation of x ?
(2)
Solution: a) This starts out as a binomial problem with n  7 and p  .6. q  1  .6  .4 For this
distribution   np  7.6  0.42 .  2  npq  7.6.4  0.42.4  .168 . This starts out as the simplest
binomial problem available, 7 tries with a constant probability of .6. Why did so many of you have to make
it more complicated?
b) We want the probability of 4 or more games won in 7 tries. It is probably easier to think of 4 to 7 wins.
Since we do not have Binomial table for probabilities above .5, we must recast the problem in terms of
failures. 4 successes in 7 tries would be 3 failures and 7 successes would be no failures. We want the
probability of 3 or fewer failures when the probability of failure is .4. According to the table
Px  3  .82080 .
c) This is a binomial problem too. But it is easier to just figure out the probability of the intersection of four
independent events. .64  .1296 .
d) This is a geometric distribution problem. Px  q x1 p  .4 2 .6  .0960
e) In order to win on the 5th game we must win exactly 3 of the first four games and then win again.. First
we find the probability of winning 3 out of 4 games. If we want to do this by hand, Px  C xn p x q n x so
P3  C 34 p 3 q 1  4.63 .4  .3456 . Using the table, we want the probability of 1 failure in 4 tries when the
chance of failure is .4. P1  Px  1  Px  0  .4752  .1296  .3456 . To finish, we must multiply this
by the .6 probability of success. .6.3456   .20736 .
f) To lose 3 games and win the series, we need 3 failures followed by 4 successes. The probability is
q 3 p 4 .43 .64  .0082944 .
g) This is the mean and standard deviation of the geometric distribution. The outline says
q
1
1
.04

 11 .1111  3.3333
 
 6.6667 .  
2
p .15
p
.06 2
14