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Unformatted text preview: a) Experiment = play the game or test the game
b) 75 of the veteran game players said that it had bad playability
65 is not a probability because the number 65 is not between 0
c) and 1, inclusive.
d) The number 1 is not between the numbers 0 and 1, inclusive.
e) After the tests were all done, 54 said they did not like it count said liked it empirical or relative frequency
80 65 0.8125 1
a) Total cards # of queens in 1 deck Probability of selecting a Queen
52
4
0.0769230769
Classical because there are n equally likely outcomes b)
2
a)
Total Children Marital Status of Parents
Divorced Married
539 333
P(Married) =
P(Divorced) =
0.6178107607 Widowed
182 24
P(Widowed) =
0.3376623377 0.0445269017 Empirical Probability or Relative Frequency because the problem uses past date to calculate a
probability that could be used for future estimation. Law of Large Numbers says that as the number of
b)
Trials the Relative Frequency will approach the true probability. 3
a) 1 because it is above 12,000. The book was written a while ago. I guess it could be either classical or empirical because 1/1 = 1 ==> 100%. You could
say its likelihood is 1, but in some ways it could be argued that it is none of the
b)
approaches to probability because it is one, known event from the past. Equal?
TRUE Two People =
Favor =
F
Against =
A
Possible Outcomes
Outcome 1
F, F
Outcome 2
F, A
Outcome 3
A, A
Outcome 4
A, F 2 Majors
Acc
Fin
Econ
Manage
Market Total a) 34 Prob Name Probability Number
P(Acc) =
0.2941176471
P(Fin) =
0.1470588235
P(Econ) =
0.0882352941
P(Manage)
0.1764705882
=
P(Market) =
0.2941176471 Total
b) # of Students Declaring Major
10
5
3
6
10 1 This is the Relative Frequency or Empirical
Approach Total Candidates
Minority
a)
b) P(Minority) =
Classical because there are n equally likely outcomes 5
2
0.4 a)
b)
c)
d) Relative Frequency or Empirical
Classical
Classical, but still extremely remote chance of winning
Empirical based on past data. Sample size =
People asked Yes or No question: 40
Oil Executives The experiment was the asking the 40 executives
a) the Yes/No environmental question
b) All 40 said No.
c) Answered Yes
P(Yes) =
10
d) Relative Frequency or Empirical
e) No, not both.
Each of the possible outcomes are not equally
likely. For example ==> 0.25 1 possible outcome is this many said yes: 2 possible outcome is this many said yes:
12 P(Yes) =
0.3
The possible outcomes are mutually exclusive
because you can't have the group have 12 say yes
and 19 say yes also. 19
P(Yes) =
0.475 Number of Violations Number of Drivers
0
1,910
1
46
2
18
3
12
4
9
5 or more
5
Total
2,000
The experiment is seeing how many speeding violations people
have from a sample of 2000 people.
One possible event is 1 speeding violation. Another example of
b)
one possible event is one or more violations.
c)
0.009
d)
Empirical or Relative Frequency
a) Table of possible probabilities
P(exactly 0 speeding violations) =
P(exactly 1 speeding violations) =
P(exactly 2 speeding violations) =
P(exactly 3 speeding violations) =
P(exactly 4 speeding violations) =
P(5 or more speeding violations) =
P(3 or more) = 0.006 + 0.0045 + 0.0025 = 0.955
0.023
0.009
0.006
0.0045
0.0025
0.013 c7466754d66532ff4fbbea246d745bff7605e657.xls  SR 53 Classifications Event
Supervisors
A
Maintenance
B
Production
C
Management
D
Secretarial
E # of Employees
120
50
1460
302
68
2000 P(X)
P(A)
P(B)
P(C)
P(D)
P(E) P(X)
0.06
0.025
0.73
0.151
0.034
1
a) i P(Maintenance or Secretarial)
ii P(Not Management) = P(B or E) =
= P(Not D) = b) Management Maintenance
Secretarial c) (a)(i)
P(B)
P(E)
Total Mutually Exclusive and not complimentary (not complimentary because they don't add up to 1)
0.025
0.034
0.059 0.025 +
1 0.034 =
0.151 = 0.059
0.849 Need corrective shoes
Need major dental work
Percent that need corrective shoes and
major dental work BOTH
a) P(Need corrective shoes OR Need major
dental work) = 0.15 + 0.08  0.03 = 8.00%
15.00%
3.00% Must subtract so you don't double count the 3.00% is BOTH 20.00% b) Need corrective shoes
Need major dental work c7466754d66532ff4fbbea246d745bff7605e657.xls  Ex13 Income After Taxes
Number of Firms Event
Under $1 million
102 A
$1 million to $20 million
61 B
$20 million or more
37 C
200 P(X)
P(A)
P(B)
P(C) P(X)
0.51
0.305
0.185
1 a) P(A)
0.51
P($1 million to $20
million or $20 million or
b) more)
=
0.3050
+
0.1850
0.0000
=
0.4900
Special Rule because the categories are mutually exclusive. However, if you used the General Rule, the joint probability would have
been zero and you would have gotten the correct answer. Profit
Break Even
Lose
Total
a)
b) P(Profit OR Break Even) = 0.5 + 0.3 =
P(Profit OR Break Even) = 1  0.2 = 50%
30%
20%
100%
80%
80% A
B 0.25
0.5 P( above C) =0.25 + 0.5 = 0.75 Two Coins are tossed
A=
B= H,H
T,T P(H,H) =
P(T,T) =
Total 0.25
0.25
0.5 The two are mutually exclusive because the two events cannot happen at the
same time. Two events, A & B, are mutually exclusive if both events, A & B,
cannot occur at the same time
a)
b) No, because they do not add up to 1, they add up to .5 All Possible Outcomes
1 H,H
2 H,T
3 T,T
4 T,H
Total 4 0.25
0.25
0.25
0.25
1 Events
A
B
A AND B
P(A OR B) = 0.2 + 0.3  0.15 = P(X)
P(X)
P(A) =
0.2
P(B) =
0.3
P(A AND B) =
0.15
0.35 P
F
D
P&F&D
P&D
F&D
P&F
a)
b)
c)
d) Southwest Grocery Stores
Had Pharmacy
Had Floral Shop
Had Deli
Had All 3
Had Pharmacy and Deli
Had Floral Shop and Deli
Had Pharmacy and Floral
P(Had Pharmacy and Had Floral Shop) =
P(Had Pharmacy and Had Deli) =
No. Because, for example, if you select a store
with a deli, it could be in a category "Deli only" or
"Deli and Pharmacy".
Joint Probability 0.4
0.5
0.7
0.1
0.3
0.25
0.2 P(F) = 0.5 0.2
0.3 P(D) = 0.7 e)
Step 1: Find P(P or F or D)
Step 2: Find 1  P(P or F or D) =
Notes:
This is to add the Probability
This is to not double count
This if for the triple count
Once you have all that you can
subtract it from 1 P(P) = 0.4 P(P OR F OR D) = P(P) + P(F) + P(D)  P(P & D) P(F & D)  P(P & F) + P(P & F & D)
0.05 0.95 =C2+C3+C4C6C7C8+C5 P(P) + P(F) + P(D)
 P(P & D)  P(F & D)  P(P & F)
+ P(P & F & D)
1  0.95 = 0.05 P(P) = 0.4
P(F) = 0.5
P(D) = 0.7 This problem was not assigned Vacationers Going To Rocky Mountain Region
Yellowstone
0.5
Tetons
0.4
Yellowstone and Tetons
0.35
P(Y) =
P(T) =
P(Y and T) = 0.5 + 0.4  0.35 = P(Y and T) =
0.5 + 0.4  0.35
= 0.35 0.50
0.40
0.55 a) P(Y and T) = 0.5 + 0.4  0.35 = 0.55
Notes:
P("visit at least one" means Y or T or
Both)
But you can't double count
b) "Joint" or "Concurrent"
No because of the overlapping circles
c) in the Venn Diagram to the right ==> P(Y) = Vacationers destinations according to National Park Service
P(X)
Compliment
Yellowstone
0.5
0.5
Tetons
0.4
0.6
Yellowstone and Tetons
0.35
Visit only Yellowstone (0.5*(10.4))
Visit only Tetons (0.4*(10.5))
Yellowstone and Tetons 0.3
0.2
0.35 Done with formulas that take more
time than they are worth
P(Y) =
P(T) =
P(Y and T) = 0.50
0.40
0.55 P(T) = P(Y and T) = 0.5 + 0.4  0.35 = 0.35 a) 4/52 = 2/26 = 1/13 ==> because the sample space has not changed
b) 3/51 because there are only 3 kings and the sample space has changed
c) P(2 straight Kings) = 4/52*3/51 = 0.0045249 4
52 3
51 a)
b)
c)
d) Derreck Lee Highest Ave in 2005 Season
0.335
Assume he hits in one game =
3
Empirical because it is based on past data and we are using it to estimate how he will do
for the game
P(3 straight hits ) = 0.335*0.335*0.335 = 0.037595375
0.0375954
P(No Hits) = (10.335)*(10.335)*(10.335) = (10.335)^3 = 0.294079625
0.2940796
P(at least 1 hit) = P(1 or 2 or 3 hits) = 1  P(No Hits) =
0.7059204
or d) P(1) = P(1 hit, No hit, No Hit) + P(No hit, 1 Hit, No Hit) + P(No hit, No Hit, 1 hit) =
P(2) = P(1 hit, 1 hit, No Hit) + P(1 Hit, No Hit, 1 hit) + P(No hit, 1 hit, 1 hit) =
P(3) = P(1 hit, 1 hit, 1hit) =
P(at least 1 hit) = P(1 or 2 or 3 hits) = 0.4444361
0.2238889
0.0375954
0.7059204 H
NH DL Average = Probability of getting a hit
DL Average = Probability of NOT getting a hit
# of at bats = # of Trials =
Because averages do not include walks, hit by
pitch, sacrifices, and other nonHits/NotHits in the
calculation, The events H and NH are mutually
exclusive and independent and the categories H
and NH are collectively exhaustive. a
b
c
d Empirical
P(H and H and H) = P(H)*P(H)*P(H) =
P(0)
P(at least one hit) = P(1 or 2 or 3) 0.335 P(H)
0.665 P(NH)
3 0.037595375
0.294079625
0.705920375 check check
0.0376 0.0376
0.29408
0.70592
# of Possible outcomes in the sample space = ss =
Possible
Outcomes # of at bats = # of Trials =
1
2
1H
H
H
2H
H
NH
3H
NH
H
4 NH
H
H
5H
NH
NH
6 NH
H
NH
7 NH
NH
H
8 NH
NH
NH 8 3 # of Hits Probability
out of 3 at
of
bats
Occurance
3
2
2
2
1
1
1
0 0.03759538
0.07462963
0.07462963
0.07462963
0.14814538
0.14814538
0.14814538
0.29407963
1 Random Variable
= X = # of Hits in
3 at bats
P(X) 0 P(0)
1 P(1)
2 P(2)
3 P(3) P(X) 0.2940796
0.4444361
0.2238889
0.0375954
1 Teton Tires
Failure hurdle
P(XB70 tire will last 60,000 before it becomes bald or fail)
An adjustment is made on any tire that does not last 60,000
We purchase this # of tires:
P(All 4 tires will last 60,000 miles) = 0.8*0.8*0.8*0.8 = 0.8^4 = 60,000
0.8
4
0.4096 Board of Directors
Men
Women
Total
Wee need to
select a
committee with 4
members at
random 8
4
12 4 Sample Space Conditional
Women Selected Changes
Probability
4
12 0.3333333333
3
11 0.2727272727
2
10
0.2
1
9 0.1111111111 a)
b) c) P(All Women) = 8/12*7/11*6/10*5/9 =
0.141414141414141 =
P(All Men) = 4/12*3/11*2/10*1/9 =
0.00202020202020202 =
No because 0.141414141414141 +
0.00202020202020202 Does not equal 1.
In addition, there are other possibilities such as, for
example, 3 women and a man.
Also, the categories are not collectively exhaustive. Men Selected Sample Space Conditional
Changes
Probability
8
12 0.6666666667
7
11 0.6363636364
6
10
0.6
5
9 0.5555555556 0.1414141414
0.0020202020 Executives were asked whether or not they would remain with the company if they received a better offer from a different
company
Length of Service
Less than
15
More than
Loyalty
1 year
years
610 years 10 years
Total
Would remain
10
30
5
75
120
Would not remain
25
15
10
30
80
35
45
15
105
200
P(Randomly selecting executive who has More than 10 years) =
a) 105/200 =
P(Randomly selecting executive who Would not remain GIVEN
b) THAT More than 10 years) = 30/105 =
P(Randomly selecting executive who has More than 10 years OR
c) Would not remain) = 105/200 + 80/200  30/200 =
P(Randomly selecting executive who Would remain AND who has
More than 10 years) =
P(Randomly selecting executive who Would not remain AND who
has More than 10 years) = 0.525000
0.285714
0.775
0.375 0.375 0.15 0.15 Visits
Yes
Often
Occasional
Never
Total a)
b)
Visits
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Often
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Occasional
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never
Never Enclosed Mall
No
Total
60
20
80
25
35
60
5
50
55
90
105 195 Because Independence means that the occurrence of one event does not affect the occurrence of another event, for our two variables "Visits" and
"Enclosed Mall", they are not independent. Using the Rule of Independence, we can show that P(NeverYes) Does Not Equal P(Never) ==> 5/90
Does Not Equal 55/195
Enclosed Mall
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
Yes
Yes
Yes
Yes
Yes
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No
No Enclosed Mall
Visits
Often
Occasional
Never
Grand Total Data
Yes
CondProb Count of Visits 2/ 3
5/ 18
1/ 18
1/ 1 4/ 13
5/ 39
1/ 39
6/ 13 JointProb
30.77%
12.82%
2.56%
46.15% Count of Visits 60
25
5
90/ 1 Data
Yes
CondProb
66.67%
27.78%
5.56%
100.00%
Data
Yes
CondProb
0.6666666667
0.2777777778
0.0555555556
1 JointProb
0.3076923077
0.1282051282
0.0256410256
0.4615384615 Count of Visits Count of Visits Enclosed Mall
Visits
Often
Occasional
Never
Grand Total No
CondProb Count of Visits 60
25
5
90
Enclosed Mall Visits
Often
Occasional
Never
Grand Total JointProb Count of Visits
60
25
5
90 JointProb 20
35
50
105 4/ 21
1/ 3
10/ 21
1/ 1 4/ 39
7/ 39
10/ 39
7/ 13 20
35
50
105/ 1 No
CondProb
19.05%
33.33%
47.62%
100.00% JointProb
10.26%
17.95%
25.64%
53.85% No
CondProb
0.1904761905
0.3333333333
0.4761904762
1 JointProb
0.1025641026
0.1794871795
0.2564102564
0.5384615385 20
35
50
105 done with
PivotTable and
Fractional
Number Format Done with
formulas and %
format Done with
PivotTable and
Custom Fraction
Number Format 25 a)
b) P(Checking)
P(Savings)
P(C & S)
P(C OR S) = 0.8 + 0.6  0.5 =
P(Not C OR S) = 1  +0.9 = 0.8
0.6
0.5
0.9
0.1 26
P(1st truck available) =
P(2nd truck available) =
P(Both 1st and 2nd truck available) =
Step 1: Find P(1st OR 2nd) = 0.75 + 0.5  0.3 =
Step 2: Find P(neither available) = 1  P(1st OR 2nd) = 1  0.95 = 0.75
0.5
0.3
Remember: P(1st OR 2nd) means 1st is working, or second is
0.95 working or Both are working.
0.05 So that means that 1  P(1st OR 2nd) means neither are working Defective Toothbrushes =
NonDefective Toothbrushes =
Total a)
b) P(first 2 brushes are not defective) =3/20*2/19 = 0.0157894736842105
P(first 2 brushes are not defective) =17/20*16/19 = 0.715789473684211 3
17
20 0.0157894737 Sample Space Changes
0.7157894737 Sample Space Changes c7466754d66532ff4fbbea246d745bff7605e657.xls  Ex29 Potential for advancement
Fair
Sales Ability Below Ave.
Ave.
Above Ave.
Totals Good
16
45
93
154 Excellent Totals
12
60
72
144 22 50
45 150
135 300
202 500 Sales Ability Fair
Below Ave.
Ave.
Above Ave. Joint Probabilities
Good
0.032
0.09
0.186 Excellent
0.024
0.12
0.144 0.044
0.09
0.27
SUM = a) b)
c) Contingency Table:
Good for Nominal
Data
P(Above Ave. Sales
Ability AND
Excellent Potential
for advancement) =
135/500 = 0.27 0.27 1 c7466754d66532ff4fbbea246d745bff7605e657.xls  Ex29 (2) Potential for advancement
Good Sales
Ability 16
45
93
154 Excellent
12
60
72
144 Totals
22 50
45 150
135 300
202 500 Sales Ability
Below Ave.
Ave.
Above Ave. Fair Joint Probabilities
Good
3.20%
9.00%
18.60% Excellent
2.40%
12.00%
14.40% 4.40%
9.00%
27.00%
SUM = Contingency Table:
Good for Nominal
Data
P(Above Ave. Sales
Ability AND
Excellent Potential
for advancement) =
135/500 = a) b)
c) PotForAd
Sales Ability
Below Ave.
Ave.
Above Ave.
Grand Total Count
16
45
93
154 0.27 0.27 Data
Fair
CondProb
32.00%
30.00%
31.00%
30.80% JointProb
3.20%
9.00%
18.60%
30.80% Count
12
60
72
144 Good
CondProb
24.00%
40.00%
24.00%
28.80% JointProb
2.40%
12.00%
14.40%
28.80% Count
22
45
135
202 Excellent
CondProb
44.00%
30.00%
45.00%
40.40% JointProb
4.40%
9.00%
27.00%
40.40% 100.00% 1
2
3
4
5
6 Sales Ability
Below Ave.
Below Ave.
Below Ave.
Below Ave.
Below Ave.
Below Ave. Potential for advancement
Fair
Fair
Fair
Fair
Fair
Fair 7 Fair
Below Ave.
Ave.
Above Ave.
Totals Below Ave. Fair 8
9
10
11
12
13
14
15
16
1
2
3
4
5
6
7
8
9
10
11
12
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
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Excellent Sweaters
1
2
3
4
5
Slacks
1
2
3
4
1
Receivers
docks
speakers
carousels 20 2 72 3
2
4
3 1 a)
b) n
5 r
3 Formula Formula
60 =C4*(C41)*D4
60 =PERMUT(C4,D4) 2 4 4 24 =PERMUT(C7,D7) 3 10 4 5040 =PERMUT(C11,D11) 4 a)
b) 8
10 3
2 56 =COMBIN(C14,D14)
45 =COMBIN(C15,D15) 5 a)
b) 50
50 3
3 117600 =PERMUT(C17,D17)
19600 =COMBIN(C18,D18) Yes Ex41&45 #41
n r
10 Combination
4 210.00 10 Permutation
10,897,286,400.00 #45
n r
15 Problems not assigned ==> P(A1B) = P(A1)*P(BA1)
P(A1)*P(BA1) + P(A2)*P(BA2) Prior Probability = The initial
Probability based on the
Event Ai (These events are
mutually Exclusive and Collectively present Level of Information
Prior Probability = P(Ai) =
Exhaustive) P(Ai)*P(BAi) P(AiB) = data in white cells is given P(A1)*P(BA1) + P(A2)*P(BA2) + … + P(An)*P(BAn) Conditional Probability =
Probability that B given that Ai
has already happened = P(BAi)
= Has Disease in Pop. = A1 0.05 Not Have Disease in Pop. = A2 0.95
1 P(B) = Joint Probability = Both are true = P(Ai and B) = Posterior Probability = A
revised Probability based on
Additional Information = P(Ai
B) = 0.9 0.045 0.24 0.15 0.1425
0.1875 0.76
1 P(A1B) = P(AiB) = P(A1)*P(BA1) P(Ai)*P(BAi) P(A1)*P(BA1) + P(A2)*P(BA2)
B1 = P(A1)*P(BA1) + P(A2)*P(BA2) + … + P(An)*P(BAn) Defective B2 = Key Concept!!!!!
Particular Joint
Probability/SUM(of joint
Probabilities) Not Defective Event Ai (These events are mutually Exclusive and Collectively Exhaustive) Prior Probability = The initial
Probability based on the
present Level of Information
Prior Probability = P(Ai) = Conditional Probability =
Probability that B1 (defective)
given that Ai has already
happened = P(BAi) = Posterior Probability = A revised Probability
based on Additional Information = P(AiB1) = Joint Probability = Both are true = P(Ai and B) = Prob. Chips bought from Hall = A1 0.3 0.03 0.009 0.2307692308 Prob. Chips bought from SS = A2 0.2 0.05 0.01 0.2564102564 Prob. Chips bought from CC = A3 0.5
1 P(B) = 0.04 0.02
0.039 0.5128205128
1 Event Ai (These events are mutually Exclusive and Collectively Exhaustive) Prior Probability = The initial
Probability based on the
present Level of Information
Prior Probability = P(Ai) = Conditional Probability =
Probability that B (notdefective) given that Ai has
already happened = P(B2Ai) = Posterior Probability = A revised Probability
based on Additional Information = P(AiB2) = Joint Probability = Both are true = P(Ai and B) = Prob. Chips bought from Hall = A1 0.3 0.97 0.291 0.3028095734 Prob. Chips bought from SS = A2 0.2 0.95 0.19 0.197710718 Prob. Chips bought from CC = A3 0.5
1 P(B) = 0.96 0.48
0.961 0.4994797086
1 Total
1 data in white cells is given Given Information:
Prior Probabilities = P(Ai)
Probability that part came from Hall =
Probability that part came from SS =
Probability that part came from CC =
= P(A1)
P(A2)
P(A3) 0.3
0.2
0.5
1 Additional Information:
Part is Defective =
Part is Not Defective =
=
Given that part came from Hall, probability that part is defective =
Given that part came from SS, probability that part is defective =
Given that part came from CC, probability that part is defective =
Given that part came from Hall, Probability that part is not defective =
Given that part came from SS, Probability that part is not defective =
Given that part came from CC, probability that part is not defective = B1
B2
Conditional Probability = P(Bi)
0.03
0.05
0.04
0.97
0.95
0.96 P(B1A1)
P(B1A2)
P(B1A3)
P(B2A1)
P(B2A2)
P(B2A3) 1
1
1 Joint Probabilities
Probability that part came from Hall AND it is defective = P(A1) * P(B1A1) = 0.3 * 0.03 =
Probability that part came from SS AND it is defective = P(A2) * P(B1A2) = 0.2 * 0.05 =
Probability that part came from CC AND it is defective = P(A3) * P(B1A3) = 0.5 * 0.04 = 0.009
0.01
0.02
0.039 Probability that part came from Hall AND it is NOT defective = P(A1) * P(B2A1) = 0.3 * 0.97 =
Probability that part came from SS AND it is NOT defective = P(A2) * P(B2A2) = 0.2 * 0.95 =
Probability that part came from CC AND it is NOT defective = P(A3) * P(B2A3) = 0.5 * 0.96 = 0.291
0.19
0.48
0.961 0.009
0.010
0.020
0.039
Joint Probabilities Because theses are all the possibilities 0.291
0.190
0.480
0.961 1 Array Formula
Posterior Prob.
Posterior Probabilities = Revised
Probabilities based on additional
Information Posterior Probabilities (Revised Probability Based On Additional Information) = Probability that
chip is from Hall, given that we have selected a defective chip = 0.009/0.039 = 9/39 =
Posterior Probabilities (Revised Probability Based On Additional Information) = Probability that
chip is from SS, given that we selected a defective chip = 0.01/0.039 = 10/39 =
Posterior Probabilities (Revised Probability Based On Additional Information) = Probability that
chip is from CC, given that we selected a defective chip = 0.02/0.039 = 20/39 = Posterior Probabilities (Revised Probability Based On Additional Information) = Probability that
chip is from Hall, given that we have selected a nondefective chip = 0.291/0.961 = 291/961 =
Posterior Probabilities (Revised Probability Based On Additional Information) = Probability that
chip is from SS, given that we selected a nondefective chip = 0.19/0.961 = 190/961 = Posterior Probabilities (Revised Probability Based On Additional Information) = Probability that
chip is from CC, given that we selected a nondefective chip = 0.48/0.961 = 480/961 = check P(A1B1) = P(A1B1) 0.2307692308 0.2307692308 0.2307692308 0.2307692308 P(A2B1) 0.2564102564 0.2564102564 0.2564102564 0.2564102564 P(A3B1)
check 0.5128205128
1 0.5128205128
1 0.5128205128 0.5128205128 P(A1B2) 0.3028095734 0.3028095734 0.3028095734 0.3028095734 0.197710718 0.197710718 0.197710718 0.197710718 0.4994797086
1 0.4994797086
1 0.4994797086 0.4994797086 P(A2B2)
P(A3B2) =
(P(A3)*P(B2
A3))/(Sum of all
Joint
Probabilities)
check Prior Probabilities
Probability that Game Played at Night
Probability that Game Played in Day N
D P(N)
P(D) 0.7
0.3
1 Conditional Probabilities
Win Given at Night
Win Given at Day
Loss Given at Night
Loss Given at Day WN
WD
LN
LD P(WN)
P(WD)
P(LN)
P(LD) 0.5
0.9
0.5
0.1 Joint Probabilities
Probability that Game Played at Night * Win
Given at Night = P(N) * P(WN) = 0.7 * 0.5
Probability that Game Played in Day * Win
Given at Day = P(D) * P(WD) = 0.3 * 0.9 Probability that Game Played at Night * Loss
Given at Night = P(N) * P(LN) = 0.7 * 0.5
Probability that Game Played in Day * Loss
Given at Day = P(D) * P(LD) = 0.3 * 0.1 check
1
1
Posterior Probability = Revised
Probability Based on Additional
Information = Particular
Joint/Sum(all joint) 0.35 P(WN) 0.564516129 P(WD) 0.435483871 0.35 P(LN) 0.9210526316 0.03
0.38 P(LD) 0.0789473684 0.27
0.62 check
1
1 F
NF PF
PNF Prior Probabilities
Probability that student finish assignments =
Probability that student not finish assignments =
Conditional Probabilities
Probability they pass class, given that they finished
assignments =
Probability they pass class, given that they not finished
assignments = P(F)
P(NF) 0.8
0.2 P(PF) 0.9 P(PNF) 0.6 Joint Probabilities
P(F) * P(PF) =
P(NF) * P(PNF) = 0.72 =
0.12 =
0.84 = Posterior Probabilities
Probability that they finished assignments given that they
passed the class = P(FP)
Probability that they not finish assignments, Given that
they passed the class = P(NFP) 0.8571428571 0.857143
0.1428571429 0.142857
1 0.72
0.12
0.84 Prior
Probabilities
Cash
C
Credit Card CC
Debt Card DC >$50.00 P(C)
P(CC)
P(DC) Sales paid for with Cash
Sales paid for with Credit Card
Sales paid for with Debt Card 50 Purchanse was for more than $50.00
Probability that Purchanse was for more than $50.00, Given that Sales
P(>$50.00C) paid for with Cash
Probability that Purchanse was for more than $50.00, Given that Sales
P(>$50.00CC) paid for with Credit Card
Probability that Purchanse was for more than $50.00, Given that Sales
P(>$50.00DC) paid for with Debt Card P(C) * P(>$50.00C)
P(CC) * P(>$50.00CC)
P(DC) * P(>$50.00DC) P(C>$50.00) Probability that Cash was paid Given that sales was >$50.00
P(CC>$50.00) Probability that Credit Card was paid Given that sales was >$50.00
P(DC>$50.00) Probability that Debt Card was paid Given that sales was >$50.00 0.3
0.3
0.4
1
Conditional
Probabilities
0.2
0.9
0.6
Joint
Probabilities
0.06
0.27
0.24
0.57
Posterior Probabilities
0.1052631579 0.105263
0.4736842105 0.473684
0.4210526316 0.421053
1
1 Prior
Leave garage open
Not Leave garage open
Conditional
Stuff Stolen Given, Leave garage open
Stuff Stolen Given, Not Leave garage open
Doors open given Stolen 0.25
0.75
0.05
0.01
0.625 Gen
der
Going To
College Root or Prior Probabilities
P(F)
P(M)
P(WCF)
P(NotWCF)
P(WCM)
P(NotWCM) c
0.8 Female
0.2 Male
Conditional Probabilities
0.9 Went To College
0.1 Did Not Go To College
0.78 Went To College
0.22 Did Not Go To College a P( select a female from whole
pool who is female AND did not
go to college) b Independence means that one
event does not have an affect
on another event. The Sample
Space is not changed when the
second event occurs.
P(NotWCF) = P(F) 0.08 Root
Prob Conditional
Prob F
M
Went To College
WC
NotWC
WC
NotWC P(F)
P(M) P(WCF)
P(NotWCF) Joint Prob 0.9 P(F) * P(WCF) = 0.8 * 0.9
0.1 P(F) * P(NotWCF) = 0.8 * 0.1 0.72
0.08 0.78 P(M) * P(WCM) = 0.2 * 0.78
0.22 P(M) * P(NotWCM) = 0.2 * 0.22 0.156
0.044 0.8
0.2
Did Not Go To College P(WCM)
P(NotWCM) c
1
It equals 1 because all the possible probabilities are included
No, because if you know
gender, it changes the
probability that they go to
college. Variable
0.9 w
0.1 Nw
4
F1
F2
F3
F4 Flight arrives within 15 minutes of scheduled arrival
Flight does not arrive within 15 minutes of scheduled arrival
We select this # of flights to study:
Flight 1
Flight 2
Flight 3
Flight 4
If you assume independence, then P(F1 w 15 AND F2 w 15 AND F3 w 15
AND F4 w 15) =
If you assume independence, then P(F1 Nw 15 AND F2 Nw 15 AND F3 Nw
15 AND F4 Nw 15) = a
b
c
Possible Outcomes Sample space = (# of outcomes)^(# of Trials) = 2^4 =
Flights and w or Nw
1 w,w,w,w
2 Nw,w,w,w
3 w,Nw,w,w
4 w,w,Nw,w
5 w,w,w,Nw
6 Nw,Nw,w,w
7 w,Nw,Nw,w
8 w,w,Nw,Nw
9 Nw,w,w,Nw
10 Nw,w,Nw,w
11 w,Nw,w,Nw
12 Nw,Nw,Nw,w
13 w,Nw,Nw,Nw
14 Nw,w,Nw,Nw
15 Nw,Nw,w,Nw
16 Nw,Nw,Nw,Nw 0.6561
0.00010000
16
# of Nw
0
0.6561
1
0.0729
1
0.0729
1
0.0729
1
0.0729
2
0.0081
2
0.0081
2
0.0081
2
0.0081
2
0.0081
2
0.0081
3
0.0009
3
0.0009
3
0.0009
3
0.0009
4 0.00010000
1 chapter 6:
Binomial?
Fixed # of trials?
Yes, n = 4
Each trial independent?
Yes, n = 4
S/F each time
Yes, n = 4
Probability stays the same
Yes pi = .9
each trial?
0.6561
0.2916
0.0486
0.0036
0.0001 0 Random Variable X = # of Nw = # of flights
not within 15 minutes
0
1
2
3
4 1 P(X = 0) =
4 P(X = 1) =
6 P(X = 2) =
4 P(X = 3) =
1 P(X = 4) =
16
P(X >= 1) =
P(X >= 1) = 0.6561
0.2916
0.0486
0.0036
0.0001 check
check
check
check
check 0.3439
0.3439 cumulative 0 = exact value, 1 = up to that value
0
0
0
0
0 >=1
1 0.3439 1 Production workers
Supervisors
Secretaries
President
a
b
c
d 57 P(Prod)
40 P(Supe)
2 P(Secr)
1 P(Pres)
100 P(Prod)
P(Prod OR Supe)
Yes
P(NOTProd OR NOTSupe)
P(NOTProd OR NOTSupe) 0.57
0.97
0.03
0.03 0.57
0.4
0.02
0.01
1 X = Random
Variable = # of
hits in 4 tries
a
b
c P(X)
4 P(4)
0 P(0)
1 P(1) Probability that missle hits its target
Probability that missle NOT hit its target
# of Outcomes =
# of Trials = # of Missles sent
Sample Space Size = # of possible
outcomes in sample space =
Outcomes^#Trials =
P(All 4 Hit) =
P(None Hit) =
P(at least 1 hit) = HT
NHT P(HT)
P(NHT)
2
4 16
0.4096
0.0016
0.9984 0.8
0.2
ch 6 check
check
0.4096
0.0016
0.9984 Cumulative: 0 = exact
value, 1 = up to that
value
0
0
1 # of Students Graduating =
# of students who are
graduating and going to college
Students picked at random to
carry flags at graduation = 90
50
2
check Probability that the student 1
selected to carry the flag is
going to college = 50/90 =
0.5555556 Root or Prior Probability
Probability that the student 2
selected to carry the flag is
going to college = 49/89 =
0.5505618 Conditional Prob.
Probability that both students
selected to carry flag are going
a to college
0.3058677 Joint Probability Probability that one of the 2
students selected to carry flag
b are going to college 0.4993758 0.5555555556
0.5505617978 selection 1
selection 2
P(X)
Possibility 1 Going to college
Not going to college
0.24968789 50/90*40/89
Possibility 2 Not going to college Going to college
0.24968789 40/90*50/89
0.49937578 Probability that a 60 year old man will survive 1 year
Probability that a 60 year old man will Not survive 1 year
# of Outcomes =
Policy offered to this many men = Trials =
Sample Space = # of Possible Outcomes =
a Probability that all 5 survive?
Probability that none survive?
b Probability that at least one does not survive? 60S1Y
60NOTS1Y P(60S1Y)
P(60NOTS1Y) 0.98
0.02 chapter 6:
Fixed # Trials?
Each trial independent?
Probability stay same each trial?
Probability of Success the same each trial? 2
5
32
0.9039207968
0.00000000320
0.0960792032 check
0.9039207968
0.00000000320
0.0960792032000
0.0960792032000 Not Survive
P(That 0 Not survive)
P(That 1 Not survive)
P(That 2 Not survive)
P(That 3 Not survive)
P(That 4 Not survive)
P(That 5 Not survive) Yes n =
Yes
Yes
Yes 0
1
2
3
4
5 5 0.9039207968000
0.0922368160000
0.0037647680000
0.0000768320000
0.0000007840000
0.0000000032000 cumulative
0
0
0
0
0
0 0.0960792032
Survive
P(That 0 Survive)
P(That 1 Survive)
P(That 2 Survive)
P(That 3 Survive)
P(That 4 Survive)
P(That 5 Survive) 0
1
2
3
4
5 0.0000000032000
0.0000007840000
0.0000768320000
0.0037647680000
0.0922368160000
0.9039207968000 cumulative
0
0
0
0
0
0 Probability that Homes contain Security System
Probability that Home does not contain Security System
# of outcomes =
# of Trials = Homes selected at random =
Sample Space = # of possible outcomes =
62
a
b
c
d P(3)
P(0)
P(1 or 2 or 3) = 1  P(0) =
Independent 0.4 success = home has SS =
Not success = home Not
0.6 have SS =
2
3
8
0.064
0.216
0.784 Probability that Homes contain Security System
Probability that Home does not contain Security System
# of outcomes =
# of Trials = Homes selected at random =
Sample Space = # of possible outcomes =
# of Homes
# of homes with SS
Probability that Homes contain Security System
Probability that Home does not contain Security System
# of outcomes =
# of Trials = Homes selected at random =
63
a P(3)
Pr(3 straight homes selected with SS) =
b P(0)
Pr(3 straight homes selected with No SS) =
c P(X>=1) P(at least 1 has a SS) = P(1 or 2 or 3) =
d 0.4 success = home has SS =
0.6 Not success = home Not have SS =
2
3
8
10
4
0.4 success = home has SS
0.6
2
3
0.0333333
0.1666667
0.8333333 0.8333333333 s
ns check 3 successes
0
0.4 Root or Prior
1 0.3333333333 Conditional Probability
2
0.25 Conditional Probability
0.0333333333 product
check 2 successes
0
0.4 s
1 0.3333333333 s
2
0.75 ns
0.1 product
check 1 successes
0
0.4 s
1 0.6666666667 ns
2
0.625 ns
0.1666666667 product
check 0 successes
0
0.6 Root or Prior
1 0.5555555556 Conditional Probability
2
0.5 Conditional Probability
0.1666666667 product
# of Possible # of Trials = Homes selected at random =
# of
1
2
3 successes
Outcomes
1s
s
s
3
2s
s
ns
2
3s
ns
s
2
4 ns
s
s
2
5s
ns
ns
1
6 ns
s
ns
1
7 ns
ns
s
1
8 ns
ns
ns
0 Prob of
occurance
0.03333333
0.1
0.1
0.1
0.16666667
0.16666667
0.16666667
0.16666667
1 # Families that did there own taxes
# Families that had pro do there taxes
# Families that had H&R Block do there taxes
Total # Families =
a
b c d P(OT)
P(OT)
P(Selecting 2nd OT  that you already selected a OT on 1st try)
P(select one OT AND then a second OT) =
P(OT)
P(Selecting 2nd OT  that you already selected a OT on 1st try)
P(Selecting 3rd OT  that you already selected a OT on first 2 tries)
P(select one OT AND then a second OT) =
P(select 2 families that did not have HR Block do Taxes) 10 OT
7 PDT
3 HRDT
20
0.5
0.5 Root
0.4736842105 Conditional
0.2368421053 Joint
0.5 Root
0.4736842105 Conditional
0.4444444444 Conditional
0.1052631579 Joint
0.7157894737 P(OT)
P(PDT)
P(HRDT) 0.5
0.35
0.15
1 total members
W women
M men 12
3 P(W)
9 P(M) # of outcomes = count s and count Ns =
# of committee members to select = # of Trials = #
of times that you choose and get either a man or a
woman
# of possible Outcomes in Sample Space = (# of
Outcomes)^(# of trials)
a
b P(3 straight men selected for committee)
P(at least one woman) = P(1W OR 2W OR 3W) =
1  P(0W) = 1  P(3 straight men selected for
committee) = # of possible Outcomes Prob on first selection
only
0.25 Success = Choose woman
0.75 not success = choose man
1 s
Ns use hypergeopmetric whne sample space changes 2 chapter 6: Hypergeometric 3 X 8 P(X) 0.3818182 Conditional Probability (Saample space changes) 0 0.00454545
1 0.12272727
2 0.49090909 0.6181818 0.6181818 3 0.38181818 # of possible Outcomes in Sample Space = (#
of Outcomes)^(# of trials)
# of Women selected
1s
s
s
2 Ns
s
s
3s
Ns
s
4s
s
Ns
5 Ns
Ns
s
6 Ns
s
Ns
7s
Ns
Ns
8 Ns
Ns
Ns Probability for
Occurance of Given
Outcome
3
0.0045454545 <++ formula
2
0.0409090909 <++ formula
2
0.0409090909
2
0.0409090909
1
0.1636363636 <++ formula
1
0.1636363636
1
0.1636363636
0
0.3818181818 <++ formula
1 Shareholders made $
Shareholders lost $
Total
a
b
c
d P( Company from list randomly selected,
CEO paid > $1 M )
P( Company from list randomly selected,
CEO paid > $1 M OR Shareholders lost $
)
P( Company from list randomly selected,
CEO paid > $1 M Given That
Shareholders lost $ )
P( 2 Companies from list randomly
selected, and both had CEO paid > $1
M) CEO paid > $1 M CEO paid < $1 M Total
2
11
4
3
6
14
0.3
0.45
0.5714285714
0.0789473684 13
7
20 Proportion of company's potential market that read the Gazette =
Proportion of company's potential market that don't read the Gazette =
Proportion of Gazette Readers who remember the company's ad =
Proportion of Gazette Readers who DO NOT remember the company's ad = a
b P(that Company's potential market reads and remembers the ad)
P(that Company's potential market reads and DOES NOT remembers the ad) 0.6
0.4
1
0.85
0.15
1
because you start with a group of readers, and then
since 85% of all readers remember the company's ad,
that means that 85% of this group will remember the
0.51
ad.
0.09 # of Presidents
# of VicePresidents
Company President has season tickets to Baseball Game
Company President always takes 1 of 4 vice presidents
President claims that he picks VicePresidents at random
Success = Being selected
Not Success = Not Being Selected
# of times not selected
What is probability of not getting selected 5 straight times? 1
4 s
Ns
5
0.2373 P(Selecting 1 Vice President)
P(Selecting of not getting selected) 0.25
0.75
1 Formatted CDs that were perfect
Formatted CDs that were usable but had bad sectors
Formatted CDs that were NOT usable
Total CDs
a
b P(CD not Perfect) =
P(CD Not Usable)) = 857 P
112 UBBS
31 NU
1000 P(P)
P(UBBS)
P(NU) 0.143
0.143
0.2167832 Conditional Probability 0.857
0.112
0.031
1 P(Bank Stock Increase) =
P(Utility Stock Increase) =
P(Bank Stock Not Increase) =
P(Utility Stock Not Increase) =
Outcome 1
Outcome 2
Outcomes =
Total Trials = 2 stocks
Sample Space =
a
b
c 0.7
0.6
0.3
0.4
Up
Down
2
2
4 P(Both Increase)
P(Bank Up and Utility Down)
P(at least one stock goes up) = P(B Up AND U Up OR B Up AND U Down OR B Down AND U Up)
Possible Outcomes s
Ns 0.42 Assuming Independence
0.28 Assuming Independence
0.88
Bank S
1 Up
2 Up
3 Down
4 Down
X Utility S
Up
Down
Up
Down
P(X)
0 P(0)
1 P(1)
2 P(2) # of Successes
2
1
1
0
P(X)
0.12
0.46
0.42
1 0.42
0.28
0.18
0.12
1 Marketing Research Company provides assessments of how well women stores will do in mall
From Past Data:
Root Prob
Conditional Prob
Store Do Good
Good P(G)
0.60 Earn Profit given that Store Do Good
Earn Profit
Good P(G)
0.60 Not Earn Profit given that Store Do Good Not Earn Profit
Store Do Fair
Fair P(F)
0.30 Earn Profit given that Store Do Fair
Earn Profit
Fair P(F)
0.30 Not Earn Profit given that Store Do Fair
Not Earn Profit
Store Do Poor
Poor P(P)
0.10 Earn Profit given that Store Do Poor
Earn Profit
Poor P(P)
0.10 Not Earn Profit given that Store Do Poor
Not Earn Profit
1.00
check
Store Do Good
Store Do Fair
Store Do Poor
Earn Profit given that Store Do Good
Earn Profit given that Store Do Fair
Earn Profit given that Store Do Poor 0.60
0.30
0.10
0.8
0.6
0.2 P(Earn Profit  G)
P(Not Earn Profit  G)
P(Earn Profit  F)
P(Not Earn Profit  F)
P(Earn Profit  P)
P(Not Earn Profit  P) Joint Prob
0.8 P(Good AND Earn Profit)
0.2 P(Good AND Not Earn Profit)
0.6 P(Fair AND Earn Profit)
0.4 P(Fair AND Not Earn Profit)
0.2 P(Poor AND Earn Profit)
0.8 P(Poor AND Not Earn Profit)
1
1
1 P(Poor  Earned a Profit) 0.0294117647
0.0294117647 0.48
0.12
0.18
0.12
0.02
0.08
1 B1
B1 MP
ST Box 1 Mesh Polo Shirts
Box 1 SuperT Shirts B2
B2 MP
ST Box 2 Mesh Polo Shirts
Box 2 SuperT Shirts 25
15
40
30
10
40 Joint Probabilities
Joint Probabilities
Joint Probabilities
Joint Probabilities 2
0.5
0.5
0.625
0.375
0.750
0.25
80
55
25
0.6875
0.3125 Box selecetd at random, then
shirt selected at random, it was
a MP.
What is P(B1  MP) Rott Probabilities
Rott Probabilities
Conditional Probabilities
Conditional Probabilities
Conditional Probabilities
Conditional Probabilities # of B1
# of B2
Total Boxes =
P(B1)
P(B2)
P(MPB1)
P(STB1)
P(MPB2)
P(STB2)
Total Shirts
Total MP Shirts
Total ST Shirts
P(MP from whole pile of Shirts)
P(ST from whole pile of Shirts) 0.625 P(B1) * P(MPB1)
P(B1) * P(STB1)
P(B2) * P(MPB2)
P(B2) * P(STB2) 1
1 0.3125
0.1875
0.3750
0.1250 checks 1
1
1 80
1 0.625
Posterior Prob
0.50000 Posterior Prob
Posterior Prob
0.5 Posterior Prob
1 P(B1  MP)
P(B2  MP)
P(B1  ST)
P(B2  ST) 0.4545
0.5455
0.6000
0.4000 1.0000
1.0000 B1 MP Box 1 Mesh Polo Shirts
B1 ST Box 1 SuperT Shirts
B2 MP Box 2 Mesh Polo Shirts
B2 ST Box 2 SuperT Shirts 25
15
40
30
10
40 Conditional
Root P(MP  B1)
P(ST  B1) P(B1) 0.5000 P(ST  B2) Joint
0.6250 P(B1) * P(MP  B1)
0.3750 P(B1) * P(ST  B1) 0.5000 P(B2) Joint for Posterior need to
be grouped by First part of
Conditional Proba
0.3125
0.1875 P(B1) * P(MP  B1)
P(B2) * P(MP  B2) 0.7500 P(B2) * P(MP  B2)
0.2500 P(B2) * P(ST  B2) 0.3750
0.1250
0.5000 P(B1) * P(ST  B1)
P(B2) * P(ST  B2) 1 0.6875 0.1875
0.1250
P(ST from whole
0.3125 pile of Shirts) 1 0.6
0.4 0.3125
0.3750
P(MP from whole
0.6875 pile of Shirts) 0.5000
P(MP  B2) check Posterior Probability
(Revised Prob given new
information)  the new
information here is Mens
Polo and now we go
backward to find prob that
it's from box 1
0.4545454545
0.5454545455 0.3125 SD Odds of free soft drink
Odds of Not winning free soft drink
P Odds of not winning free pizza
Odds of Not winning free pizza
Outcomes
Trials =
Sample space = Outcomes ^ Trials
P(Win P) =
P(Win SD) =
a
b
c
d P(Win SD or P)
P(Not Win)
P(Not Win in 3 tries)
P(at leat one prize on next three visits) P(SD)
P(NOT WIN SD)
P(P)
P(NOT WIN P) 0.1 10 10 100 0.1000 0.1 TRUE 50 2
12 100 0.0200 0.02 TRUE 0.9
0.02 0.98
2 Outcomes = win or not win
1
2 prob 0.1
0.02 AND Or AND
P((Win P AND NotWin SD)) OR
0.1160 =1/50*9/10+1/10*49/50 P(P)*P(NotSD)+P(S)*P(NotP) (Win SD AND NotWin P))
0.8840
0.6908
3
0.3092
Possible outcomes
1 P(W)
2 P(Not W) 0.1160
0.8840 0.1200 All Possible =
Subgroup size = Select 3 numbers from 10 possible
n
r 10
3 In this lottery, if a number is selected for one position in sub group, it cannot
be used for another position in the subgroup  so 337 could not be used
because there are 2 3s, but 307 or 370 could be used.
Combination (2,1 is not different from 1,2) = n!/(r!*(nr)!)
Permutation (2,1 is different from 1,2) = n!/((nr)!
a
b
c Permutations =
Permutations =
P(win) = # tickets bought =
P(win when you buy 3 tickets) =
P(not win after buying 3 ticket) = 720
720
0.0013888889
3 0.0041666667
0.9958333333 Ways to make hamburger
Items avaiable on Hamberger:
Mustard
Ketchup
Onion
pickel
tomato
relish
mayonnaise
lettuce
Count of items:
You may choose to have, or ommit, any combination of items, but
Onion,Relish is the same as Relish, Onion. Thus use Combination.
Order does not matter 256 8 =n Combinations
r
0 items on Burger r0
1 items on Burger r1
2 items on Burger r2
3 items on Burger r3
4 items on Burger r4
5 items on Burger r5
6 items on Burger r6
7 items on Burger r7
8 items on Burger r8
Total The advertisement is correct because if you find the combination for each possible number from 0 (no items on burger) to 8 (all
items on burger), and then add them up you get 256.
Testbook:
Outcomes =
Trials = n
Sample Space = (# of Outcomes)^(#of trials) = 2
8 256 0
1
2
3
4
5
6
7
8 1
8
28
56
70
56
28
8
1
256 Beijing
Xi'an B
X P(B)
P(X)
P(B AND X)
P(at least 1 site) = P(B or X or Both) = P(B OR X) P(B) = 0.6 P(B AND X) = 0.3
0.4 P(X) = 0.6 P(B) = 0.6
0.4 P(X) = 0.4
0.3 P(B AND X) = 0.3 0.7 P(B AND X) = 0.3 P(X) = 0.4 Stop Smoking Gum
S
NS
# in Group of smokers that try to quit = # of Trials = n =
# of Outcomes = Stop or Not Stop =
Total Possible Outcomes =
P(0 stop smoking) =
P(at least 1 stop smoking) = P(1 stop OR 2 stop OR 3 stop OR 4 stop) = S w SSG
NS w SSG P(S w SSG)
P(NS w SSG) 0.6 0.4 4
2 16
0.0256
0.9744 check 0.9744 # of people
Possible outcomes # in Group of smokers that try to quit = # of Trials = n =
Probability
that STOP
1st person 2nd person 3rd person 4th person
1S
2 NS
3S
4S
5S
6 NS
7 NS
8 NS
9S
10 S
11 S
12 S
13 NS
14 NS
15 NS
16 NS S
S
NS
S
S
NS
S
S
NS
NS
S
NS
S
NS
NS
NS S
S
S
NS
S
S
NS
S
NS
S
NS
NS
NS
S
NS
NS S
S
S
S
NS
S
S
NS
S
NS
NS
NS
NS
NS
S
NS 4
3
3
3
3
2
2
2
2
2
2
1
1
1
1
0 0.1296
0.0864
0.0864
0.0864
0.0864
0.0576
0.0576
0.0576
0.0576
0.0576
0.0576
0.0384
0.0384
0.0384
0.0384
0.0256
1 # of Exterior possibilities
# of Interior possibilities
# of different ways to offer Exterior and Interior plans to Customers = 5
3
15 Breaks Do Not work
Breaks Do work
Steering Does Not Work
Steering Does Work
Assume events are Independent B
BW
S
SW P(B)
P(BW)
P(S)
P(SW) If 1 or the other problem is
present (but not both) car is called
a:
Lemon P(B OR S) = P(B)*P(SW)+P(S)*P(BW)
If Both Problems are present Car
Hazard P(B AND S) = P(B) * P(S)
is called a: 0.15 0.85 1 0.05 0.95 1 You must do it this way because there are 2 possibilities that must
be added, but each possibility is a "AND"jointmultiplying situation.
The first possibility is that the breaks don't work AND the Steering
works (the AND implies multiplying) OR the steering does not work
0.185
AND the Breaks do work (the AND implies multiplying). 0.0075 how ma ny license pla tes a re possible?
# of pla ces tha t License Pla te numbers a re a llowed:
# of pla ces tha t License Pla te letters a re a llowed:
Number of Different Numbers
Number of Different Letters
# of license pla tes 3
3
10
26
17,576 ,000 = 10*10*10*26 *26 *26 = 10^3*26 ^3 10 PO SSIBLE CHARACTERS HERE = = > 10 PO SSIBLE CHARACTERS HERE = = > 10 PO SSIBLE CHARACTERS HERE = = >
0 26 PO SSIBLE CHARACTERS HERE = = > 000 AAA
001 AAA
002 AAA
003 AAA
004 AAA
005 AAA
006 AAA
007 AAA
008 AAA
009 AAA
010 AAA
011 AAA
012 AAA
013 AAA
014 AAA
015 AAA
016 AAA
017 AAA
018 AAA
019 AAA
020 AAA
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023 AAA
024 AAA
025 AAA
026 AAA
027 AAA
028 AAA
029 AAA
030 AAA
031 AAA
032 AAA
033 AAA
034 AAA
035 AAA
036 AAA
037 AAA
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039 AAA
04 0 AAA
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050 AAA
051 AAA
052 AAA
053 AAA
054 AAA
055 AAA
056 AAA
057 AAA
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059 AAA
06 0 AAA
06 1 AAA
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070 AAA
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09 0 AAA
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100 AAA
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14 0 AAA
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150 AAA
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16 0 AAA
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200 AAA
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24 0 AAA
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26 0 AAA
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29 0 AAA
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300 AAA
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34 0 AAA
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350 AAA
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359 AAA
36 0 AAA
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370 AAA
371 AAA
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384 AAA
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39 0 AAA
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4 00 AAA
4 01 AAA
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4 4 0 AAA
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4 50 AAA
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4 6 0 AAA
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4 70 AAA
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4 9 0 AAA
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4 9 9 AAA
500 AAA
501 AAA
502 AAA
503 AAA
504 AAA
505 AAA
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508 AAA
509 AAA
510 AAA
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525 AAA
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527 AAA
528 AAA
529 AAA
530 AAA
531 AAA
532 AAA
533 AAA
534 AAA
535 AAA
536 AAA
537 AAA
538 AAA
539 AAA
54 0 AAA
54 1 AAA
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550 AAA
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556 AAA
557 AAA
558 AAA
559 AAA
56 0 AAA
56 1 AAA
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570 AAA
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59 0 AAA
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6 00 AAA
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6 31 AAA
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6 33 AAA
6 34 AAA
6 35 AAA
6 36 AAA
6 37 AAA
6 38 AAA
6 39 AAA
6 4 0 AAA
6 4 1 AAA
6 4 2 AAA
6 4 3 AAA
6 4 4 AAA
6 4 5 AAA
6 4 6 AAA
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6 4 8 AAA
6 4 9 AAA
6 50 AAA
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6 53 AAA
6 54 AAA
6 55 AAA
6 56 AAA
6 57 AAA
6 58 AAA
6 59 AAA
6 6 0 AAA
6 6 1 AAA
6 6 2 AAA
6 6 3 AAA
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6 6 5 AAA
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6 6 8 AAA
6 6 9 AAA
6 70 AAA
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6 88 AAA
6 89 AAA
6 9 0 AAA
6 9 1 AAA
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6 9 6 AAA
6 9 7 AAA
6 9 8 AAA
6 9 9 AAA
700 AAA
701 AAA
702 AAA
703 AAA
704 AAA
705 AAA
706 AAA
707 AAA
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709 AAA
710 AAA
711 AAA
712 AAA
713 AAA
714 AAA
715 AAA
716 AAA
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720 AAA
721 AAA
722 AAA
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725 AAA
726 AAA
727 AAA
728 AAA
729 AAA
730 AAA
731 AAA
732 AAA
733 AAA
734 AAA
735 AAA
736 AAA
737 AAA
738 AAA
739 AAA
74 0 AAA
74 1 AAA
74 2 AAA
74 3 AAA
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74 5 AAA
74 6 AAA
74 7 AAA
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74 9 AAA
750 AAA
751 AAA
752 AAA
753 AAA
754 AAA
755 AAA
756 AAA
757 AAA
758 AAA
759 AAA
76 0 AAA
76 1 AAA
76 2 AAA
76 3 AAA
76 4 AAA
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76 6 AAA
76 7 AAA
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76 9 AAA
770 AAA
771 AAA
772 AAA
773 AAA
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778 AAA
779 AAA
780 AAA
781 AAA
782 AAA
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785 AAA
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788 AAA
789 AAA
79 0 AAA
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800 AAA
801 AAA
802 AAA
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820 AAA
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830 AAA
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832 AAA
833 AAA
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835 AAA
836 AAA
837 AAA
838 AAA
839 AAA
84 0 AAA
84 1 AAA
84 2 AAA
84 3 AAA
84 4 AAA
84 5 AAA
84 6 AAA
84 7 AAA
84 8 AAA
84 9 AAA
850 AAA
851 AAA
852 AAA
853 AAA
854 AAA
855 AAA
856 AAA
857 AAA
858 AAA
859 AAA
86 0 AAA
86 1 AAA
86 2 AAA
86 3 AAA
86 4 AAA
86 5 AAA
86 6 AAA
86 7 AAA
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86 9 AAA
870 AAA
871 AAA
872 AAA
873 AAA
874 AAA
875 AAA
876 AAA
877 AAA
878 AAA
879 AAA
880 AAA
881 AAA
882 AAA
883 AAA
884 AAA
885 AAA
886 AAA
887 AAA
888 AAA
889 AAA
89 0 AAA
89 1 AAA
89 2 AAA
89 3 AAA
89 4 AAA
89 5 AAA
89 6 AAA
89 7 AAA
89 8 AAA
89 9 AAA
9 00 AAA
9 01 AAA
9 02 AAA
9 03 AAA
9 04 AAA
9 05 AAA
9 06 AAA
9 07 AAA
9 08 AAA
9 09 AAA
9 10 AAA
9 11 AAA
9 12 AAA
9 13 AAA
9 14 AAA
9 15 AAA
9 16 AAA
9 17 AAA
9 18 AAA
9 19 AAA
9 20 AAA
9 21 AAA
9 22 AAA
9 23 AAA
9 24 AAA
9 25 AAA
9 26 AAA
9 27 AAA
9 28 AAA
9 29 AAA
9 30 AAA
9 31 AAA
9 32 AAA
9 33 AAA
9 34 AAA
9 35 AAA
9 36 AAA
9 37 AAA
9 38 AAA
9 39 AAA
9 4 0 AAA
9 4 1 AAA
9 4 2 AAA
9 4 3 AAA
9 4 4 AAA
9 4 5 AAA
9 4 6 AAA
9 4 7 AAA
9 4 8 AAA
9 4 9 AAA
9 50 AAA
9 51 AAA
9 52 AAA
9 53 AAA
9 54 AAA
9 55 AAA
9 56 AAA
9 57 AAA
9 58 AAA
9 59 AAA
9 6 0 AAA
9 6 1 AAA
9 6 2 AAA
9 6 3 AAA
9 6 4 AAA
9 6 5 AAA
9 6 6 AAA
9 6 7 AAA
9 6 8 AAA
9 6 9 AAA
9 70 AAA
9 71 AAA
9 72 AAA
9 73 AAA
9 74 AAA
9 75 AAA
9 76 AAA
9 77 AAA
9 78 AAA
9 79 AAA
9 80 AAA
9 81 AAA
9 82 AAA
9 83 AAA
9 84 AAA
9 85 AAA
9 86 AAA
9 87 AAA
9 88 AAA
9 89 AAA
9 9 0 AAA
9 9 1 AAA
9 9 2 AAA
9 9 3 AAA
9 9 4 AAA
9 9 5 AAA
9 9 6 AAA
9 9 7 AAA
9 9 8 AAA
9 9 9 AAA
000 AAB
001 AAB
002 AAB
003 AAB
004 AAB
005 AAB 0A 26 PO SSIBLE CHARACTERS HERE = = > 0
Sta rt of wha t full list would look like 26 PO SSIBLE CHARACTERS HERE = = > Use the mul ti pl icatio n Rul e beca use du pli cates are o k. Where as with Co mbin atio ns and Pe rmutation s, you ca n't
re pea t the sa me chara cte r m ore th an on ce (unl ess it is orig ina ll y l isted more th an once , li ke Blu e, Bl ue, R ed).
If you used the two permuta tion formula s, multiplying them together would not give you a ll the possibilities
Permuta tions of Numbers
720
Permuta tions of Letters
15,6 00
product
11,232,000 A A 289 AAC
29 0 AAC
29 1 AAC
29 2 AAC
29 3 AAC
29 4 AAC
29 5 AAC
29 6 AAC
29 7 AAC
29 8 AAC
29 9 AAC
300 AAC
301 AAC
302 AAC
303 AAC
304 AAC
305 AAC
306 AAC
307 AAC
308 AAC
309 AAC
310 AAC
311 AAC
312 AAC
313 AAC
314 AAC
315 AAC
316 AAC
317 AAC
318 AAC
319 AAC
320 AAC
321 AAC
322 AAC
323 AAC
324 AAC
325 AAC
326 AAC
327 AAC
328 AAC
329 AAC
330 AAC
331 AAC
332 AAC
333 AAC
334 AAC
335 AAC
336 AAC
337 AAC
338 AAC
339 AAC
34 0 AAC
34 1 AAC
34 2 AAC
34 3 AAC
34 4 AAC
34 5 AAC
34 6 AAC
34 7 AAC
34 8 AAC
34 9 AAC
350 AAC
351 AAC
352 AAC
353 AAC
354 AAC
355 AAC
356 AAC
357 AAC
358 AAC
359 AAC
36 0 AAC
36 1 AAC
36 2 AAC
36 3 AAC
36 4 AAC
36 5 AAC
36 6 AAC
36 7 AAC
36 8 AAC
36 9 AAC
370 AAC
371 AAC
372 AAC
373 AAC
374 AAC
375 AAC
376 AAC
377 AAC
378 AAC
379 AAC
380 AAC
381 AAC
382 AAC
383 AAC
384 AAC
385 AAC
386 AAC
387 AAC
388 AAC
389 AAC
39 0 AAC
39 1 AAC
39 2 AAC
39 3 AAC
39 4 AAC
39 5 AAC
39 6 AAC
39 7 AAC
39 8 AAC
39 9 AAC
4 00 AAC
4 01 AAC
4 02 AAC
4 03 AAC
4 04 AAC
4 05 AAC
4 06 AAC
4 07 AAC
4 08 AAC
4 09 AAC
4 10 AAC
4 11 AAC
4 12 AAC
4 13 AAC
4 14 AAC
4 15 AAC
4 16 AAC
4 17 AAC
4 18 AAC
4 19 AAC
4 20 AAC
4 21 AAC
4 22 AAC
4 23 AAC
4 24 AAC
4 25 AAC
4 26 AAC
4 27 AAC
4 28 AAC
4 29 AAC
4 30 AAC
4 31 AAC
4 32 AAC
4 33 AAC
4 34 AAC
4 35 AAC
4 36 AAC
4 37 AAC
4 38 AAC
4 39 AAC
4 4 0 AAC
4 4 1 AAC
4 4 2 AAC
4 4 3 AAC
4 4 4 AAC
4 4 5 AAC
4 4 6 AAC
4 4 7 AAC
4 4 8 AAC
4 4 9 AAC
4 50 AAC
4 51 AAC
4 52 AAC
4 53 AAC
4 54 AAC
4 55 AAC
4 56 AAC
4 57 AAC
4 58 AAC
4 59 AAC
4 6 0 AAC
4 6 1 AAC
4 6 2 AAC
4 6 3 AAC
4 6 4 AAC
4 6 5 AAC
4 6 6 AAC
4 6 7 AAC
4 6 8 AAC
4 6 9 AAC
4 70 AAC
4 71 AAC
4 72 AAC
4 73 AAC
4 74 AAC
4 75 AAC
4 76 AAC
4 77 AAC
4 78 AAC
4 79 AAC
4 80 AAC
4 81 AAC
4 82 AAC
4 83 AAC
4 84 AAC
4 85 AAC
4 86 AAC
4 87 AAC
4 88 AAC
4 89 AAC
4 9 0 AAC
4 9 1 AAC
4 9 2 AAC
4 9 3 AAC
4 9 4 AAC
4 9 5 AAC
4 9 6 AAC
4 9 7 AAC
4 9 8 AAC
4 9 9 AAC
500 AAC
501 AAC
502 AAC
503 AAC
504 AAC
505 AAC
506 AAC
507 AAC
508 AAC
509 AAC
510 AAC
511 AAC
512 AAC
513 AAC
514 AAC
515 AAC
516 AAC
517 AAC
518 AAC
519 AAC
520 AAC
521 AAC
522 AAC
523 AAC
524 AAC
525 AAC
526 AAC
527 AAC
528 AAC
529 AAC
530 AAC
531 AAC
532 AAC
533 AAC
534 AAC
535 AAC
536 AAC
537 AAC
538 AAC
539 AAC
54 0 AAC
54 1 AAC
54 2 AAC
54 3 AAC
54 4 AAC
54 5 AAC
54 6 AAC
54 7 AAC
54 8 AAC
54 9 AAC
550 AAC
551 AAC
552 AAC
553 AAC
554 AAC
555 AAC
556 AAC
557 AAC
558 AAC
559 AAC
56 0 AAC
56 1 AAC
56 2 AAC
56 3 AAC
56 4 AAC
56 5 AAC
56 6 AAC
56 7 AAC
56 8 AAC
56 9 AAC
570 AAC
571 AAC
572 AAC
573 AAC
574 AAC
575 AAC
576 AAC
577 AAC
578 AAC
579 AAC
580 AAC
581 AAC
582 AAC
583 AAC
584 AAC
585 AAC
586 AAC
587 AAC
588 AAC
589 AAC
59 0 AAC
59 1 AAC
59 2 AAC
59 3 AAC
59 4 AAC
59 5 AAC
59 6 AAC
59 7 AAC
59 8 AAC
59 9 AAC
6 00 AAC
6 01 AAC
6 02 AAC
6 03 AAC
6 04 AAC
6 05 AAC
6 06 AAC
6 07 AAC
6 08 AAC
6 09 AAC
6 10 AAC
6 11 AAC
6 12 AAC
6 13 AAC
6 14 AAC
6 15 AAC
6 16 AAC
6 17 AAC
6 18 AAC
6 19 AAC
6 20 AAC
6 21 AAC
6 22 AAC
6 23 AAC
6 24 AAC
6 25 AAC
6 26 AAC
6 27 AAC
6 28 AAC
6 29 AAC
6 30 AAC
6 31 AAC
6 32 AAC
6 33 AAC
6 34 AAC
6 35 AAC
6 36 AAC
6 37 AAC
6 38 AAC
6 39 AAC
6 4 0 AAC
6 4 1 AAC
6 4 2 AAC
6 4 3 AAC
6 4 4 AAC
6 4 5 AAC
6 4 6 AAC
6 4 7 AAC
6 4 8 AAC
6 4 9 AAC
6 50 AAC
6 51 AAC
6 52 AAC
6 53 AAC
6 54 AAC
6 55 AAC
6 56 AAC
6 57 AAC
6 58 AAC
6 59 AAC
6 6 0 AAC
6 6 1 AAC
6 6 2 AAC
6 6 3 AAC
6 6 4 AAC
6 6 5 AAC
6 6 6 AAC
6 6 7 AAC
6 6 8 AAC
6 6 9 AAC
6 70 AAC
6 71 AAC
6 72 AAC
6 73 AAC
6 74 AAC
6 75 AAC
6 76 AAC
6 77 AAC
6 78 AAC
6 79 AAC
6 80 AAC
6 81 AAC
6 82 AAC
6 83 AAC
6 84 AAC
6 85 AAC
6 86 AAC
6 87 AAC
6 88 AAC
6 89 AAC
6 9 0 AAC
6 9 1 AAC
6 9 2 AAC
6 9 3 AAC
6 9 4 AAC
6 9 5 AAC
6 9 6 AAC
6 9 7 AAC
6 9 8 AAC
6 9 9 AAC
700 AAC
701 AAC
702 AAC
703 AAC
704 AAC
705 AAC
706 AAC
707 AAC
708 AAC
709 AAC
710 AAC
711 AAC
712 AAC
713 AAC
714 AAC
715 AAC
716 AAC
717 AAC
718 AAC
719 AAC
720 AAC
721 AAC
722 AAC
723 AAC
724 AAC
725 AAC
726 AAC
727 AAC
728 AAC
729 AAC
730 AAC
731 AAC
732 AAC
733 AAC
734 AAC
735 AAC
736 AAC
737 AAC
738 AAC
739 AAC
74 0 AAC
74 1 AAC
74 2 AAC
74 3 AAC
74 4 AAC
74 5 AAC
74 6 AAC
74 7 AAC
74 8 AAC
74 9 AAC
750 AAC
751 AAC
752 AAC
753 AAC
754 AAC
755 AAC
756 AAC
757 AAC
758 AAC
759 AAC
76 0 AAC
76 1 AAC
76 2 AAC
76 3 AAC
76 4 AAC
76 5 AAC
76 6 AAC
76 7 AAC
76 8 AAC
76 9 AAC
770 AAC
771 AAC
772 AAC
773 AAC
774 AAC
775 AAC
776 AAC
777 AAC
778 AAC
779 AAC
780 AAC
781 AAC
782 AAC
783 AAC
784 AAC
785 AAC
786 AAC
787 AAC
788 AAC
789 AAC
79 0 AAC
79 1 AAC
79 2 AAC
79 3 AAC
79 4 AAC
79 5 AAC
79 6 AAC
79 7 AAC
79 8 AAC
79 9 AAC
800 AAC
801 AAC
802 AAC
803 AAC
804 AAC
805 AAC
806 AAC
807 AAC
808 AAC
809 AAC
810 AAC
811 AAC
812 AAC
813 AAC
814 AAC
815 AAC
816 AAC
817 AAC
818 AAC
819 AAC
820 AAC
821 AAC
822 AAC
823 AAC
824 AAC
825 AAC
826 AAC
827 AAC
828 AAC
829 AAC
830 AAC
831 AAC
832 AAC
833 AAC
834 AAC
835 AAC
836 AAC
837 AAC
838 AAC
839 AAC
84 0 AAC
84 1 AAC
84 2 AAC
84 3 AAC
84 4 AAC
84 5 AAC
84 6 AAC
84 7 AAC
84 8 AAC
84 9 AAC
850 AAC
851 AAC
852 AAC
853 AAC
854 AAC
855 AAC
856 AAC
857 AAC
858 AAC
859 AAC
86 0 AAC
86 1 AAC
86 2 AAC
86 3 AAC
86 4 AAC
86 5 AAC
86 6 AAC
86 7 AAC
86 8 AAC
86 9 AAC
870 AAC
871 AAC
872 AAC
873 AAC
874 AAC
875 AAC
876 AAC
877 AAC
878 AAC
879 AAC
880 AAC
881 AAC
882 AAC
883 AAC
884 AAC
885 AAC
886 AAC
887 AAC
888 AAC
889 AAC
89 0 AAC
89 1 AAC
89 2 AAC
89 3 AAC
89 4 AAC
89 5 AAC
89 6 AAC
89 7 AAC
89 8 AAC
89 9 AAC
9 00 AAC
9 01 AAC
9 02 AAC
9 03 AAC
9 04 AAC
9 05 AAC
9 06 AAC
9 07 AAC
9 08 AAC
9 09 AAC
9 10 AAC
9 11 AAC
9 12 AAC
9 13 AAC
9 14 AAC
9 15 AAC
9 16 AAC
9 17 AAC
9 18 AAC
9 19 AAC
9 20 AAC
9 21 AAC
9 22 AAC
9 23 AAC
9 24 AAC
9 25 AAC
9 26 AAC
9 27 AAC
9 28 AAC
9 29 AAC
9 30 AAC
9 31 AAC
9 32 AAC
9 33 AAC
9 34 AAC
9 35 AAC
9 36 AAC
9 37 AAC
9 38 AAC
9 39 AAC
9 4 0 AAC
9 4 1 AAC
9 4 2 AAC
9 4 3 AAC
9 4 4 AAC
9 4 5 AAC
9 4 6 AAC
9 4 7 AAC
9 4 8 AAC
9 4 9 AAC
9 50 AAC
9 51 AAC
9 52 AAC
9 53 AAC
9 54 AAC
9 55 AAC
9 56 AAC
9 57 AAC
9 58 AAC
9 59 AAC
9 6 0 AAC
9 6 1 AAC
9 6 2 AAC
9 6 3 AAC
9 6 4 AAC
9 6 5 AAC
9 6 6 AAC
9 6 7 AAC
9 6 8 AAC
9 6 9 AAC
9 70 AAC
9 71 AAC
9 72 AAC
9 73 AAC
9 74 AAC
9 75 AAC
9 76 AAC
9 77 AAC
9 78 AAC
9 79 AAC
9 80 AAC
9 81 AAC
9 82 AAC
9 83 AAC
9 84 AAC
9 85 AAC
9 86 AAC
9 87 AAC
9 88 AAC
9 89 AAC
9 9 0 AAC
9 9 1 AAC
9 9 2 AAC
9 9 3 AAC
9 9 4 AAC
9 9 5 AAC
9 9 6 AAC
9 9 7 AAC
9 9 8 AAC
9 9 9 AAC
1000 AAD
1001 AAC
1002 AAC
1003 AAC
1004 AAC
1005 AAC
1006 AAC
1007 AAC
1008 AAC
1009 AAC
1010 AAC
1011 AAC
1012 AAC
1013 AAC
1014 AAC
1015 AAC
1016 AAC
1017 AAC
1018 AAC
1019 AAC
1020 AAC
1021 AAC
1022 AAC
1023 AAC
1024 AAC
1025 AAC
1026 AAC
1027 AAC
1028 AAC
1029 AAC
1030 AAC
1031 AAC
1032 AAC
1033 AAC
1034 AAC
1035 AAC
1036 AAC
1037 AAC
1038 AAC
1039 AAC
104 0 AAC
104 1 AAC
104 2 AAC
104 3 AAC
104 4 AAC
104 5 AAC
104 6 AAC
104 7 AAC
104 8 AAC
104 9 AAC
1050 AAC
1051 AAC
1052 AAC
1053 AAC
1054 AAC
1055 AAC
1056 AAC
1057 AAC
1058 AAC
1059 AAC
106 0 AAC
106 1 AAC
106 2 AAC
106 3 AAC
106 4 AAC
106 5 AAC
106 6 AAC
106 7 AAC
106 8 AAC
106 9 AAC
1070 AAC
1071 AAC
1072 AAC
1073 AAC
1074 AAC
1075 AAC
1076 AAC
1077 AAC
1078 AAC
1079 AAC
1080 AAC
1081 AAC
1082 AAC
1083 AAC
1084 AAC
1085 AAC
1086 AAC
1087 AAC
1088 AAC
1089 AAC
109 0 AAC
109 1 AAC
109 2 AAC
109 3 AAC
109 4 AAC
109 5 AAC
109 6 AAC
109 7 AAC
109 8 AAC
109 9 AAC
1100 AAC
1101 AAC
1102 AAC
1103 AAC
1104 AAC
1105 AAC
1106 AAC
1107 AAC
1108 AAC
1109 AAC
1110 AAC
1111 AAC
1112 AAC
1113 AAC
1114 AAC
1115 AAC
1116 AAC
1117 AAC
1118 AAC
1119 AAC
1120 AAC
1121 AAC
1122 AAC
1123 AAC
1124 AAC
1125 AAC
1126 AAC
1127 AAC
1128 AAC
1129 AAC
1130 AAC
1131 AAC
1132 AAC
1133 AAC
1134 AAC
1135 AAC
1136 AAC
1137 AAC
1138 AAC
1139 AAC
114 0 AAC
114 1 AAC
114 2 AAC
114 3 AAC
114 4 AAC
114 5 AAC
114 6 AAC
114 7 AAC
114 8 AAC
114 9 AAC
1150 AAC
1151 AAC
1152 AAC
1153 AAC
1154 AAC
1155 AAC
1156 AAC
1157 AAC
1158 AAC
1159 AAC
116 0 AAC
116 1 AAC
116 2 AAC
116 3 AAC
116 4 AAC
116 5 AAC
116 6 AAC
116 7 AAC
116 8 AAC
116 9 AAC
1170 AAC
1171 AAC
1172 AAC
1173 AAC
1174 AAC
1175 AAC
1176 AAC
1177 AAC
1178 AAC
1179 AAC
1180 AAC
1181 AAC
1182 AAC
1183 AAC
1184 AAC
1185 AAC
1186 AAC
1187 AAC
1188 AAC
1189 AAC
119 0 AAC Total
> 60 years of age
Female
Female and > 60 years of age
Male
Male and > 60 years of age
< 60 years of age Total # of people being considered =
> 60 years of age # of people being considered =
Female # of people being considered =
Female and > 60 years of age # of people being considered =
Male # of people being considered =
Male and > 60 years of age # of people being considered =
< 60 years of age # of people being considered = 4
3
2
1
2
2
1 P(> 60 years of age)
P(Female)
P(Female and > 60 years of age)
P(Male)
P(Male and > 60 years of age)
P(< 60 years of age) < 60 years of age  Male
Female  > 60 years of age < 60 years of age  Male # of people being considered =
Female  > 60 years of age # of people being considered = 0
1 P(< 60 years of age  Male)
P(Female  > 60 years of age) 0.75
0.5
0.25
0.5
0.5
0.25
0
0.333 0.25 H
N
a # of Holly Land tracts purchased =
# of Newburg Land tracts purchased = 2 P(next 2 tracts sold are N) =
0 4 Trials P(next 4 tracts sold 0 are H) =
b
P(next 4 tracts at least 1 is H) =
P(next 4 tracts at least 1 is H) =
c
Dependent
Success = Sell H
Sell H
Not Sell H
# of Outcomes =
# of Trials
Sample Space = Outcomes ^ (# of Trials) 4 P(H)
6 P(N)
10 0.4
0.6 0.3333
= 6/10*5/9
0.0714285714
0.92857143 0.92857143 1 SH
1 NSH
2
4
16
Try to sell Tract (S =
SH)
Count of Successes (S = SH) Probability of Occurrence
1
2
3
4
1 SH SH SH SH
4
0.00476190
2 NSH SH SH SH
3
0.02857143
3 SH NSH SH SH
3
0.02857143
4 SH SH NSH SH
3
0.02857143
5 SH SH SH NSH
3
0.02857143
6 NSH NSH SH SH
2
0.07142857
7 SH NSH NSH SH
2
0.07142857
8 SH SH NSH NSH
2
0.07142857
9 NSH SH NSH SH
2
0.07142857
10 NSH SH SH NSH
2
0.07142857
11 SH NSH SH NSH
2
0.07142857
12 SH NSH NSH NSH
1
0.09523810
13 NSH SH NSH NSH
1
0.09523810
14 NSH NSH SH NSH
1
0.09523810
15 NSH NSH NSH SH
1
0.09523810
16 NSH NSH NSH NSH
0
0.07142857
1.00000000 Possible # Outcomes 6 POSSIBLE CHARACTERS HERE ==>
456,976 6 POSSIBLE CHARACTERS HERE ==> 4
26 6 POSSIBLE CHARACTERS HERE ==> Characters available for password =
Letters available for each of the 4 characters =
Each Character can be used more than once
# of Passwords Available 6 POSSIBLE CHARACTERS HERE ==> # # # # Total Items =
Size of Sub Group =
Conditional Probability = X/Y*(X1)/(Y1) n
r 5
3 Permutations = 60 5*4*3*2*1
3*3*1
53 = 2 …… 2*1 120
6
2 120/2 = (5*4*3*2*1)/(53 = 2 …… 2*1)
3/5*2/4*1/3 120
6 60
0.1000 0.1000 Total cans in case
Cans that are contaminated
1 Trial = Pull 1 can
# of Trials = 24 n
1
3r # of combinations of cans that could be selected 2,024 P(Can Not Selected during 3 trials)
P(contaminated can is selected during Experiment) 0.875
0.125 SC, NSC, NSC
NSC, SC, NSC
NSC, NSC, SC
P(contaminated can is selected during Experiment) 0.0416667
0.0416667
0.0416667
0.125 ...
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This note was uploaded on 01/10/2012 for the course BUSN 210 taught by Professor Girvin during the Winter '09 term at Des Moines CC.
 Winter '09
 Girvin

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