251y0721 3/23/07
ECO251 QBA1
SECOND HOUR EXAM
March 28, 2007
Name: _____KEY______________
Student Number: _____________________
Hour of Class Registered: Circle MWF10, MWF11
There will be a penalty if you do not provide all this information!
Part I: (3 points) 2 point penalty for not trying.
A set of 5 speed tests were made for a new automobile model, with the following results.
115 116 114 113 114
Find the standard deviation of the speed. For your convenience the sum of the first four numbers is
4
4

x  458 and the sum of the first four numbers squared is
x
2
 52446 .
i 1
i 1
Solution: This can be written out as the tableau below.
x2
x  x 
x  x 2
13225
13456
12996
12769
12996
65442
0.6
1.6
-0.4
-1.4
-0.4
0.0
0.36
2.56
0.16
1.96
0.16
5.20
x
1
2
3
4
5
115
116
114
113
114
572
4

You were given
x  458 , so
i 1
x
2
n 5,

x  458  114   572 . You were also given
4
x
2
 52446 , so
i 1
 52446  114 2  65442 . If you used these numbers, the computations above were not needed.
 x  572 and  x
2
 65442 , So x 
 x  572  114 .4 . If you have wasted your time
n
5
 x  x   0 (a check) and  x  x 
1
1
 x  n  x  65442  5 572 
computing the x  x  and x  x  columns, you have
2
In any case, s

 x  x 
n 1
2


2

x 2  nx 2
n 1
65442  5114 .42


4
2
2
2
n 1
 5.20 .
2

4
5.20
 1.30 and s  1.30  1.140 .
4
1
251y0721 3/23/07
Part II: (50 points) Do most of the following: All questions are 2 points each except as marked. Exam is
normed on 50 points including take-home. [Bracketed numbers are a point total.] If you answer ‘None of
the above’ in most questions, you should provide an alternative answer and explain why. You may receive
credit for this even if you are wrong. This section is long and few people will finish. You only need 35
points for a perfect score, so look them over! As always any extra points on this exam wrap around.
1. If events A and B are independent and their probabilities are not zero, the following cannot be true. (3)
a) A and B are complements.
b) A and B are mutually exclusive.
c) P A  B   0
d) All of the above could be true
e) *None of the above can be true
f) Not enough information
Explanation: A and B are complements if P A  B   1 and A and B are mutually exclusive. If A and B
are mutually exclusive, P A  B   0 . But we have seen in class that if events A and B are independent,
P A  B   P APB  . But this product can only be zero if P A or PB  is zero.
Exhibit 1: Consider the Following table. It refers to employees of a corporation and their absences. The
events are: A1 'No Absences', A2 '1-2 Absences', A3 '3-4 Absences', A4 'Over 4 Absences', B1 'Age below
Event
B1 B 2 B3

.250 .125 .075 
A1


A2
.080 .140 .065 
21', B2 'Age 21-35', B3 'Age above 35'.
.040 .060 .060 
A3


.030 .025

A4


2. Fill in the missing probability inside the table.
We know that the numbers in the table must total 1. If we look at the row totals, 1 - .450 - .285 - .160 =
.105. To make the total of A4 equal to .105, the missing number must be .105 - .030 - .025 = .050. We can
check this by finishing the column totals.
Event
B1 B 2 B3
Event
B1 B 2 B3

.250 .125 .075  .450
.250 .125 .075  .450
A1
A1




A2
A2
.080 .140 .065  .285
.080 .140 .065  .285
.040 .060 .060  .160
.040 .060 .060  .160
A3
A3




.030 .025

A4
A4
.
030
.
025
.
050

 .105



.400 .350 .250 1.000
.400 .350
3. Find the joint probability of A1 and B1 in exhibit 1.
a. P A1 B1  .250


b.* PA1 B1  .250
c. PA1 B1  .625
d. PA1 B1  .250


e. P A1 B1  .625
f. PA1 B1  .600
g. None of the above – fill in correct answer.
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251y0721 3/23/07
Exhibit 1: Consider the Following table. It refers to employees of a corporation and their absences. The
events are: A1 'No Absences', A2 '1-2 Absences', A3 '3-4 Absences', A4 'Over 4 Absences', B1 'Age below
21', B2 'Age 21-35', B3 'Age above 35'. I have put in the table with missing items added.
Event
B1 B 2 B3
.250 .125 .075  .450
A1


A2
.080 .140 .065  .285
.040 .060 .060  .160
A3


A4
.030 .025 .050  .105
.400 .350 .250 1.000
4. Find the conditional probability of A1 given B1 in exhibit 1.
a. P A1 B1  .250


b. PA1 B1  .250
c. PA1 B1  .625
d. PA1 B1  .250


e. * P A1 B1  .625
f. PA1 B1  .600
g. None of the above – fill in correct answer.
 
Explanation: The Multiplication Rule says P A B 
PA1 B1  .250 so PA1 B1 
P A  B 
. Because this is a joint probability table,
P B 
P A1  B1 .250

 .625
PB1
.400
5. Find the probability of A1 or B1 in exhibit 1.
a. P A1 B1  .250


b. PA1 B1  .250
c. PA1 B1  .625
d. PA1 B1  .250


e. P A1 B1  .625
f. * PA1 B1  .600
g. None of the above – fill in correct answer.
Explanation: The Addition Rule says PA1 B1  P A1  PB1  P A1  B1  .450  .400  .250 .
6. What is the probability that someone above 35 has no absences in exhibit 1 – fill in a number. [12]
Solution: The events are: A1 'No Absences', A2 '1-2 Absences', A3 '3-4 Absences', A4 'Over 4 Absences',
P A1  B3 .075
B1 'Age below 21', B2 'Age 21-35', B3 'Age above 35'. So we want P A1 B3 

 .300 .
PB3
.250


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251y0721 3/23/07
Exhibit 2: 20 per cent of Airplane crashes involve bad weather and 45 per cent of crashes involve
mechanical failures. Among crashes that involve bad weather, the proportion that also involves mechanical
failure is 80 per cent.
7. In exhibit 2, the proportion of crashes that involve both mechanical failure and bad weather is:
a. .80
b. .49
c. * .16
d. .09
e. None of the above. Fill in a correct solution.
Explanation: Let W= 'Bad Weather" and F= 'Equipment Failure'. The problem says 3
things: PW   .20, PF   .45, PF W   .80 . The problem asks for PF  W  . By the Multiplication Rule,
P F  W   P F W  PW    .80 .20 .16
8. In Exhibit 2, the proportion of crashes that involve neither bad weather nor mechanical failure is:
a. *.51
b. .44
c. .84
d. .49
e. None of the above. Fill in a correct solution.
[16]
Explanation: If we use the notation above, the problem asks for P F  W . We have shown in class that






the event F  W is the complement of F  W  , so P F  W  1  PF  W 
 1  PF   PW   PF  W   1  .45.20.16  1.49 .51 . A diagram would help. At the end of
Question 7, we had
W
F F
.16
W
.45
W
W
F F
.16 .04
.20
. If we just fill in the numbers so that they add up, we get
1.00
.20
.29 .51 .80
.45 .55 1.00
9. In Exhibit 2, what is the proportion of crashes that involve mechanical failure that also involve bad
weather. (3)
[19]
We were given PF W   .80 . The question is asking for PW F  . We can treat this as a Bayes’ Rule
problem and say PW F  
PF W PW 
P F 

.80 .20 
 .2844 or we can look at the table and say
.45
PW  F  .16

 .2844 .
PF 
.45
10. The number of samples of 6 that can be taken from a population of 10 (Assuming that order is not
important) is:
a. 5040
b. 1 million
c.* 210
d. 10
e. None of the above – fill in correct answer.
[21]
10
!
10

9

8

7
5040


 210
Explanation: C 610 
4!6! 4  3  2 1
24
PW F  
4
251y0721 3/23/07
Exhibit 3: Mansfield gives us the following data. x is the weight of a car in thousands of pounds and y is
the miles per gallon.
Row
1
2
3
4
5
6
x
2.6
3.6
3.8
2.9
3.2
3.6
19.7
y
26
18
23
38
21
22
148
From these numbers, I have computed the following sums and sample statistics.
x
2
 65 .77 ,
y
2
 x  19 .7,  y  148 ,
 3898 , x  3.2833 , y  24.6667 , s x2  0.217667 and s 2y  49 .466667 . Please
do not waste time duplicating these calculations.
 xy . (3)
11. Using the data in Exhibit 3, Compute
Solution: Row
1
2
3
4
5
6
x
2.6
3.6
3.8
2.9
3.2
3.6
19.7
y
xy
26
18
23
38
21
22
148
67.6
64.8
87.4
110.2
67.2
79.2
476.4
[24]
So
 xy  476 .4
12. Compute the sample covariance between weight and miles per gallon. What conclusion can you draw
19 .7
xy  476 .4 . I did the following for you. Then x 
from it? (2) Solution: We just found
 3.2833 ,
6

y
148
 24 .6667 , s x2 
6
s 2y 
y
2
 ny
n 1
2

x
2
 nx 2
n 1

65 .77  63.2833 2
 0.2179 (Minitab says 0.217667) and
5
3898  624 .6667 2
 49 .4647 (Minitab says 49.466667). Now you can compute
5
 x  x  y  y    xy  nx y
476 .4  63.2833 24 .6667 
 1.9058 . This is negative, so x
5
n 1
n 1
tends to fall as y rises. No conclusion about the strength of the relationship is possible without variances.
[26]
13. Compute the sample correlation between weight and miles per gallon. What can you say about its
strength? (3)
[29]
s xy
 1.9058

 0.5808 . The square of this is .3373 on a zero to one
Solution: rxy 
sx s y
0.217667 49 .466667
scale. This is not very strong, but neither is it negligible.
s xy 

5
251y0721 3/23/07
14. If the cost of a car in thousands of dollars is described by C  11x  0.5 , using only the numbers I
computed and your results in questions 11-13, find: (4)
[33]
Explanation: We know that if w  ax  b and v  cy  d , Varw  a 2Varx , Covw, v   acCovx, y  ,
and Corr w, v   Sign(ac)Corr x, y  . If our w  C  11x  0.5 and v  1y  0 , a  11 and c  1 .
a. s C2 , the sample variance of cost.
s x2  0.217667 . So Var C   112 Var x   1210.217667   26 .3377 .
b. s Cy , the covariance between cost and miles per gallon.
s xy  1.9058 . So CovC, y   111Covxy   111.9058   20.9638 .
c. rCy , The correlation between cost and miles per gallon.
rxy  0.5808. CorrC, y   Sign(111)Corrx, y   10.5808   0.5808 .
Please do not waste our time by writing out a column of costs and using it and the column of miles
per gallon to compute the statistics in question 14.
15. A machine has three components. Each has a 30 per cent chance of failure during the next month.
(Assume that failure of one component is independent of failure of another.) If the device will work as long
as at least one of the three components works, its chance of failure during the month is (rounded to two
decimal places): (2.5)
a. 30%
b. *3%
c. 90%
d. 66%
e. 72%
f. None of the above.
Explanation: The machine only fails if all components fail. Because of the independence assumption, we
can multiply the probabilities to get .30 3  .027
16. A machine has three components. Each has a 30 per cent chance of failure during the next month.
(Assume that failure of one component is independent of failure of another.) If all three components must
work to get the device through the month, its chance of failure during the month is (rounded to two decimal
places): (2.5)
[38]
a. 30%
b. 3%
c. 90%
d. * 66%
e. 72%
f. None of the above.
Explanation: The machine only works if all components work. If there is a 30% chance of a component
failing, there is a 70% chance of the component working. Because of the independence assumption, we can
multiply the probabilities to get the probability that all three components work is .70 3 . The probability of
failure is thus 1  .70 3  .657 .
6
251y0721 3/23/07
17. Assume that x , y , and z are independent random variables. If we have the following information:
y
x
z
Mean
7
9
11
Standard
3
4
8
deviation
Let v  x  y  z . Find the mean and standard deviation of v. (3) Show your work!
[41]
Solution: We can add means and we can add variances of independent random variables. So the mean is
Ev   E x   E  y   E z   7  9  11  27 . The variance is
Varv  Varx  Var y   Varz   32  4 2  8 2  9  14  64  89 . The standard deviation is
89  9.4340 .
x
0 7

Exhibit 4:
1 .12 .18 
y


9 .28 .42 
18. Check the joint probability table in exhibit 4 for independence (2). Solution: We can expand this as
x
y
1
9
Px 
xPx 
0 7
.12 .18 


.28 .42 
.40 .60
0 4.2
P y  yP y  y 2 P y 
.30
.3
.3
.70
6.3
56 .7 . But note that .12 = .30(.40) and so on. All the numbers
1.00
6.6
57 .0
4.2
x 2 Px 
0 29 .4 29 .4
inside the table are products of the probabilities outside the table, so we can say that the two random
variables are independent.
19. Compute E  y  and Var y  in exhibit 4. (2.5)
 Px  1 ,   Ex   xPx  4.2 E x    x
 E  y    yP y   6.6 and E y    y P y   57 .0 . Then 
2
To summarize
y
2
x
2
2
y
 
Px   29 .4
 Ey 
 y2  E y 2   y2  57.0  6.62  13.44 .
 
 P y   1 ,
 yP y   6.6 and
 
20. Compute Covx, y  or  xy and Corr x, y  or  xy in exhibit 4 (2.5)
Since x and y are independent, their covariance and correlation must be zero. If you really like to work,
 
2
however,  x2  E x 2   x2  29 .4  4.2  11 .76 .
E xy  
 .12 01
 xyPxy    .2809
 .187 1 0  1.26 

 27 .72 .
 .42 7 9 0 26 .46 
This means that  xy  Covxy   Exy   x  y  27.72  4.26.6  0 and
 xy 
 xy
 x y

0
0 .
11 .76 13 .44
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251y0721 3/23/07


21. Find Px  y  8 and P x  0 x  y  8 in exhibit 4. (2)
x
0 7

1 .12 .18 
y


9 .28 .42 
[50]
x
0 7

.The values of x  y are
1 1 8
y


9 9 16 
, so the first probability
sought is Px  0, y  1  Px  7, y  1  .12  ..18  .30. To do the second part, remember the
 
Multiplication Rule says P A B 
Px  0  x  y  8
.12
P A  B 
 .40 .
. 
=
.30


P
x

y

8
P B 
8
251y0721 3/23/07
ECO251 QBA1
SECOND EXAM
March 28, 2007
TAKE HOME SECTION
Name: _________________________
Student Number: _________________________
Throughout this exam show your work! Neatness counts! Please indicate clearly what sections of the
problem you are answering and what formulas you are using.
Part II. Do all the Following (12 Points) Show your work!
1. Seymour Butz’s student number is 976500
Take the following set of numbers:
100
200
300
400
500
600
700
and use your Student Number to provide the last digit of the last six numbers, so that, for Seymour, the
numbers would become 100, 209, 307, 406, 505, 600, 700. Compute a sample standard deviation for the
resulting numbers. Do not invent any probabilities. (2- 2point penalty for not doing)
Solution: If you are in a hurry compute columns (1) and (2) then find the sample mean and use the
x
1417792  7403 .857 2
 46014 .6108 . If time is not
n 1
6
important, compute columns (1), (3) and (4) and use the definitional formula
computational formula. s 2 
s
2
 x  x 

n 1
(1)

(3)
2
 x  x 2
-303.857 92329.1
-194.857 37969.3
-96.857
9381.3
2.143
4.6
101.143 10229.9
196.143 38472.1
296.143 87700.7
0.001 276087.0
 x  2827 ,  x
 x  x 

(4)
xx
2
100
10000
209
43681
307
94249
406 164836
505 255025
600 360000
700 490000
2827 1417791
n  7,
 nx 2
276087 .665
 46014 .6108
6
(2)
x
x
1
2
3
4
5
6
7
2
2
 276087 .0 s 2 
2
 1417792 so x 
x
2
 nx 2
n 1

2827
 403 .857 ,
7
 x  x   0
276087 .665
1417792  7403 .857 2
 46014 .6108

6
6
s  46014.6108  214.51
9
251y0721 3/23/07
2. Take exactly the same numbers as before and assume that they have the following probabilities: .05, .07,
.08, .10, .10, .12, .48. Show that this represents a valid distribution and compute the population standard
deviation from the distribution. (2). Find Px  500  and the median of the distribution. (1)
Seymour would now have:
x
P x 
100
.05
209
.07
307
.08
406
.10
505
.10
600
.12
700
.48
Solution: If you are in a hurry compute columns (1), (2), (3) and (4) then find the mean
  E x  
xPx   543 .29 and use the computational formula.
2

 E x  
2
2
 331484  543 .29 2  36319 .9759 . If time is not important, compute columns (1), (2),
(3), (5), (6) and (7) and use the definitional formula  2  E x   2 
(1)
x
1
2
3
4
5
6
7
100
209
307
406
505
600
700
(2)
(3)
(4)
P x 
xPx 
x P x 
0.05
0.07
0.08
0.10
0.10
0.12
0.48
1.00
5.00
14.63
24.56
40.60
50.50
72.00
336.00
543.29
500
3058
7540
16484
25503
43200
235200
331484
2
(5)
x   
-443.29
-334.29
-236.29
-137.29
-38.29
56.71
156.71
(6)
 x   
2
Px  36319.7 .
(7)
x   Px x   2 Px 
-22.1645
-23.4003
-18.9032
-13.7290
-3.8290
6.8052
75.2208
0.0000
9825.3
7822.5
4466.6
1884.9
146.6
385.9
11787.9
36319.7
Note that to show that this is a valid distribution, we should check to see that all probabilities are between
zero and one and that the sum of column (2) is one.
xPx   543 .29 , E x 2 
x 2 Px   331484 ,
Px   1 ,   E x  


 x   Px   0 ,   Ex   
 E x    331484  543 .29   36319 .9759  
E x    
2
  
  x   
2
2
2
Px  36319.7
36319.9  190.578
Px  500   .05  .07  .08  .10  .30 . The median is 600, since there are approximately equal
probabilities on either side.
[5]
2
2
2
10
251y0721 3/23/07
3. As everyone knows, a jorcillator has two components, a phillinx and a flubberall. It seems that the
jorcillator only works as long as one component works (so that it fails in the first month only if both
components fail).
The probability of the phillinx failing is given by a continuous uniform distribution between c1 and
d1 with c1  1 and d1  4  0.2 g , where g is the last digit of your student number. For example, if
the life of the phillinx is represented by x1 , then the chance of the phillinx failing in the 3 rd month is
P2  x1  3 and the probability of it failing after the third month is Px1  3 . For example: Ima
Badrisk has the number 375292, so the top of her distribution is d1  4  0.22  4.4 . Depending on
your number d1 could go up to 5.
The probability of the flubberall failing is given by the continuous uniform distribution between c2
and d 2 with c2  .9  0.1g and d 2  4 . For example: Ima Badrisk has the number 375292, so the
bottom of her distribution is c 2  .9  0.052  .8 . If x2 represents the life of the flubberall, the
probability of the flubberall failing in the first month is P0  x2  1 , the probability of the flubberall
failing in the second month is P1  x2  2 etc. We will divide time into four periods with the last
period being ‘beyond the third month.’ The jorcillator is guaranteed to fail in one of the four periods.
Failure of components is assumed to be independent, so if the probability of the phillinx failing in the
first month is .1, and the probability of the flubberall failing in the first month is .7, the probability of
both components failing in the first month is (.1) (.7) =.07 (This is not the necessarily the probability
that the jorcillator will fail in the first month!)
In order to maintain my sanity, use the following events.
Failure of the phillinx in period 1, 2, 3, 4 are events A1, A2 , A3 , and A4 .
Failure of the flubberall in period 1, 2, 3, 4 are events B1, B2 , B3 , and B4 .
Failure of the jorcillator in period 1, 2, 3, 4 are events C1, C2 , C3 , and C 4 .
a) What is the probability that the phillinx will fail in month 1? Month 2? Month 3? After Month
3? (1.5)
b) What are the mean and standard deviation of the philinx’s failure time? (1.5)
c) What is the probability that the jorcillator will fail in the first month? (2)
d) What is the probability that the jorcillator will fail in the second month? (1)
e) What is the probability that the jorcillator will fail in the third month? (1)
f) What is the probability that the jorcillator will last beyond 3 months? (1)
g) Find the probability that the jorcillator and the Phillinx both fail in the second month (1)
h) Find the probability that the phillinx fails in the second month, given that the jorcillator fails in
the second month i.e. P A C (1)
 
i) Demonstrate Bayes’ rule by showing how to get the probability that the jorcillator fails in the
second month, given that the phillinx fails in the second month, from your result in h). (1)
Solution: Phillinx has a Continuous Uniform distribution between 1 and 4.4. Flubberall has a
Continuous Uniform distribution between .8 and 4
Each component has probabilities that add to 1. The 16 possible joint events must have probabilities that
add to 1.
For the Phillinx, make a diagram. Show a box between 1 and 4.4 with an area of 1 and a height of
1
1
c  d 1  4.4

 .2941 . b) According to “Great Distributions”  

 2.7 ,
4 . 4  1 3. 4
2
2
2 
d  c2  4.4  12
 0.9633 . So   0.9633  0.9815
12
12
Divide the box into the four events by putting vertical lines at 1, 2 and 3. Note that the first area has no
height so that its probability is zero.
11
251y0721 3/23/07
a), c), d), f).
Events
A1 P0  x  1  0
A2 P1  x  2  1.2941   .2941
A3 P2  x  3  1.2941   .2941
A4 . Px  3  4.4  3.2941   .4117 These add to .9999
For the flubberall, make a diagram. Show a box between .8 and 4 with a height of
1
1

 .3125 .
4  0. 8 3 . 2
Divide the box into the four events by putting vertical lines at 1, 2 and 3.
B1 P0  x  1  1  .8.3125  = .0625
B 2 P1  x  2  1.3125   .3125
B 3 P2  x  3  1.3125   .3125
B4 Px  3  4  3.3125   .3125
Joint Event
A1 B1
Probability
0
When Fails
Period 1
A1
A1
A1
A2
A2
A2
A2
A3
A3
A3
A3
A4
A4
A4
A4
0
Period 2
0
Period 3
0
Period 4
B2
B3
B4
B1
B2
B3
B4
B1
B2
B3
B4
B1
B2
B3
B4
.2941  .0625  = .01838
.2941  .3125  = .09191
.2941  .3125  = .09191
.2941  .3125  = .09191
.2941  .0625  = .01838
.2941  .3125  = .09191
.2941  .3125  = .09191
.2941  .3125  = .09191
.4117  .0625  = .02573
.4117  .3125  = .12866
.4117  .3125  = .12866
.4117  .3125  = .12866
Period 2
Period 2
Answer to g)
Period 3
Period 4
Period 3
Period 3
Period 3
Period 4
Period 4
Period 4
Period 4
Period 4
c) What is the probability that the jorcillator will fail in the first month? (2)
PC1   P A1  B1   0
d) What is the probability that the jorcillator will fail in the second month? (1)
This event is the union of A1 B 2 , A2 B1 and A2 B 2 . Its probability is PC 2   0 + .01838 +
.09191 = .11029.
e) What is the probability that the jorcillator will fail in the third month? (1)
This event is the union of A2 B 3 , A3 B1 , A3 B 2 and A3 B 3 . Its probability is PC3 
= .09191 + .01838 + .09191 + .09191 = .29411
f) What is the probability that the jorcillator will last beyond 3 months? (1) If either component
lasts for 4 months, the machine will last for four months P A4  B4   .4117  .3125  .12866
 .59554 This is also the union of A2 B4 , A3 B4 , A4 B1 , A4 B 2 , A4 B 3 and A4 B4
PC 4   .09191 + .09191 + .02573 + .12866 + .12866 + .12866 = .59553.
12
251y0721 3/23/07
g) Find the probability that the jorcillator and the Phillinx both fail in the second month (1)
We already know that failure in the second month is the union of A1 B 2 , A2 B1 and A2 B 2 , and
its probability is PC 2   0 + .01838 + .09191 = .11029. A2 is the event that contains failure of
the Phillinx in the second month. It is contained in the joint events A2 B1 and A2 B 2 the
probability of the union of these two events is P A2  C 2   P A2  B1   P A2  B2 
= .01838 + .09191= .11029.
h) Find the probability that the phillinx fails in the second month, given that the jorcillator fails in
P A2  C 2 
the second month i.e. P A2 C 2 (1) Use the multiplication rule PA2 C 2  
1
PC 2 
i) Demonstrate Bayes’ rule by showing how to get the probability that the jorcillator fails in the
second month, given that the phillinx fails in the second month, from your result in h). (1)
P A2   .2941 , PC 2   .11029 and P A2 C2  1 We can use the multiplication rule to say



PC 2 A2  


P A2  C 2  .11029

 .37501 . According to Bayes’ Rule P C 2 A2 
P A2 
.2941
PA2 C 2 PC 2 
P A2 



1.11029 
 .37501
.2941
\
13