Statistics
Probability
Probability
1 Events, Sample Spaces and Probability
Experiment: The observation of activity or the act of taking some measurement.
Outcome: A particular result of an experiment.
Sample Space : The collection of all possible outcomes for an experiment.
Event: A subset of a sample space.
(1) Simple event: An event containing only one element of the sample space.
(2) Compound event: The union of some simple events.
Classical Probability: A value between 0 and 1 inclusive that represents the chance that a
particular event will happen.
The number of outcomes in event A
P( A ) 
The number of outcomes in the sample space
Note: P() = 1.
Basic concepts
Complementary Events: The complement of event A is A or  – A.
Complement Rule: P( A ) = 1 – P(A)
Ex.1 A fair coin is tossed twice. Determine the following:
1.1. Sample Space: __{HH, HT, TH, TT}______________________________________________
1.2. Events: ___ HH, HT, TH, and TT ___________________________________________
_______________________________________________________________________
1.3. Probabilities:
Event
Favorable outcomes
{ TT }
1.4.1 No head occurs.
1.4.2 One head and one tail occur.
Probability
1
4
2 1

4 2
3
4
3
4
{ TH, HT }
1.4.3 At least one head occurs.
{ HT, TH, HH }
1.4.4 At most one head occurs.
{ HT, TH, TT }
Pr and Con Pr
Ex.2 Select a student at random.
Status
gender
First year
Second year
Third year
Fourth year
Worksheet Answer
© 2010-1 Dr. Min Aung
male
female
Total
19
10
4
2
35
24
9
5
35
73
54
34
13
7
108
6 pages
Statistics
2
Probability
What is the probability that the student is
2.1. A second year student?
P(Y2) =
Y2 34
 = 0.3148 (31.48%)
S 108
Ans.: 0.3148
2.2. A female student?
P(F) =
F 73
 = 0.6759
S 108
Ans.: 0.6759
2.3. A second year female student?
P(Y2F) =
Y2  F 24
 = 0.2222
S
108
Ans.: 0.2222
2.4. A second year student given that the selected one is a female student?
Y  F 24
 = 0.3288
P(Y2 / F) = 2
F
73
Y2
24
F
73
Ans.: 0.3288
2.5. A female student or a second year student or both?
P(FY2 ) =
F  Y2 F  Y2  F  Y2 73  34  24 83


 = 0.7685. Ans.: 0.7685
S
S
108
108
2 Probability of Unions
Two events A and B are mutually exclusive if A  B = .
Special Rule of Addition: If two events A and B are mutually exclusive, then
Special sum
P(A or B) = P(A  B) = P(A) + P(B)
Ex.3 The probability of a student getting an A in Statistics class is 0.15 and that of getting a B is
0.28. What is the probability that a randomly selected student from this class will get an A or
a B?
P(AB) = P(A) + P(B) = 0.15 + 0.28 = 0.43
15%
Ans.: 0.43
28%
General Rule of Addition:
P(A or B) = P(A  B) = P(A) + P(B) – P(A  B)
3 Conditional Probability and Joint Probability
The probability of an event B occurring, given that another event A has occurred is
P( A  B )
P( B / A ) 
, which implies the
P( A )
Worksheet Answer
© 2010-1 Dr. Min Aung
6 pages
Statistics
3
Probability
General Rule of Multiplication: P( A and B)  P( A  B)  P( A)P(B / A)
Joint probability: A probability that measures the likelihood that two or more events will happen
concurrently.
Ex.4 A Hamburger chain finds that 75% of customers use mustard, 80% of customers use
Ketchup, and 65% of customers use both. What is the probability that a particular customer
will use
4.1. At least one of these?
P(MK) = P(M) + P(K) – P(MK) = 0.75 + 0.80 – 0.65 = 0.90
0.75
M
Ans.: 0.90
0.80
K
0.65
4.2. Neither mustard nor ketchup?
P(M  K) = 1 – P(MK) = 1 – 0.90 = 0.10
0.9
Ans.: 0.10
K
M
1
0.1
4.3. Mustard given that the customer uses Ketchup?
P(M  K) 0.65
=
= 0.8125
0.80
P(K)
P(M / K) =
M
0.65
K
Ans.: 0.8125
81.25% of K is in M.
0.8
Q: What proportion of people who use Ketchup uses Mustard too?
4.4. Ketchup provided that the customer does not use Mustard?
P(K/ M ) =
=
Worksheet Answer
P(K  M ) P(K)  P(K  M) 0.80  0.65
=
=
1  0.75
P( M )
1 P(M)
0.15
= 0.60
0.25
Ans.: 0.60
© 2010-1 Dr. Min Aung
6 pages
Statistics
4
Probability
Ex.5 According to the 7-year records on entrance examinations of a certain university, 40% of the
candidates pass Mathematics test, 50% passes English test, while 50% of the candidates
who pass Mathematics test also pass English test. What is the probability that an entrance
taker.
5.1 Passes both tests?
P(ME) = P(M)P(E/M) = 0.40  0.50 = 0.20
M
M
Total
E
50% of 40%
M E
50%
E
ME
ME
50%
Ans.: 0.20
Total
40%
60%
100%
5.2 Passes at least one test?
P(ME) = P(M) + P(E) – P(ME) = 0.50 + 0.40 – 0.20 = 0.70
Ans.: 0.7
5.3 Fails both tests?
P( M E ) = 1 – P(ME) = 1 – 0.70 = 0.30
Ans.: 0.30
5.4 Pass Mathematics test provided that the candidate passes English test?
P(M / E) =
P(M  E) 0.20
=
= 0.40
0.50
P(E)
Ans.: 0.40
5.5 Pass Mathematics but not English?
P(M  E ) = P(M) – P(ME) = 0.40 – 0.20 = 0.20
Worksheet Answer
© 2010-1 Dr. Min Aung
Ans.: 0.20
6 pages
Statistics
5
Probability
5.6 Passes only one test?
P(ME) – P(ME) = 0.7 – 0.2 = 0.50
Ans.: 0.50
5.7 Passes Mathematics test given that the candidate fails English test?
P(M/ E ) =
P(M  E ) 0.20
= 0.40

P( E )
0.50
Ans.: 0.40
4 Independent Events
Event A is independent of event B if the occurrence B has no effect on the probability of the
occurrence of A. Hence, we call A independent of B if P(A / B)  P(A) or equivalently, if the
following rule holds.
Special rule of multiplication: For two independent events A and B,
P(A and B) = P(A  B) = P(A) P(B)
Daniel76-12
adapted
Note: If A is independent of B, then it also holds that B is independent of A. Moreover either of the
events A or A is independent of either of the events B or B .
Ex.6 In a certain population of women 4 percent have had breast cancer, 20 percent are
smokers, and 3 percent are smokers and have had breast cancer. Determine whether
having breast cancer and being a smoker are independent.
P(BC) = 0.03, P(B)P(C) = 0.04  0.20 = 0.008
P(BC) ≠ P(B)P(C)
Worksheet Answer
© 2010-1 Dr. Min Aung
Ans.: Not independent.
6 pages
General product
Statistics
6
Probability
Ex.7 A company is considering to market two new products A and B. The probability that
marketing A is successful is 0.7 and the probability for B to be successful is 0.8. Assume
that the success of one product is independent of another. Determine the probability that
7.1. Both products are successful.
P(A  B) = P(A)P(B) = 0.7  0.8 = 0.56
Ans.: 0.56
7.2. At least one product is successful.
P(A  B) = P(A) + P(B) – P(AB) = 0.7 + 0.8 – 0.56 = 0.94
Ans.: 0.94
7.3. Neither is successful.
P( A  B ) = 1 – P(AB) = 1 – 0.94 = 0.06
Ans.: 0.06
7.4. Product A is successful but not B.
P(A B ) = P(A) – P(AB) = 0.7 – 0.56 = 0.14
Ans.: 0.14
******************************************************************************************************************
Ex.8 A committee of 6 is to be formed from 5 boys and 3 teachers. If the members of the
committee are chosen at random, what is the probability that
8.1 There will be a majority of boys on the committee?
Total number of ways = 8C6 = 28
Number of favorable ways = 5C5 3C1 + 5C4 3C2 = 18
18 9
9
Probability = = = 0.6429
Ans.: or 0.6429
28 14
14
8.2 There are at least two teachers given that there are at least two boys on the committee?
Total number of forming committee containing at least two boys = 8C6 – 5C0 3C6 –
5
C1 3C5 = 28 – 0 – 0 = 28
Number of favorable ways = 5C4 3C2 + 5C3 3C3 = 25
25
25
Probability =
Ans.: or 0.8929
28
28
Worksheet Answer
© 2010-1 Dr. Min Aung
6 pages
Statistics
Worksheet Answer
7
© 2010-1 Dr. Min Aung
Probability
6 pages