251y0622 3/10/06
ECO251 QBA1
SECOND HOUR EXAM
March 22 and 23, 2006
Name: _____KEY____________
Student Number: _____________________
Hour of Class Registered: Circle MWF1, MWF2, TR12:30, TR2:00
There will be a penalty if you do not provide all this information!
Part I: (3 points) 2 point penalty for not trying.
The scores of a sample of statistics exams are (with extra credit included) as below. Find the sample
standard deviation. Show your work!
96 85 105 86 103 101 97 94 75 98
Solution: Row
x
x2
1
96
9216
2
85
7225
3 105 11025
4
86
7396
5 103 10609
6 101 10201
7
97
9409
8
94
8836
9
75
5625
10
98
9604
Sum 940 89146
x
 x  940  94.00
n
s2 
x
10
2
 nx 2
=
89146  10 94 2
=
9
n 1
786
 87 .3333 s  87.3333  9.3452
9
————— 3/10/2006 4:37:45 PM ————————————————————
Welcome to Minitab, press F1 for help.
MTB > describe x
Descriptive Statistics: x
Variable
N N*
Mean
x
10
0 94.00
MTB > let c2 = c1*c1
MTB > sum c1
Sum of x
Sum of x = 940
SE Mean
2.96
StDev
9.35
#How to fake this on Minitab.
Minimum
75.00
Q1
85.75
Median
96.50
Q3
101.50
Maximum
105.00
#Square x
MTB > sum c2
Sum of xsq
Sum of xsq = 89146
MTB > let k1 = stdev(x)
MTB > let k1 = k1*k1
MTB > print k1
Data Display
K1
87.3333
#Square k1 to get variance.
MTB > print c1 c2
Data Display
Row
x
xsq
1
96
9216
2
85
7225
3 105 11025
4
86
7396
5 103 10609
6 101 10201
7
97
9409
8
94
8836
9
75
5625
10
98
9604
1
251y0622 3/10/06
Part II: (35 points+) Do all 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 any question, you should provide an alternative answer and explain why. You may receive credit
for this even if you are wrong.
1.
If events A and B are independent and their probabilities are not zero, the following cannot be true.
a) A and B are mutually exclusive.
b) A and B are complements.
c) *Neither of the above can be true.
d) Both of the above can be true.
Explanation: For independent events P A  B   P A PB  . For mutually exclusive events
P A  B   0 . Since it is impossible for both of these to be true a) cannot be true. If A and B
are complements, however, they must be mutually exclusive. So if a) is false, b) must be false
too.
2.
v1. If events A and B are independent and P A  .5 and PB   .3 what is P A  B  ?
a) .80.
b) .15
c) .20
d) *.65
e) None of the above.
f) Not enough information to tell.
Explanation: For independent events P A  B   P A PB  . By the addition rule
P A  B   P A  PB   P A  B   .5  .3  .5.3  .8  .15  .65.
3.
 
If P A  .3, PB   .5 and P B A  .4 what is P A  B  ?
a) .80.
b) .15
c) *.12
d) .20
e) None of the above.
[6]
Explanation: By the multiplication rule PA  B  PB  A  P B A  P A  .4.3  .12 .
 
Can you explain why any answer to a problem like 2 and 3 that says P A  B   P A ,
P A  B   PB  , P A  B   P A or P A  B   PB  is unreasonable?
2
251y0622 3/10/06
Questions 4-7 are based on exhibit 1.
Exhibit 1: A fast food restaurant has its outlets distributed as follows. (Dummeldinger). A restaurant is to
be chosen at random to test market a new menu item.
City
Population
4.
Below
10000 A
10000100000 B
Above
100000 C
NE
.05
Region
SE
.06
.15
.20
.03
SW
.03
NW
.00
.13
.05
.10
.05
Fill in the missing number (1) Solution: If we add up all the rows and columns and then put in a
NE SE SW NW total
A
.05 .06 .03 .00
.14
number that makes the probabilities add to 1, we get
.48 . So the
B
.15 .15 .13 .05
C
.20 .03 .10 .05
.38
total
.40 .24 .26 .10
1.00
missing number is .15.
5.
What is the probability that, if we select a restaurant at random, we pick a restaurant that is in a
city of above 100000 and is in the Northeast. (2.5)
a) * PC  NE 
b) PC  NE 
c)
d)
e)
6.
PC NE 
PNE C 
None of the above.
And the probability is ___.20______.
What is the probability that, if we select a restaurant at random, we pick a restaurant that is in a
city of above 100000 or one that is in the Northeast. (2.5)
a) PC  NE 
b) * PC  NE 
c)
PC NE 
PNE C 
e) None of the above.
And the probability is _____.58__________. Explanation: By the addition rule
P A  B   P A  PB   P A  B  . So PC  NE   PC   PNE   PC  NE 
 .38  .40  .20  .58. But it is also true that
PC  NE   PC  NE   PB  NE   P A  NE   PC  SE   PC  SW   PC  NW 
 .20  .15  .05  .03  .10  .05  .58 .
d)
7.
What is the probability that, if we select a restaurant in a city in the northeast at random, we pick a
restaurant that is in a city above 100000. (2.5)
a) PC  NE 
b)
PC  NE 
c)
* P C NE
d)

PNE C 

And the probability is ___.5____ Explanation: By the multiplication rule
PC  NE  .20
P A  B 

 .5
. So PC NE  
P AB 
PNE 
.40
P B 
 
[14.5]
3
251y0622 3/10/06
8.
You and two of your friends are shooting arrows at a target. Each of you has an independent
probability of hitting the target of 80%. What is the probability that all three of you will hit the
target? ? Solution: By the multiplication rule for independent events P A  B   P A PB  ,
so that P A  B  C   P A PB  PC  . If P A1   P A2   P A3  then
P A1  A2  A3   P A1 3 If these probabilities are the probabilities of your posse hitting the
target. P A1  A2  A3   .83  .5120 .
9.
You and two of your friends are shooting arrows at a target. Each of you has an independent
probability of hitting the target of 80%. What is the probability that at least one of you will hit the

  3  1  P A1 3 .
target? Solution: By same logic as above P A1  A2  A3  P A1


Since
1  P A1   1  .8  .2 , we can say P A1  A2  A3  .2  .0080 . The compliment of this is the

3

probability that at least one hits the target. 1  P A1  A2  A3  1  .23  1  .0080  .9920 .
If you have a lot of time you could use the extended addition rule. P A  B  C 
 P A  PB   PC   P A  B   P A  C   PB  C   P A  B  C  . This means that
P A1  A2  A3 
 P A1   P A2   P A3   P A1  A2   P A1  A3   P A2  A3   P A1  A2  A3 
 3.8  3.82  .83  2.4  1.920  .5120  .9920
[18.5]
10. You and two of your friends are shooting arrows at a target. Each of you has an independent
probability of hitting the target of 80%. Let x be the number of individuals that hit the target. Do
the following:
a) Find the distribution of x . (Show what values x can take, what the probability of each
value is, and demonstrate that we have a valid distribution.) (6) If you cannot do this
problem you may make up a reasonable distribution to do b), c) and d). Warning – this
problem could take a while to do. If time is a constraint for you, skip it and come
back to it later.
b) Find F 2 , the cumulative probability ( Px  2 ) (2)
c) Find the expected value and variance of x . (3)
d) Using the mean and variance that you found in c) , If you and your buddies as a group
paid $8 to shoot and the operator of the game gives your group $3 for each hit on the
target, what is the expected value and variance of your winnings? (3) [32.5]
Solution: This solution was edited from problems J1 and J2. a) Let H be a hit and M be a miss. First,
using the multiplication rule for independent events, we know that the probability of a pattern with all hits
is, from Problem 8, PHHH   P A1  A2  A3   .83  .5120 . The probability of a pattern with all


misses was done in Problem 9 PMMM   P A1  A2  A3  .23  .0080 . If we continue to use the
multiplication rule, the probability of a pattern with only two hits like HHM is .8.8.2  .1280 . For this
particular pattern there are two hits, so x  2 . The probability of a pattern with one hit like HMM is
.8.2.2  .0320 . For this particular pattern there is one hit, so x  1 . But there are other ways to get 1 or
2. We have to find out how many. There are two ways to do this.
The easy way: We have to replace one or two locations in MMM
3!
 3 . The number of ways we can do
with H . The number of ways we can do this for one hit is C13 
2!1!
this for two is C 23 
3!
 3 , so we must multiply the probabilities we found above by 3.
1!2!
4
251y0622 3/10/06
The hard way: List all the possible patterns and add their probabilities. There are 8 patterns.
Regardless of how we do it, we get the following:
Pattern x Probabilit y
Probability
x Number of ways to
HHH 3
.5120
get this value
HHM 2
.1280
P0  1.0080 
3!
3
HMH 2
.1280
0 C 0  3! 0!  1
 .0080
HMM 1
.0320
P1  3 .0320 
3!
3
MHH 2
.1280
1 C1  2!1!  3
 .0960
MHM 1
.0320
P2  3.1280 
3!
3
MMH 1
.0320
2 C 2  1! 2!  3
 .3840
MMM 0
.0080
P3  1 .5120 
3!
3
You did P0 and P3 in questions 8 and 9!
3 C 3  1!3!  1
 .5120
Note that these probabilities add to one.
b) To find F 2 , the cumulative probability ( Px  2 ) , simply add the probabilities of 0, 1, 2
and 3 hits. F 2 = .0080 + .0960 + .3840 = 1 - .5120 = .4880.
c) Now we find the expected value and variance of x .
x
0
1
2
3
From this table
P x 
.0080
.0960
.3840
.5120
1.0000
x Px 
0.0000
0.0960
0.7680
1.5360
2.4000
x 2 P x 
0.0000
0.0960
1.5360
4.6080
6.2400
 Px   1.00 (valid distribution!),   E x    xPx   2.40.
 
Varx   x2  E x 2   2 
x
2
Px   2  6.24  2.402  0.48
(  x  0.48  0.6928 )
d) Now, using the mean and variance that we found in c) , if our posse as a group paid $8 to shoot and the
operator of the game gives your group $3 for each hit on the target, we must the expected value and
variance of your winnings? What this says is the winnings are y  3x  8 . Part of the table in your outline
appears below.
Rule
If y 
E y    y 
Var  y    y2 
Std.Dev. y    y 
4)
ax  b
aEx   b  a x  b
a 2Varx  a 2 x2
So a  3 and b  8 . E  y    y  3E x   8  32.4  8  0.80
Var  y    y2  32 Varx   3 2 0.48   4.32
 a Std.Dev.x   a  x
(  y  4.32  2.0785 )
5
251y0622 3/10/06
11. You are dealt 7 cards from a deck of 53 cards containing 4 each of 2s, 3s, 4s, 5s, 6s, 7s, 8s, 9s, 10s,
jacks, queens kings and aces, 13 each of hearts, diamonds, spades and clubs and one joker. What is
the probability that your hand contains the joker? (4) You must work this out all the way to get full
credit and at least show how many possible hands there are.
Solution: There are C753 possible hands. There is one way to get the joker and C652 ways to get the
52!
46!6! 52!46!7! 7
other cards.



 .1321 . There is another way to do this. Without using
53!
53!46!6! 53
46!7!
52
Cs, find the probability that you will not get a joker and subtract it from 1. On the first try this is
. If
53
51
. If we follow this through, we
you do not get a joker on the first try, the (conditional) probability is
52
52  51  50  49  48  47  46
46 7
1
  .1321 .
get 1 
53  52  51  50  49  48  47
53 53
C11C 652
C 753
53! 53  52  51  50  49  48  47

 154143080 , which may be useful below.
46!7!
7  6  5  4  3  2 1
12. You are dealt 7 cards from a deck of 53 cards containing 4 each of 1s, 2s, 3s, 4s, 5s, 6s, 7s, 8s, 9s,
10s, jacks, queens kings and aces, 13 each of hearts, diamonds, spades and clubs and one joker.
What is the probability that your hand contains three hearts? (2.5) This can be left in factorial
form.
Solution: There are C753 possible hands. There are C 313 ways to get the hearts and and C440 ways to get
13! 40!
C 313C 440 10!3! 36!4!
the other cards.
is an acceptable answer. If you want the whole megillah

53!
C 753
46!7!
13! 40!
13  12  11 40  39  38  37
C 313C 440 10!3! 36!4!
13  12  11  40  39  38  37 7  6  5
4  3  2 1

 3  2 1

53!
53  52  51  50  49  48  47 53  52  51  50  49  48  47 3  2  1
C 753
46!7!
7  6  5  4  3  2 1
 .004845 35   .169567
13. You are dealt 7 cards from a deck of 53 cards containing 4 each of 1s, 2s, 3s, 4s, 5s, 6s, 7s, 8s, 9s,
10s, jacks, queens kings and aces, 13 each of hearts, diamonds, spades and clubs and one joker.
What is the probability that your hand contains at least one heart? (2.5) This can be left in factorial
form.
[40.5]
Solution: There are C753 possible hands. ‘At least one’ means ‘not zero.’ There are C013  1 ways to get
40!
13! 40!
13 40
C C
13!0! 33!7!
33!7!
no hearts and and C740 ways to get the other cards. 1  0 537  1 
is an
1
53
!
53!
C7
46!7!
46!7!
acceptable answer.
Note that C753 
6
251y0622 3/10/06
If you want the whole megillah
13! 40!
40  39  38  37  36  35  34
C 013C 740 13!3! 33!7!
40  39  38  37  36  35  34
7  6  5  4  3  2 1



 .120950
53!
53  52  51  50  49  48  47 53  52  51  50  49  48  47
C 753
46!7!
7  6  5  4  3  2 1
So 1 
C 013C 640
C 653
 1  .167195  .8328048 .
14. Assume that z is a standardized variable and that y  3z  17 . The standard deviation of y is:
(2.5)
[43]
a) 14.
b) -3
c) 3
d) 9
e) 51
f) Not enough information.
g) None of the above.
x
Solution: A standardized variable is manipulated by the formula z 
so that E z    z  0 and

 z2  Varz   1.
Part of the table in your outline appears below.
Rule
If y 
E y    y 
4)
ax  b
aEx   b  a x  b
Var  y    y2 
a 2Varx  a 2 x2
Std.Dev. y    y 
 a Std.Dev.x   a  x
If y  3z  17 , a  3 and b  17 . I did nor ask for E  y    y   3E z   17  30  17  17 . What I
wanted was Var  y    y2   32 Varz    32 1  9
and  y  9  3 . If you want to try the
 
 
other formula, Std.Dev.z    z  1, so that Std.Dev. y    y   3 Std.Dev.z    3 1  3.
7
251y0622 3/10/06
ECO251 QBA1
SECOND EXAM
March 22 and 23, 2006
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. (5 points)
row
x Qx 
1
2
3
4
5
6
7
-4
-3
-2
-1
0
1
2
Make yourself an individualized distribution as
follows. Take the six digits of your student
number and add them to the numbers in Qx  .
Divide the result by 100 to get a column of
numbers and fill in the missing number to get a
valid distribution.
5
5
10
..
10
5
5
 stu 
  Q' x 
no 
x Qx 
row
1
2
3
4
5
6
7
-4
-3
-2
-1
0
1
2
5
5
10
..
10
5
5
9
9
9
.
9
9
9
14
14
19
..
19
14
14
For example, assume that my student number is
999999. I have written it in the column labeled
‘stu no’ and then added it to Qx  to get the
P x 
0.14
0.14
0.19
....
0.19
0.14
0.14
column Q' x  . Then I divided the result by 100
and filled in the missing number in the Px 
column.
a) Use the numbers in the x column and the probabilities in the Px  to compute the population standard
deviation of x . You will not get credit for this if you do not show your work.
b) I got the numbers below, which I will call y , by multiplying by -9 (negative 9) and subtracting 3 from
the result.
33
24
15
6
-3
-12
-21
y
Without using the numbers in y , find the expected value and standard deviation of y . You will not get
credit for this unless you show what formulas you use to get the results.
Solution:
a) row x
1
2
3
4
5
6
7
-4
-3
-2
-1
0
1
2
Px 
0.14
0.14
0.19
0.06
0.19
0.14
0.14
1.00
 x  E x  

x
2
x P x  x 2 P  x 
-0.56
-0.42
-0.38
-0.06
0.00
0.14
0.28
-1.00
2.24
1.26
0.76
0.06
0.00
0.14
0.56
5.02
 xPx   1.00
 
Varx    x2  E x 2  E x 2
Px   x2  5.02   1.002  4.02 .
 x  4.02  2.00499
8
251y0622 3/10/06
b) Since y  9x  3 . We can let a  9 and b  3 and use the formula table in the outline.
Rule
If y 
E y    y 
Var  y    y2 
Std.Dev  y    y 
4)
aEx   b  a x  b
ax  b
 a Std.Dev.x   a  x
a 2Varx  a 2 x2
So  y  E y   9Ex  3  91.00  3  9  3  6
Var y    y2   92 Varx   814.02  325.62  y  325.62  18.0447
Or Std.Dev  y    y   9  x  92.00499   18 .0449
2. (Bassett et. al.). (10 Points) The stock of a warehouse consists of bulbs of Type A, Type B and Type C in
separate unlabeled boxes. Let the last three digits of your student number be a, b, and c. The proportions of
the bulbs in our stock and the proportion that are defective are given by the table below.
Type
Proportion in stock
Proportion of this type that is defective.
A
.2
(0 + a)%
B
.4
(10 + b)%
C
.4
(20 + c)%
So if my student number is 123456, the proportions defective will be (0 + 4)% = .04, (10 + 5)% = .15 and
(20 + 6)% = .26. Hint: You might want to start with 1000 taken from 1000 boxes of bulbs. Your events
should be (types) A , B , C . Let the event that the bulb is defective be D. As above, you must show your
work if you want any credit.
a) What proportion of 1000 bulbs would be defective?
b) If I open a box and find a defective bulb, what is the probability that it comes from a box of type A?
Type B? Type C?
c) If I open a box and find a satisfactory bulb, what is the probability that it comes from a box of type A?
Type B? Type C?
d) Actually this problem as originally written asked about the probabilities above in the event we found two
satisfactory bulbs. So assume that the bulbs’ satisfactoriness are independent of one another, what is the
probability of finding two satisfactory bulbs of Type A, of type B and of Type C.
e) Now, assuming that we open a box and find two satisfactory bulbs, what is the probability that they come
from a box of type A?
Version 0: Assume that the student number is xxx000. We have P A  .2, PB   .4 and PC   .4 .
 
 
 
We also have P D A  .0, P D B  .10 and P D C  .20 . Let SS be the event that we find two
satisfactory bulbs.
The problem asks for: a) PD  b) P A D , P B D and P C D ; c) P A D P B D and P C D ;
   
d) PSS A, PSS B  and PSS C  and e) PA SS .


   
 
A B C total
The easiest way to do a) b) and c) is the box method. Assume 1000 bulbs.
D
D
total
A
B
D
Using P A  .2, PB   .4 and PC   .4 , divide up the bulbs.
D
total 200
C
1000
total
.
400
400 1000
9
251y0622 3/10/06
 
 
 
If P D A  .0, P D B  .10 and P D C  .20 , Then 0 200  0 of Type A, .10  400  40 of Type B and
.20  400  80
A
0
of Type C are defective. Fill these into the table and we have
B
40
C
80
total
A
B
C
total
D
D
0
40
80
120
. Now make the table add up.
.
D
D
200 360 320
880
total 200 400 400 1000
total 200 400 400 1000
120
3
0
40 1

 .1200 b) P A D  
 0, P B D  
  .3333 and
So we have for a) PD  
1000 25
120
120 3
80
2
200
5
360
9
P C D  
  .6667 . ; c) P A D 

 .2273 P B D 

 .4091 and
120 3
880 22
880 22
320
8
PCD 

 .3636 .
880 22
To do d) we must remember that if PD A  .0, PD B  .10 and PD C   .20 , and the event S is one
 
 
 
 
 
   
PS C   1  PD C   1  .20  .80. Because of the independence assumption,
PSS A  12  1, PSS B   .90 2  .81 and PSS C   .80 2  .64 .
satisfactory bulb, P S A  1  P D A  1, P S B  P D B  1  .10  .90 and
Now we can do e). Start out with our original box diagram, but change the Ds to SSs .
A
B
C
total
SS
. Apply P SS A  12  1, P SS B  .90 2  .81 and P SS C  .80 2  .64 .
SS
total 200 400 400 1000

SS
SS
total
A
B
200 324
C
256
total
780
200
400
1000
400



So P A SS  
.


200
 .2564 .
780
 
 
If you wish to do this formally, recall that P A  .2, PB   .4 , PC   .4 , P D A  .0, P D B  .10
 
 PD A P A  PD B PB  PD C  PC 
and P D C  .20 . Then a) PD  P A  D  PB  D  PC  D
 0.2  .10 .4  .20.4  .1200 .
 
We must use Bayes’ Rule and the fact that P A  1  P A to do the rest of the problem.
b) PA D  
PC D  
P D AP A
P D 
PD C PC 
P D 


PD B PB  .10 .4 
0.2 
 0, PB D  

 .3333 and
.1200
P D 
.1200
.20 .4
 .6667 ;
.1200
 
 
 
 
 
and PD   1  PD   1  .1200  .8800 .
PD AP A 1.2
PD B PB  .90 .4
PA D  

 .2273 , PB D  

 .4091
.8800
.8800
PD 
PD 
PD C PC  .80 .4
PC D  

 .3636 .
.8800
PD 


c) P D A  1  P D A  1  0  1, P D B  1  P D B  1  .10  .90, P D C  1  P D C  1  .20  .80
and
10
251y0622 3/10/06
 
 
PD C   .20 , and the event S is one satisfactory bulb, PS A  1  PD A  1,
PS B  PD B  1  .10  .90 and PS C   1  PD C   1  .20  .80. Because of the independence
assumption, PSS A  12  1, PSS B   .90 2  .81 and PSS C   .80 2  .64 .
PSS   PSS  A  PSS  B  PSS  C   PSS A P A  PSS BPB  PSS C PC 
To do e) formally we must still remember that we found in d) if P D A  .0, P D B  .10 and
 1.2  .81.4  .64 .4  .2  .3240  .2560  .7800
PA SS  
PSS AP A
PSS 

1.2 
 .2564
.7800
Version 999: Assume that the student number is xxx999. We have P A  .2, PB   .4 and PC   .4 .
 
 
 
We also have P D A  .09, P D B  .19 and P D C  .29 . Let SS be the event that we find two
satisfactory bulbs.
The problem asks for: a) PD  b) P A D , P B D and P C D ; c) P A D P B D and P C D ;
   
d) PSS A, PSS B  and PSS C  and e) PA SS .


   
 
A B C total
The easiest way to do a) b) and c) is the box method. Assume 1000 bulbs.
D
D
total
A
B
D
Using P A  .2, PB   .4 and PC   .4 , divide up the bulbs.
D
total 200
 
 
 
C
1000
total
.
400
400 1000
If P D A  .09, P D B  .19 and P D C  .29 , Then .09  200  18 of Type A, .19  400  76 of Type B
and .29  400  116 of Type C are defective. Fill these into the table and we have
A
B
C
total
A
B
C
total
D
18 76 116
D
28 76 116
220
. Now make the table add up.
.
D
D
172 324 284
780
total 200 400 400 1000
total 200 400 400 1000
220
11
28
7
76 19

 .2200 b) P A D  

 .1273 , P B D  

 .3455
So we have for a) PD  
1000 50
220 55
220 55
116 29
172
43
324
81

 .5273 . ; c) P A D 

 .2205 P B D 

 .4154 and
and P C D  
220 55
780 195
780 195
284
71
PCD 

 .3641 .
780 195
To do d) we must remember that if P D A  .09, P D B  .19 and P D C  .29 , and the event S is one
 
 
 
 
 
 
satisfactory bulb, PS A  1  PD A  1  .09  .91, PS B  PD B  1  .19  .81 and
PS C   1  PD C   1  .29  .71. Because of the independence assumption, PSS A  .91 2  .8281 ,
PSS B   .81 2  .6561 and PSS C   .71 2  .5041 .
11
251y0622 3/10/06
Now we can do e). Start out with our original box diagram, but change the Ds to SSs .
A
B
C
total
SS
. Apply P SS A  .8281 , P SS B  .6561 and P SS C  .5041
SS
total 200 400 400 1000

A
B
C
SS 165 .22 262 .44 201 .64
SS
total 200 .00 400 .00 400 .00
total
631 .33



So P A SS  
.


165 .22
 .2617 .
631 .33
1000 .00
3. Extra Credit (Bassett et. al.). This is a stinker but there is a way to do it fairly rapidly.
Three quarters of the members of a club are adults. Three quarters of the adults and three fifths of the
children are male. Half of the adult males, one third of the adult females, four fifths of the boys and four
fifths of the girls use the pool. Find the following Probabilities:
a) A member of the club uses the swimming pool.
b) A user of the swimming pool is male.
c) A member of the club is female.
d) A user of the swimming pool is female.
e) A male user of the swimming pool is a child.
f) A non-user of the swimming pool is either female or an adult.
Hint: Start out with the following events: A adult; C child; F female; M male; S user of the swimming
pool. Assume that the club has 800 members. Identify the probabilities in terms of these events.
A C
total
M
..... .....
Fill in the following table.
Use the results of the table to fill in a second table.
F
..... .....
total
800
A  M A  F C  M C  F total
S
.....
.....
.....
.....
You should now have the numbers you need to solve the
S
.....
.....
.....
.....
total
800
problem.
Solution: Three quarters of the members of a club are adults. P A  34 , and PC   14 . Three quarters of
the adults and three fifths of the children are male. PM A  34 , P M C   53 . We can thus divide our 800
members into 600 adults and 200 children. The conditional probabilities tell us that 450 of the adults and
120 of the children are male. We can fill in the female members to make this add up.
A C
total
M 450 120 570
Now we can put our numbers of adult males, adult females, boys and girls at the
F 150 80 230
600 200 800
bottom of the second table. Half of the adult males, one third of the adult females, four fifths of the boys
and four fifths of the girls use the pool. P S A  M  12 , PS A  F   13 , PS C  M   54 ,
PS C  F  

4
. If
5

you apply these to the column totals and fill in the non-swimmers you have your table.
12
251y0622 3/10/06
A M A F C  M C  F
total
S
225
50
96
64
435
S
225
100
24
16
365
450
150
120
80
800
a) A member of the club uses the swimming pool - PS  . 435 of the 800 membeers use the pool.
435
800
 .544 .


b) A user of the swimming pool is male - P M S . There are 435 users of the pool. Of these 225 are adult
males and 96 are boys.
225 96
435

321
435
 .738
c) A member of the club is female - PF  . 230 of the 800 members are female.
d) A user of the swimming pool is female - P F S  

50 64
435


230
800
 .2875
. This is 1  P M S  1  .738  .262

e) A male user of the swimming pool is a child - P C M  S  . There are 225 + 96 = 321 male swimmers,
96 are children.
96
321
 .299 .


f) A non-user of the swimming pool is either female or an adult - P F  A S . There are 365 non
swimmers. 225 are adult males, 100 are adult females, and 16 are girls.
22510016
365
341
 365
 .934 or
24
1  365
 .934 .
13