Introduction to Probability and Random Variables and Discrete probability Distributions

6.2 Joint, Marginal, and Conditional Probability The probability of an event : The probability assigned to the simple events in A.. Mutual fund outperforms the market B1 Mutual fund do

6.1 Assigning probabilities to Events

2

• Flip a coin Heads and Tails

• The marks of a statistics test Numbers between 0 and 10

• The time to assemble Non-negative numbers

a computer

•Repeated random experiments may result in different outcomes One can only refer to the experiment

outcome in terms of the probability of each outcome

to occur

• To determine the probabilities we need to define

the sample space,

– which is exhaustive (list of all the possible

outcomes)

– in which the outcomes are mutually exclusive

Example 1: Build the sample space for the two

random experiments described below

4

The sample space: {B, R}

Case 2: Two balls are randomly selected from a

bag that contains blue and red balls An outcome is considered the colors of the two balls drawn

The sample space: {BB, BR, RB, RR}

Case 1: A ball is randomly selected from a bag

containing blue and red balls An outcome is

considered the ball color

Example 2: The random experiment: randomly

select two numbers from the set 1, 2, 3, 4, 5 (a number cannot be selected twice) An outcome

is defined by the sum of the two numbers

The sample space: {3, 4, 5, 6, 7, 8, 9}

1+3 1+5, 2+4 3+5

Example 3: A dice is rolled once Specify the

sample points that belong to each event

Event A = the number facing up is 6

Event B = The number facing up is odd

Event C = The number facing up is not

greater than 4

Solution: A = {6};B = {1, 3, 5}; C = {1, 2, 3)

Example 4: A dice is rolled once

Event A = the number facing up is 6

Event B = The number facing up is odd

Event C = The number facing up is less than 4 The outcome is 3 Which event takes place?

Solution: Event B and event C take place

because the outcome ‘3’ belongs to both

events

8

6.1 Assigning probabilities to

Events

Given a sample space S={O 1 ,O 2 ,…,O k},

The probability of a simple event P(O i): the

following characteristics must hold:

1

) i

P(O 2.

i each for

1

) i

P(O 0

1

) i

P(O 2.

i each for

1

) i

P(O 0

1.

6.2 Joint, Marginal, and

Conditional Probability

The probability of an event : The probability

assigned to the simple events in A.

Understanding relationships among events may reduce the computational effort in determining the probability of events

The intersection

and B” ( ), is the event that both A and B occur.

Two events are said to be mutually exclusive if the occurrence of one precludes the occurrence of the other one.

If A and B are mutually exclusive, by definition, the probability of their intersection is equal to zero.

Example: When rolling a dice once the event “The

number facing up is 6” and the event “The number facing up is odd” are mutually exclusive.

The Joint Probability

The Probabilities of Joint Events

The probability of the intersection of A and B is

called also the joint probability of A and B = P(A

and B)

14

Example 9: A potential investor examined the relationship between the performance of

mutual funds and the school the fund manager earned his/her MBA The following table

describes the joint probabilities

Mutual fund outperforms the market

(B1)

Mutual fund doesn’t outperform the market

(B2) Top 20 MBA program

A A and C B and C B

The probability of event C can be calculated as the sum of the two joint probabilities.

18

A C B

P(C) = P(A and C) + P(B and C))

Applying this concept to the table of joint

determining by adding joint probabilities across rows and columns

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

P(Ai) Top 20 MBA program (A1)

Not top 20 MBA program (A2)

+ = P(A 1) P(A 1 and B1) P(A 1 and B2)

P(A2 and B1) + P(A2 and B2) = P(A2)

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

P(Ai)

Not top 20 MBA program (A2) 06 54 60

Marginal Probability P(Bj)

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

P(Ai)

Marginal Probability P(Bj)

P(A1 and B 1)

+ P(A2 and B 1

= P(B 1)

P(A1 and B 2)

+ P(A2 and B 2

= P(B 2)

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

P(Ai)

Not top 20 MBA program (A2) 06 54 60 Marginal Probability P(Bj) 17 83

+ +

Conditional Probability

Frequently, information about the occurrence

of event A changes the probability of event B

S

B

The sample space is S.

Now, assume event ‘A’

takes place before event B.

Note that event B is contained in event A.

{The area of B} {The area of A}.

P(B) is not equal to P(B given A)

A

This is event B

Specifically, the conditional probability of event

B given that event A has occurred is calculated

as follows:

P(A and B)

P(A) P(B|A) =

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

a “Top 20 MBA Program graduate”, given that

it did not outperform the market

P(A 1 |B 2 ) = P(A 1 and B 2 ) = 29 = 3949

P(B2) .83

are called “dependent events”.

Two events A and B are said to be independent

if

P(A|B) = P(A) or

P(B|A) = P(B)

affected by the occurrence of the other event

28

Example 13 –Let us check A2 and B2.

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

P(Ai)

Not top 20 MBA program (A2) 06 54 60

Marginal Probability P(Bj) 17 83

•It is denoted “A or B” (A U B)

30

Example 5: A fair dice is rolled once Find the

probabilities of the following events:

Event A = the number facing up is 6

Event B = The number facing up is odd

Event C = The number facing up is less than 4

Solution:

P(A) = P{6} = 1/6;

P(B) = P(1 or 3 or 5) = 3/6

Example 6: A dice is rolled twice: Define the

Example 9: Determine the probability that a

randomly selected fund outperforms the

market or the manager graduated from a top

20 MBA Program

Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the

market (B2) Top 20 MBA program (A1) 11 29

Not top 20 MBA program (A2) 06 54

6.3 Probability Rules and Trees

Three rules assist us in determining the probability of complex events

•The complement rule

•The multiplication rule

•The addition rule

34

Multiplication Rule

•For any two events A and B

•When A and B are independent P(B|A)

•Example 14: What is the probability that two female students will be selected at random from a class of seven males and three female students?

•Solution

A – the first student selected is a female

B – the second student selected is a femaleP(A and B) = P(A)P(B|A)

= (3/10)(2/9) = 6/90 = 067

Example 15: What is the probability that a

female student will be selected at random in

each of two classes of seven males and three female students?

Solution

A – the student selected in one class is a female

B – the student selected in the other class is a female

P(A and B) = P(A)P(B)

= (3/10)(3/10) = 9/100 = 09

38

Addition Rule

For any two events A and B

P(A or B) = P(A) + P(B) - P(A and B)

P(A or B) = P(A) + P(B) - P(A and B)

When A and B are mutually exclusive,

40

P(A or B) = P(A) + P(B) – P(A and B)

P(A or B) = P(A) + P(B) – P(A and B)

A

P(A and B) = 0

B

Example 11: The circulation departments of two newspapers in a large city report that 22% of the city’s households subscribe to the Sun, 35% subscribe to the Post, and 6% subscribe to both.

What proportion of the city’s household subscribe to either

newspaper?

•Solution

A = the household subscribes to the Sun

B = the household subscribes to the Post

P(A or B) = P(A) + P(B) – P(A and B)

= 22+.35 -.06 = 51

Example 12: Using the addition rule determine the probability that a randomly selected fund

outperforms the market or the manager

graduated from a top 20 MBA Program in

Example 9

outperfor

ms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Margi nal Prob P(Ai)

Top 20 MBA program (A1)

.11 29 40

Not top 20 MBA program (A2) .06 .54 .60Marginal

Probability P(Bj) .17 .83 42

Example: A local photocopy shop has three

black-and white (BW) copy machines and two color copiers (CC) It is known that a BW is

down 10% of the time for repairs and a CC is

down 20% Assume machines break down

independently

What proportion of the time a customer cannot run

a color photocopy job?

If a customer wants both a color copy and a black and white copy now, but this time a CC can perform also a BW job, what is the

probability he can complete the two jobs?

Answer:

• Example 14 revisited (dependent events).

Find the probability of selecting two female

students (without replacement), if there are 3

female students in a class of 10

Joint probabilities

F F

M F

F M

•Example 15 – revisited (independent events)

Find the probability of selecting one female student in each of two classes if there are 3 female students in a class of 10

P(F) = 3/1

0

P( M) = 7/10

Second selection

Second selection

P(MM)=(7/10)(7/10)

= P(F) =

= P(F) =

= P(M) = = P(M) =

•Example 16 (conditional probabilities)

The pass rate of first-time takers for the bar exam at a certain jurisdiction is 72%

Of those who fail, 88% pass their second

50

Baye’s Law

Baye’s law is to find the conditional probability

of a possible cause for an event that is known to have occurred

P )

B

| A (

Example 17

Medical tests can produce false-positive or false-negative results.

A particular test is found to perform as follows:

1.Correctly diagnose “Positive” 94% of the time 6% of the time ill person is

diagnosed “negative”.

2.Correctly diagnose “Negative” 98% of the time 2% of the time healthy person is diagnosed “positive”.

3.It is known that 4% of men in the general population suffer from the illness.

What is the probability that a men is suffering from the illness, if the test result were positive?

•Solution: Define the following events

•D = Has the disease

•PT = Positive test results

•NT = Negative test results

•Build a probability tree

) = 96

P(D) =

04

| D ( P

6620

0568

0376

.





) = 96

P(D) =

04

) PT

| D (

0568

The prior probability of event ‘D’ was 4%, but because event ‘PT’ occurred the probability the cause had been ‘D’

increased to 66.2%.

7.1 Random Variables and Discrete

Probability Distribution

When the simple events in the sample space can

be assigned numerical values, a random

There are two types of random variables:

•A random variable is discrete if it can

assume only a countable number of

values

•A random variable is continuous if it can assume an uncountable number of values.

A table, formula, or graph that lists all

members of the sample space, together

with the probabilities associated with each

distribution.

Discrete Probability Distribution

1)

x(f

2

xallfor

1)

f(x0

1

i x all i

i i

x(f

2

xallfor

1)

f(x0

1

i

x all i

i i

To calculate the probability that the random

variable X assumes the value x, P(X = a),

events for which X is equal to ‘a’, or

tree diagram),

(½)(½)=1/4 (½)(½)=1/4 (½)(½)=1/4 (½)(½)=1/4

H

T

Example 1: Find the probability

distribution of the random variable

describing the number of heads that

turn-up when a coin is flipped twice.

•Solution

Example 2: A survey reveals the following

frequencies for the number of colored TV per household.

Example 3: The number of cars a dealer is

selling daily were recorded in the last 100 days This data was summarized as

selling more than 2 cars

a day.

Example 4: A mutual fund sales person

knows that there is 20% chance of closing

a sale on each call she makes.

What is the probability distribution of the number of sales if she plans to call three customers?

Developing a Probability Distribution

Solution

Use probability rules and probability tree

Define event A = {a sale is made in the iit phone

P(S C )=.8

P(S C )=.8

P(S C )=.8 P(S C )=.8

Sales Call 1 Sales Call 2 Sales Call 3

The Expected Value (Mean)

Given a discrete random variable X with

)X(

i

x all xi p(xi )

)X(EThe population mean is the weighted average of all of

Let X be a discrete random variable

The variance of X is defined by

2

2 E ( X ) ( x ) p ( x ) )

X (

2

) X ( V

The Population Variance

2

is deviation dard

tan s The





is deviation dard

tan s The







Example 5: Find the mean the variance

and the standard deviation for the

population of the number of colored TV per household in example 2

X ( E )

X ( V

2 2

2 2

2 x

2 i

2 2

X ( E )

X ( V

2 2

2 2

2 x

2 i

2 2

It is the weighted average of the squared

deviations from the mean

Laws of Expected Value…

E(c) = c

The expected value of a constant (c) is just the value of the constant E(X + c) = E(X) + c

E(cX) = cE(X)

Example: Monthly sales have a mean of $25,000 and a standard deviation of

$4,000 Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6,000 Find the mean monthly profit.

E(Sales) = 25,000

Profit = 30(Sales) – 6,000

E(Profit) =E[.30(Sales) – 6,000]

=E[.30(Sales)] – 6,000 [by rule #2]

=.30E(Sales) – 6,000 [by rule #3]

=.30(25,000) – 6,000 = 1,500

Laws of Variance…

V(c) = 0

V(X + c) = V(X)

V(cX) = c 2 V(X)

Example : Monthly sales have a mean of $25,000 and a standard deviation of

$4,000 Profits are calculated by multiplying sales by 30% and subtracting fixed costs of $6,000 Find the standard deviation of monthly profits.

V(Sales) = 4,000 2 = 16,000,000

Profit = 30(Sales) – 6,000

V(Profit) =V[.30(Sales) – 6,000]

=V[.30(Sales)] [by rule #2]

=(.30) 2 V(Sales) [by rule #3]

=(.30) 2 (16,000,000) = 1,440,000

Again, standard deviation is the square root of variance,

7.4 The Binomial Distribution

characteristics:

success or a failure.

trials.

•The binomial random variable

counts the number of successes in

n trials of the binomial experiment.

•The possible values of this count

are 0,1, 2, …,n, and therefore the

binomial variable is discrete.

Binomial Random Variable

The Binomial Probability Distribution

) x ( p )

x X

x ( p )

x X

P(FSS)=(1-p)p 2

P(FSF)=(1-p)P(1-p) P(FFS)=(1-p) 2 p

Since the outcome of each trial is independent of the previous outcomes, we can replace the conditional

probabilities with the unconditional probabilities.

Since the outcome of each trial is independent of the previous outcomes, we can replace the conditional

probabilities with the unconditional probabilities.

Example 10

Pat takes a course in statistics, and

intends to rely on luck to pass the next

quiz.

The quiz consists on 10 multiple choice

questions with 5 possible choices for each question, only one is the correct answer Pat will guess the answer to each question Find the following probabilities

•Pat gets no answer correct

•Pat gets two answer correct

•Pat fails the quiz

Checking the conditions

• Only one out of two outcomes can occur

(An answer can be either correct or incorrect)

• There is a fixed finite number of trials

(There are 10 questions in the test, n=10)

• Each answer is independent of the others.

• The probability p of a correct answer does not

change from question to question

(20% chance that an answer is correct)

Let X = the number of correct answers

.1074 (.80)

(.20) 0)!

(10 0!

(.20) 2)!

(10 2!