Statistics_Chapter 3_Lecture 7
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CONTENTS LECTURE 7
Chapter 3 - Introduction to Probability (Events, Sample spaces, probability; Unions and intersections; Conditional
Probability)
Chapter 4 - Probability Distributions (Types of Variables, The Normal Distribution, The Binomial Distribution)
Chapter 5 – Inferences Based on a Single Sample: Estimation with Confidence Intervals (Large-sample confidence
interval for population mean, Small-sample confidence interval for population mean, Large-sample confidence
interval for population proportion, Sample size)
Chapter 6 – Inferences Based on a Single Sample: Hypothesis Testing (Large-sample test of hypothesis about a
population mean, p-values, Small-sample test of hypothesis about a population mean, Large-sample test of
hypothesis about a population proportion)
Chapter 7 – Inferences Based on Two Samples: Confidence Intervals and Tests of Hypotheses (Target parameter,
Comparing two population means, Comparing two population proportions, Sample size)
Chapter 8 - Analysis of Variance (ANOVA)
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Chapter 3 - Introduction to
Probability
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3. Introduction to Probability
By the end of this chapter, you will be able to . . .
1 Understand the meaning of an experiment, an outcome, an event, and a sample space.
2 Describe the classical method of assigning probability.
3 Explain the Law of Large Numbers and the relative frequency method of assigning probability.
Sometimes, the amount of uncertainty in our daily lives is so great that there appears to be no order to the world
whatsoever. However, if you look closely, there are patterns in randomness. In this chapter, we learn to become
better decision-makers by becoming acquainted with the tools of probability in order to quantify many of the
uncertainties of everyday life.
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3. Introduction to Probability
We try to cope with uncertainty by estimating the chances that a particular event will occur. We are daily called
on to make intelligent decisions about probabilities. Consider the following scenarios and think about how the
italicized words all refer to uncertainty:
• What is the chance that there will be a speed trap on this stretch of A7 on a particular day?
• What is the likelihood that this lottery ticket will make me rich?
• What is the probability that this throw of the dice will come up a six?
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Chapter 3 - Introduction to Probability
1.
Most managerial decisions are made in the face of some uncertainty regarding the possible
consequences of these decisions. The decision-maker is unsure of the outcome. Determining
the chances of the various possible outcomes occurring in advance should help in making the
correct decision.
2.
Probability is a measure of uncertainty.
3.
Probability theory deals with predicting how likely it is that something will happen.
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3.1 Experiment, Sample Space, Event
1.
Experiment
In probability, an experiment is any activity for which the outcome is uncertain. An outcome is the result
of a single performance of an experiment.
Consider the stock market, for example. Suppose you own 100 shares of Consolidated Widgets and are
interested in the share price at the end of trading tomorrow. Will the share price increase or decrease? The actual
result is uncertain.
2.
Sample point is the outcome of an experiment
3.
Sample space (S) is the collection of all possible outcomes of a random experiment
4.
Event (E) – any subset of outcomes from the sample space
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3.1 Experiment, Sample Space, Event
Examples
Experiment
Sample Space
• Toss a Coin
{Head, Tail}
• Toss 2 Coins
{HH, HT, TH, TT}
• Play a Football Game
{Win, Lose, Tie}
• Quality control check
{Defective, Good}
• Observe Gender
{Male, Female}
Note: H – Head; T - Tail
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3.1 Experiment, Sample Space, Event
Sample Space Properties
1.
Experiment: Observe Gender
Mutually Exclusive
2 outcomes can not occur at the same time
— Male & Female in same person
2.
Collectively Exhaustive
One outcome in sample space must occur.
— Male or Female
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3.1 Experiment, Sample Space, Event
Examples
Experiment: Toss 2 Coins
Sample Space:
HH, HT, TH, TT
Event
Outcomes in Event
1 Head & 1 Tail
Head on 1st Coin
At Least 1 Head
Heads on Both
HT, TH
HH, HT
HH, HT, TH
HH
Experiment: Roll a die
Sample space: {1, 2, 3, 4, 5, 6}
Event E: roll an even number = {2, 4, 6}
Event L: roll a 4 or larger = {4, 5, 6}
Note: H – Head; T - Tail
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3.1 Experiment, Sample Space, Event
Example
Let the Sample Space be the collection of all possible outcomes of rolling
one die:
S = [1, 2, 3, 4, 5, 6]
Let A be the event “Number rolled is even”
Let B be the event “Number rolled is at least 4”
Then
A = [2, 4, 6] and B = [4, 5, 6]
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3.2 Probability
The probability of an outcome represents the chance or likelihood that the
outcome will occur.
1. Numerical measure of the likelihood that an event will occur is denoted
1
Certain
P(Event) or P(A)
2.
pi represents the probability of sample point i. All sample point
probabilities must lie between 0 and 1, inclusive (0 ≤ pi ≤ 1).
3.
.5
Sum of sample points is 1. The probabilities of all sample points
within a sample space must sum to 1 ( pi = 1).
0
Impossible
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3.2 Probability
There are three approaches to assessing the probability of an uncertain event:
1. Classical method of assigning probability
2. Relative frequency method
3. Subjective method
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3.2 Probability
3.2.1Classical Probability
Assumes all outcomes in the sample space are equally likely to occur.
Equally likely outcomes are outcomes that have the same probability of occurring.
For example, if you toss a fair coin, the probability of observing either of the outcomes’ heads or tails is the same.
The classical method of assigning probabilities is used when an experiment has equally likely outcomes.
Classical probability of event A:
P(A) =
N
number of outcomes in event A
=
N
total number of outcomes in the sample space
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3.2 Probability
3.2.1Classical Probability
Example: The object of a game is to roll a 6 on a single roll of a single fair die. If you do so, you win $5. It costs $1 to
play the game. What is the probability of winning?
Solution
The sample space for a single die toss consists of six outcomes, {1, 2, 3, 4, 5, 6}. When the six outcomes are equally
likely, we say that the die is fair. If the outcomes are not equally likely, then the die is loaded or defective. If we
assume the die is fair, then, since the sum of the probabilities of the n = 6 outcomes must equal 1, the probability of
any particular outcome must equal 1/6, using the classical method.
The probability of winning: P(W ) = 1/6
P A =
number of outcomes in event A
1
=
total number of outcomes in the sample space 6
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3.2 Probability
3.2.1Classical Probability
Example: Find the probability of obtaining one head and one tail when a fair coin is tossed twice.
Solution
We assume that the N(S ) = 4 outcomes in the sample space {HH, HT, TH, TT} are equally likely. Recall that the sum
of the probabilities of all the outcomes in the sample space must equal 1. Thus, each of the four outcomes must have a
probability of 1/4.
Let E be the event that one head and one tail is obtained. Then E = {HT, TH}, so N(E ) = 2. Thus,
P A =
N
number of outcomes in event A
2
1
=
= =
N
total number of outcomes in the sample space 4
2
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3.2 Probability
3.2.1Classical Probability
Counting the Possible Outcomes
• Use the Combinations formula to determine the number of combinations of n items
taken k at a time;
• The number of combinations of k objects chosen from n is the number of possible
selections that can be made
Cnk 
• where
n!
k! (n  k)!
• n! = n(n-1)(n-2)…(1)
• 0! = 1 by definition
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Example
(continued)
Suppose that two letters are to be selected from A, B, C, D. How many
combinations are possible (i.e., order is not important)?
• Solution
The number of combinations is
C 24 
4!
6
2! (4  2)!
The combinations are
AB (same as BA)
BC (same as CB)
AC (same as CA)
BD (same as DB)
AD (same as DA)
CD (same as DC)
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3.2 Probability
Law of Large Numbers
As the number of times that an experiment is repeated increases, the relative frequency (proportion) of a
particular outcome tends to approach the probability of the outcome.
• For quantitative data, as the number of times that an experiment is repeated increases, the mean of the
outcomes tends to approach the population mean.
• For categorical (qualitative) data, as the number of times that an experiment is repeated increases, the
proportion of times a particular outcome occurs tends to approach the population proportion.
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3.2 Probability
3.2.2 Relative frequency probability
If we can’t use the classical method for assigning probabilities, then the Law of Large Numbers gives us a hint about
how we can estimate the probability of an event.
Relative frequency can be used to estimate the probability of the event. The probability of event A is approximately
equal to the relative frequency of event A.
P(A) =
n
frequency of event A
=
n
total number of events
3.2.3 Subjective probability refers to the assignment of a probability value to an outcome based on personal
judgment.
There are cases where the outcomes are not equally likely (so the classical method does not apply) and there has been
no previous research (so the relative frequency approach does not apply). The subjective method should be used when
the event is not repeatable.
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3.2 Probability
3.2.2 Relative frequency probability
Example: A recent study found that 35% of all online teen girls are bloggers, compared to 20% of online teen boys.
Suppose that the 35% came from a random sample of 100 teen girls who use the Internet, 35 of whom are bloggers. If
we choose one teen girl at random, find the probability that she is a blogger.
Solution: Define the event.
Event B: The online girl is a blogger.
We use the relative frequency method to find the probability of event B:
P(B) =
=
event
=
total number of events
= 35%
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3.3 Unions & Intersections
Compound events: Composition of two or more other events.
Can be formed in two different ways.
1. Union
•
Outcomes in either events A or B or both
•
‘OR’ statement
•
Denoted by  symbol (i.e., A  B)
2. Intersection
•
Outcomes in both events A and B
•
‘AND’ statement
•
Denoted by  symbol (i.e., A  B)
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3.3 Unions & Intersections
• Union of Events – If A and B are two events in a sample space S, then the union, A U B, is the set of
all outcomes in S that belong to either A or B.
• Venn diagram
S
A
B
The entire shaded area
represents
AUB
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3.3 Unions & Intersections
Examples
• S = {1,2,3,4,5,6}
• E = {1,3,5}
• F = {1,2,3}
 E U F = {1,2,3,5}
• A = {H}
• B = {T}
 A U B = {H, T}
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3.3 Unions & Intersections
• Intersection of Events – If A and B are two events in a sample space S, then the intersection, A ∩ B, is
the set of all outcomes in S that belong to both A and B.
• Venn diagram
S
A
AB
B
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3.3 Unions & Intersections
• Two events are said to be mutually exclusive if the events have no outcomes in common, events that
do not occur simultaneously. Two events are mutually exclusive if, when one event occurs, the other
cannot occur.
• Two events are mutually exclusive if, when one event occurs, the other cannot occur.
• Mutually Exclusive Events are also said to be Disjoint Events (i.e., the set A ∩ B is empty)
Sample
Space S
Event A
Event B
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3.3 Unions & Intersections
Example
S = [1, 2, 3, 4, 5, 6]
A = [2, 4, 6]
B = [4, 5, 6]
• Mutually exclusive:
• A and B are not mutually exclusive
• The outcomes 4 and 6 are common to both
• Collectively exhaustive:
• A and B are not collectively exhaustive
• A U B does not contain 1 or 3
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3.3 Unions & Intersections
Example:
1.
P(A) = 6/10
2.
P(D) = 5/10
3.
P(C  B) = 1/10
4.
P(A  D) = 9/10
5.
P(B  D) = 3/10
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
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Event Probability Using
Two–Way Table (2x2 Table)
Event
Event
B1
B2
Total
A1
P(A 1  B1) P(A 1  B 2) P(A 1)
A2
P(A 2  B1) P(A 2  B 2) P(A 2)
Total
Joint Probability
P(B 1)
P(B 2)
1
Marginal (Simple) Probability
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3.4 Complementary Events
Complement of Event A
• The event that A does not occur
• All events not in A
• Denote complement of A by AC or 𝑨
The sum of the probabilities of complementary events equals 1:
A
P(A) + P(AC) = 1
AC
S
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3.4 Complementary Events
Example
S = [1, 2, 3, 4, 5, 6]
A = [2, 4, 6]
B = [4, 5, 6]
Complements:
A = [1, 3, 5]
B = [1, 2, 3]
Intersections:
A ∩ B = [4, 6]
Unions:
A ∩ B = [5]
A ∪ B = [2, 4, 5, 6]
A ∪ A = [1, 2, 3, 4, 5, 6] = S
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3.5 The Additive Rule
1.
Used to get compound probabilities for union of events
2.
P(A OR B) = P(A  B)
= P(A) + P(B) – P(A  B)
3.
If events A and B are mutually exclusive, P(A  B) = 0. The additive rule (addition law) for
mutually exclusive events is:
P(A OR B) = P(A  B) = P(A) + P(B)
P(A or B or C) = P(A) + P(B) + P(C)
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3.5 The Additive Rule
Example: Using the additive rule, the probabilities are:
1. P(A  D) = P(A) + P(D) – P(A  D) =
=
6
5
2
9
+
–
=
10 10 10 10
2. P(B  C) = P(B) + P(C) – P(B  C)
=
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
4
5
1
8
+
–
=
10 10 10 10
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3.6 Conditional Probability
• Partial knowledge about the outcome of an experiment
• Probability that the share price of IBM increased today?
• Dow Jones Industrial Average rose 20 points
• Conditional probability that the price of IBM share increased, given that the DJIA rose 20 points?
For two related events A and B, the probability of B given A is called a conditional probability and denoted P(B
u A).
Conditional probabilities can often be interpreted as percentages of some subset of a population. For example,
the conditional probability that a customer responded, given that the customer has a credit card on file, may be
interpreted as the percentage of customers with credit cards who responded.
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3.6 Conditional Probability
The probability of an event given that another event
has occurred is called a conditional probability.
The conditional probability of A given B is denoted
by P(A|B).
A conditional probability is computed as follows :
P(A | B) = P(A and B) = P(A  B)
P(B)
P(B)
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3.6 Conditional Probability
Example: Using the table then the formula, what’s the probability?
P C B ) 
P C  B ) 110 1


4
P B )
4
10
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
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3.6 Conditional Probability
Event M = Marrley Oil Profitable
Event C = Cullins Mining Profitable
P(C | M ) = Cullins Mining Profitable
given Marrley Oil Profitable
We know: P(M C) = 0.36, P(M) = 0.70
Thus:
𝑃(𝐶|𝑀) =
𝑃(𝐶 ∩ 𝑀) 0.36
=
= 0.5143
𝑃(𝑀)
0.70
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3.7 Multiplication Rule
1.
Used to get compound probabilities for the intersection of events
2.
The rule is written as: P(A and B) = P(A  B) = P(A)  P(B|A) = P(B)  P(A|B)
3.
For Independent Events:
P(A and B) = P(A  B) = P(A)  P(B)
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3.7 Multiplication Rule
Example: Using the multiplicative rule, the probabilities are:
P C  B )  P C ) P B C ) 
P B  D )  P B ) P D B ) 
5 1 1
 
10 5 10
4 3 6
 
10 5 25
P A  B )  P A ) P B A )  0
Event
A
Event
C
D
4
2
Total
6
B
1
3
4
Total
5
5
10
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Independent Events
Since having a credit card increased the probability of a customer responding from 0.1667 to
0.2818, we can therefore say that the probability of responding depends in part on whether the
customer has a credit card. In other words, events R and C are dependent events.
On the other hand, if the probability of responding had been unaffected by whether the customer
had a credit card, then we would have said that R and C were independent events. That is, R and C
would have been independent events had P(R | C ) equaled P(R ). In general, if the occurrence of an
event does not affect the probability of a second event, then the two events are independent.
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Independent Events
If the probability of event A is not changed by the existence of
event B, we would say that events A and B are independent.
Two events A and B are independent if:
P(A|B) = P(A)
P(A  B) = P(A)  P(B)
or
P(B|A) = P(B)
Tests for independence
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Example
Keep Kool Inc. manufactures window air conditioners in both a deluxe model and a standard model.
An auditor collected 200 invoices for a month, some of which were sent to wholesalers and the
remainder to retailers. Of the 140 retail invoices, 28 are for the standard model. Of the wholesale
invoices, 24 are for the standard model. Find the following probabilities:
a.
The invoice selected is for the deluxe model.
b.
The invoice selected is a wholesale invoice for the deluxe model.
c.
The invoice selected is either a wholesale invoice or an invoice for the standard model.
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Solution
Display the data in a cross-classification table:
MODEL
Wholesale (W)
Retail (R)
Total
Deluxe (D)
36
112
148
Standard (S)
24
28
52
Total
60
140
200
The sample space consists of 200 invoices.
The events of interest are as follows:
W: Wholesale invoice is selected
R: Retail invoice is selected
D: Invoice for the deluxe model is selected
S: Invoice for the standard model is selected
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Solution
a. The invoice selected is for the deluxe model.
• Each invoice has the same chance of being selected, therefore the probability that any particular
invoice will be selected is 1/200.
• There are 148 invoices for the deluxe model
• 𝑃 𝐷 =
event D contains 148 simple events.
= 0.74
• The events of interest are as follows:
W: Wholesale invoice is selected
R: Retail invoice is selected
D: Invoice for the deluxe model is selected
S: Invoice for the standard model is selected
MODEL
Wholesale
(W)
Retail
(R)
Total
Deluxe (D)
36
112
148
Standard (S)
24
28
52
Total
60
140
200
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Solution
b.. The invoice selected is a wholesale invoice for the deluxe model
• There are 36 wholesale invoices for the deluxe model
• 𝑃 𝐷∩𝑊 =
the event 𝐷 ∩ 𝑊 contains 36 simple events.
= 0.18
• The events of interest are as follows:
W: Wholesale invoice is selected
R: Retail invoice is selected
D: Invoice for the deluxe model is selected
S: Invoice for the standard model is selected
MODEL
Wholesale
(W)
Retail
(R)
Total
Deluxe (D)
36
112
148
Standard (S)
24
28
52
Total
60
140
200
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Solution
c. The invoice selected is either a wholesale invoice or an invoice for the standard model
• The number of invoices that are either wholesale invoices or invoices for the standard model is 36
+ 24 + 28 = 88
• The event 𝑊 ∪ 𝑆 contains 88 simple events.
= 0.44
• 𝑃 𝑊∪𝑆 =
• The events of interest are as follows:
W: Wholesale invoice is selected
R: Retail invoice is selected
D: Invoice for the deluxe model is selected
S: Invoice for the standard model is selected
MODEL
Wholesale
(W)
Retail
(R)
Total
Deluxe (D)
36
112
148
Standard (S)
24
28
52
Total
60
140
200
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Example
• Suppose that the invoice selected by the auditor is a
wholesale invoice (W).
• We may determine the probability that this invoice is for
the deluxe model (D), making use of the knowledge that it
is a wholesale invoice. In other words, we are seeking the
MODEL
Wholesale
(W)
Retail
(R)
Total
Deluxe (D)
36
112
148
Standard (S)
24
28
52
Total
60
140
200
conditional probability that D will occur, given that W
occurred (P(D|W)).
• Wholesale invoice was selected restricts our inquiry to the
first column; the size of the sample space is now 60 simple
events.
• P(D|W) =
• P(D|W) =
=
.
.
=
= 0.60
( ∩ )
( )
= 0.60
Joint probabilities
MODEL
Wholesale
(W)
Retail
(R)
Total
Deluxe (D)
0.18
0.56
0.74
Standard (S)
0.12
0.14
0.26
Total
0.30
0.70
1.00
Marginal
probabilities
Marginal probabilities
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Dependent events
Example
P(D) = 0.74
P(D|W) = 0.60
P(D) ≠ P(D|W)
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Example
Event M = Marrley Oil Profitable
Event C = Cullins Mining Profitable
M C = Marrley Oil Profitable
or Cullins Mining Profitable
We know: P(M) = 0.70, P(C) = 0.48, P(M C) = 0.36
Thus: P(M  C) = P(M) + P(C)  P(M  C)
= 0.70 + 0.48  0.36
= 0.82
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Example
Event M = Marrley Oil Profitable
Event C = Cullins Mining Profitable
M C = Marrley Oil Profitable
and Cullins Mining Profitable
We know: P(M) = 0.70, P(C|M) = 0.5143
Thus: P(M  C) = P(M)P(M|C)
= (0.70)(0.5143)
= 0.36
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Multiplication Rule
for Independent Events
Event M = Marrley Oil Profitable
Event C = Cullins Mining Profitable
Are events M and C independent?
DoesP(M  C) = P(M)P(C) ?
We know: P(M  C) = 0.36, P(M) = 0.70, P(C) = 0.48
But: P(M)P(C) = (0.70)(0.48) = 0.34, not 0.36
Hence: M and C are not independent.
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prof. dr. Laura Asandului
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Statistics_Chapter 3_Lecture 7
4/11/2023
Type of
Compound Event
Union
(A or B)
Mutually Exclusive
P(A or B) =
P(A  B) = P(A) +
P(B)
Intersection
Complementary
(A and B)
(not A)
Not Mutually Exclusive Independent
P(A or B) =
P(A and B) =
P(A  B) = P(A) + P(A  B) = P(A)
 P(B)
P(B) – P(A  B)
Dependent
P(A and B) = P(A
B)= P(A) 
P(B|A)
= P(B)  P(A|B)
P(A) + P(AC) =
1
Conditional
(A given B)
Independent
P(A | B) = P(A)
Dependent
P(A | B) = P(A and
B)
P(B)
= P(A  B)
P(B)
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Example
A CONTINGENCY TABLE is a table used to classify sample observations according to two or
more identifiable characteristics
E.g. A survey of 150 adults classified each as to gender and the number of movies
attended last month. Each respondent is classified according to two criteria—the
number of movies attended and gender.
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Statistics_Chapter 3_Lecture 7
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What is the probability of randomly selecting a male person
P(B1)?
1.
2.
3.
4.
5.
70/150
80/150
60/150
20/150
70/80
Answer Now
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What is the probability of randomly selecting a person who
would have seen 2 movies (A3) and was male (B1)?
1.
2.
3.
4.
5.
20/150
10/20
10/150
20/10
10/70
Answer Now
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prof. dr. Laura Asandului
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Statistics_Chapter 3_Lecture 7
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Example
1) What is the probability of randomly selecting a person who
would have seen 2 movies (A3) and was male (B1)?
 20  10  10
P  A3  B1 )  P A3 )PB1 | A 3 )  
 0.06(6)
  
 150  20  150
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Example
1) What is the probability of randomly selecting a person who
would not have seen any movie (A1) and was female (B2)?
 60  40  40
P  A1  B2 )  P  A1 )P B2 | A 1 )  
 0.26(6)
  
 150  60  150
2) What is the probability of randomly selecting a person who
would not have seen any movie (A1) or was female (B2)?
 60   80   40 
P  A1  B2 )  P  A1 )  PB2 )  P  A1  B2 )  


  0.66(6)
 150   150   150 
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prof. dr. Laura Asandului
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Statistics_Chapter 3_Lecture 7
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prof. dr. Laura Asandului
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