1
Chapter 4
Introduction to Probability
4.1 Experiments, Counting Rules and Assigning Probabilities
Example
•
Rolling a dice you can get the values: S = {1, 2, 3, 4, 5, 6}
S is called the sample space.
•
Experiment: Rolling a dice
•
The possible experimental outcomes:
1, 2, 3, 4, 5, 6
(Experimental outcomes are also known as sample points)
•
The number of experimental outcomes: 6
•
Defining the outcomes:
E1 = 1
E2 = 2
E3 = 3
E4 = 4
E5 = 5
E6 = 6
•
Value on the dice is a one
Value on the dice is a two
Value on the dice is a three
Value on the dice is a four
Value on the dice is a five
Value on the dice is a six
Then:
P(E1) =
P(E2) =
P(E3) =
P(E4) =
P(E5) =
P(E6) =
•
Note P(E1) + P(E2) + P(E3) + P(E4) + P(E5) + P(E6) = + + + + + = = 1
•
More examples of experiments:
Coin:
Select a part for inspection:
Play a football game:
Counting Rules, Combinations and Permutations
Question:
How many experimental outcomes are there?
Copyright Reserved
1
2
Multiple-Step Experiments
Example:
•
•
Throwing 2 coins can be thought of as a two-step experiment in which step 1 is the throwing of
the first coin and step 2 is the throwing of the second coin.
•
Sample Space (S):
The number of experimental outcomes:
Tree Diagram:
A graphical representation that is helpful in visualizing a multiple-step experiment is a tree diagram.
A tree diagram for the experiment of tossing 2 coins:
Copyright Reserved
2
3
Counting Rule for Multiple-Step Experiments:
•
If an experiment can be described as a sequence of k steps with
•
n1 = possible outcomes on step 1.
n2 = possible outcomes on step 2.
nk = possible outcomes on step k.
•
Then the total number of experimental outcomes is:
(n1)(n2) … (nk)
•
The coin example:
Example:
How many different routes can a parcel follow if
• 2 starting venues
• 3 cars
• 4 delivery points
.
Copyright Reserved
3
4
Treediagram:
Copyright Reserved
4
5
Factorials !
There are 3 people at a bus stop. How many different ways can the 3 people be arranged?
Note:
0! = 1
3 factorial is: 3! = (3)(2)(1) = 6
5 factorial is: 5! = (5)(4)(3)(2)(1) = 120
Calculator: n! or x!
Examples:
1.
6! =
2.
10! =
Combinations
Order of selection is NOT important
The number of combinations of N objects taken n at a time is:
!
= =
! − !
Example:
Consider a quality control process in which an inspector selects 2 of 5 parts, to inspect for defects. In a group
of 5 parts, how many combinations of 2 parts can be selected?
Example:
Lotto: Use numbers 1 to 49
Choose a group of 6 numbers
Question: How many experimental outcomes are there?
Calculator:
or
Examples:
1. =
2. =
3. =
Copyright Reserved
5
6
Permutations
The counting rule for permutations allows one to compute the number of experimental outcomes when n
objects are to be selected from a set of N objects where the order of selection is important.
The number of permutations of N objects taken n at a time is given by:
!
= ! =
− !
Note: Order of selection is important
Example:
Consider a quality control process in which an inspector selects 2 of 5 parts to inspect for defects. How
many permutations may be selected?
Calculator: or
Assigning Probabilities
The 3 approaches most frequently used:
1.
2.
3.
Classical Method
Relative Frequency Method
Subjective Method
Classical Method
•
The classical method of assigning probabilities is appropriate when all the experimental outcomes are
equally likely.
If n experimental outcomes are possible, a probability of is assigned to each experimental outcome.
•
Examples:
•
(i) Rolling a dice: P(Ei) =
(ii) Throwing a coin: P(Ei) =
i = 1, 2, 3, 4, 5, 6
i = 1, 2
Copyright Reserved
6
7
Relative Frequency Method
It’s appropriate when data are available to estimate the proportion of the time the experimental outcome will
occur if the experiment is repeated a large number of times.
Example:
Consider a study of waiting times in the X-ray department for a local hospital. The number of patients
waiting for service at 9am was recorded for 20 successive days. The following results were obtained:
Number waiting
0
1
2
3
4
Then:
Frequency
2
5
6
4
3
20
P(0) =
P(1) =
P(2) =
P(3) =
P(4) =
Note: The bigger your sample, the closer your probabilities will be to the “true” probabilities.
Subjective Method
i.
It is most appropriate when it is unrealistic to assume that the experimental outcomes are equally
likely (Can’t use Classical Method)
OR
ii.
when little relevant data are available (can’t use Relative Frequency Method)
Example: A ranger thinks that the probability of seeing a lion is 0.3. Then the probability of not seeing a
lion is 0.7.
Copyright Reserved
7
8
4.3 Some Basic Relationships of Probability
Figure 4.4 Complement of event A is shaded
P(A) + P(AC) = 1
Figure 4.5 Union of events A and B is shaded
∪ Figure 4.6 Intersection of events A and B is shaded
∩ Addition law: ∪ = + − ∩ Copyright Reserved
8
9
Figure 4.7 Mutually exclusive events
∩ = Addition law for mutually exclusive events:
∪ = + − ∩ with ∩ = 0
Therefore, ∪ = + 4.4 Conditional Probability
Example 1:
Gender
Promoted
Male
Female
Total
Yes
No
288
672
36
204
324
876
Total
960
240
1200
Define the events: A = Promoted and M = Male. Then AC = Not promoted and MC = Female.
1. The probability that an employee is promoted? Note: This is a marginal probability.
2. The probability that an employee is not promoted? Note: This is a marginal probability.
3. The probability that an employee is male? Note: This is a marginal probability.
Copyright Reserved
9
10
4. The probability that an employee is female? Note: This is a marginal probability.
5. The probability that a randomly selected employee is male and promoted? Note: This is a joint
probability.
6. The probability that a randomly selected employee is female and promoted? Note: This is a joint
probability.
7. The probability of getting a promotion, given the employee is male. Note: This is a conditional
probability.
8. The probability of getting a promotion, given the employee is female. Note: This is a conditional
probability.
9. The probability of being male, given that the employee was promoted. Note: This is a conditional
probability.
10. The probability of being female, given that the employee was not promoted. Note: This is a
conditional probability.
Copyright Reserved
10
11
Independent events:
Two events A and B are independent if
P(A|B) = P(A)
or
P(B|A) = P(B)
Use this definition to test for dependence / independence:
We have already calculated all the appropriate marginal and conditional probabilities. In order to test for
dependence / independence between gender and whether of not an employee gets promoted, any one of the
following marginal and conditional probabilities may be compared to one another:
Option 1:
P(Male) =
P(Male | Promoted) =
| ≠ ∴ The two events are dependent
Option 2:
P(Female) =
P(Female | Not promoted) =
| ≠ ∴ The two events are dependent
Option 3:
P(Promoted) =
P(Promoted | Female) =
$
≠ ∴ The two events are dependent
There are other options, for example, comparing P(Not promoted) to P(Not promoted | Female) ect. Here
we’ve only shown 3 possible ways to test for dependence / independence between gender and whether of not
an employee gets promoted.
Copyright Reserved
11
12
Alternative definition for independence:
Two events A and B are independent if
∩ = Use this definition to test for dependence / independence:
We have already calculated all the appropriate marginal and joint probabilities. In order to test for dependence
/ independence between gender and whether of not an employee gets promoted, any one of the following
marginal and joint probabilities may be compared to one another:
Option 1:
P(Male) =
P(Promoted) = 0
= . &. '( = . ')*
P(Male ∩ Promoted) =
∩ = ∴ The two events are dependent
Option 2:
P(Female) =
P(Promoted) =
P(Female ∩ Promoted) =
= . '. '( = . +,
∩ = ∴ The two events are dependent
There are other options. Here we’ve only shown 2 possible ways to test for dependence / independence
between gender and whether of not an employee gets promoted.
Example: Independent events
Given: P(-) = P(Seeing a lion) = 0.3 and P() = P(Seeing a cheetah) = 0.08. Suppose these two events are
independent.
Question: Calculate - ∪ :
Answer:
It is given that events - and are independent, therefore - ∩ = -
- ∪ = - + − - ∩ = - + − -
= 0.3 + 0.08 − 0.30.08
= . 0+*
Addition law
Because of independence
Copyright Reserved
12
13
Typical exam questions
Questions 1 to 3 are based on the following information:
A political analyst knows from previous experience that the probability that political party A will win the
election is 0.4 and the probability that political party B will win the election is 0.35.
Define the events: = Political party A wins the election.
= Political party B wins the election.
= Political party C wins the election.
Assume: There are only 3 political parties running for election.
Note: Only one party can win the election.
Question 1
The probability that political party C will win the election is:
Answer 1
P(Party C will win) = 1 – P(Party A will win) – P(Party B will win) = 1 – 0.4 – 0.35 = 0.25.
Question 2
Events A and B are:
(A) Mutually exclusive since | ≠ 0.
(B) Mutually exclusive since ∩ = .
(C) Not mutually exclusive since | = 0.
(D) Not mutually exclusive since ∩ = 0.
(E) Mutually exclusive since ∪ = Question 3
The probability that political party A or political party B wins the election is:
(A) 0.14
(C) 0.75
(E) 0.61
(B) 0.05
(D) 0.67
Answer 3
∪ = + − ∩ = 0.4 + 0.35 − 0 = 0.75.
Copyright Reserved
13
14
Questions 4 to 7 are based on the following information:
Suppose we have a sample space with five equally likely experimental outcomes, namely: E1, E2, E3, E4
and E5.
Let
A = {E1, E2}
B = {E3, E4}
C = {E2, E3, E5}
Question 4
Calculate P(A), P(B), P(C), and P(AC), P(BC), P(CC):
Answer 4
P(A) = = 0.4 and using the fact that P(A) + P(AC) = 1 we find P(AC) = 1 - P(A) = 1 – 0.4 = 0.6.
P(B) = = 0.4 and using the fact that P(B) + P(BC) = 1 we find P(BC) = 1 - P(B) = 1 – 0.4 = 0.6.
5
P(C) = = 0.6 and using the fact that P(C) + P(CC) = 1 we find P(CC) = 1 - P(C) = 1 – 0.6 = 0.4.
Another way to calculate P(AC), P(BC) and P(CC):
5
AC = {E3, E4, E5} and consequently P(AC) = = 0.6.
C
5
C
B = {E1, E2, E5} and consequently P(B ) = = 0.6.
CC = {E1, E4} and consequently P(CC) = = 0.4.
Question 5
Calculate ∪ :
Answer 5
5
∪ = + − ∩ = + − = 0.8
Question 6
Calculate |:
Answer 6
67∩8
9
| = 68 = = 0
:
Question 7
Calculate ∩ :
Answer 7
∩ = = 0.2
Copyright Reserved
14
15
Questions 8 to 10 are based on the following information:
A survey of undergraduate students in Commerce at the XYZ University revealed the following
regarding the gender and majors of the students.
Major
Gender
Total
Accounting
Economics
130
70
Male
200
50
70
Female
120
Total
180
140
320
A student is selected at random from the 320 students.
Define the events as follows: M: Male and E: Economics
Question 8
The probability of selecting a female who majors in Economics is:
Answer 8
?9
<= ∩ > = 59 = 0.22.
Question 9
The probability of selecting an Accounting major, given that the person selected is a male, is:
Answer 9
> = |< =
6@A B ∩CD
6C
59:
= 99:59 = 0.65
59
Question 10
Events M and E are not mutually exclusive because:
(A) P(M ∩ E) = 0
(C) P(M ∪ E) = 0
(E) P(M) P(E) ≠ 0
(B) P(E | M) ≠ 0
(D) P(M ∩ E) ≠ 0
Questions 11 and 12 are based on the following information:
Suppose that we have two events, A and B, with P(A) = 0.5, P(B) = 0.6 and ∩ = 0.4.
Question 11
| =
Answer 11
67∩8
9.F
| = 68 = 9. = 0. 6G.
Question 12
The events A and B are:
(A) dependent because | ≠ (B) dependent because | = (C) independent because | = (D) dependent because | ≠ (E) independent because | = Answer 12
| = 0. 6G and = 0.5. The events are dependent, because | ≠ . Answer = D.
Copyright Reserved
15