INTRODUCTION TO STATISTICS & PROBABILITY

Chapter 4: Probability: The Study of Randomness

(Part 4)

Dr. Nahid Sultana

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Chapter 4 Probability: The Study of Randomness

4.1 Randomness

4.2 Probability Models

4.3 Random Variables

4.4 Means and Variances of Random Variables

4.5 General Probability Rules*

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4.5 General Probability Rules 3

Rules of Probability

General Addition Rules

Conditional Probability

General Multiplication Rules

Bayes’s Rule

Independence

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Probability Rules

Return to the laws of probabilities:

Rule 1. The probability P(A) of any event A satisfies 0 ≤ P(A) ≤ 1.

Rule 2. If S is the sample space in a probability model, then P(S) = 1.

Rule 3. If A and B are disjoint, P(A or B) = P(A) + P(B).

Rule 4. For any event A, P(AC) = 1 – P(A).

Rule 5. If A and B are independent, P(A and B) = P(A)P(B).

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The General Addition Rule

Addition Rule for Unions of Two Events

The probability that A occurs, B occurs, or both occurs is:

P(A or B) = P(A) + P(B) – P(A and B)

Addition Rule for Disjoint Events: If A, B, and C are disjoint in the sense that no two have any in common, then:

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

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Conditional Probability

Rule 5. If A and B are independent, P(A and B) = P(A)P(B). The probability an event can change if we know that some other event has occurred. This idea is the key to many applications of probability. The conditional probability reflects how the probability of an event can change if we know that some other event has already occurred.

When P(A) > 0, the conditional probability of B given A is

)() and ()|(

APBAPABP =

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The General Multiplication Rule

The probability that any two events A and B both occur is

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

This is the general multiplication rule, where P(B | A) is the conditional probability that event B occurs given that event A has already occurred.

Conditional probability rule for the probability that both of two event occur.

Note: If Two events A and B are independent, P(A and B) = P(A)P(B). Therefore, If A and B that both have positive probability are independent if:

P(B|A) = P(B)

)() and ()|(

APBAPABP =

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The General Multiplication Rule

Example: Suppose 29% of Internet users download music files, and 67% of downloaders say they don’t care if the music is copyrighted. What is the percent of Internet users who download music and don’t care about copyright.

Example:

Employed Unemployed Total

Male 460 40 500

Female 140 260 400 Total 600 300 900

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Probability Trees

Conditional probabilities can get complex, and it is often good strategy to build a probability tree that represents all possible outcomes graphically and assigns conditional probabilities to subsets of events..

Consider flipping a coin twice. What is the probability of getting two heads?

Sample Space: HH HT TH TT So, P(two heads) = P(HH) = 1/4

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Probability Trees: Example

The Pew Internet and American Life Project finds that 93% of teenagers (ages 12 to 17) use the Internet, and that 55% of online teens have posted a profile on a social-networking site. What percent of teens are online and have posted a profile?

55.0)online|profile(93.0)online(

==

PP

51.15% of teens are online and have posted a profile.

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

5115.0)55.0)(93.0(

online)|(profile(online)profile) have and (online

==

⋅= PPP

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Probability Trees: Example

29% of adult Internet users are 18 to 29 years old (event A1), another 47% are 30 to 49 (event A2), and the remaining 24% are 50 and over (event A3).

47% of the 18 to 29 age group chat, as do 21% of the 30 to 49 age group and just 7% of those 50 and over. What is the probability that a randomly chosen user of the Internet participates in chat rooms?

P(C) = 0.1363 + 0.0987 + 0.0168 = 0.2518

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Bayes’s Rule

If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she does have breast cancer?

Cancer

No cancer

Mammography

Positive

Negative

Positive

Negative

Disease incidence

Diagnosis sensitivity

Diagnosis specificity

False negative

False positive

0.0004

0.9996

0.8

0.2

0.1

0.9

Incidence of breast cancer among women ages 20–30

Mammography performance

Example

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Bayes’s Rule (Cont…) Example

If a woman in her 20s gets screened for breast cancer and receives a positive test result, what is the probability that she does have breast cancer?

A1 is cancer, A2 is no cancer, C is a positive test result.

%3.0)9996.0(1.0)004.0(8.0

)0004.0(8.0)cancer no()cancer no|pos()cancer()cancer|pos(

)cancer()cancer|pos()pos|cancer(

≈+

=

+=

PPPPPPP

Cancer

No cancer

Mammography

Positive

Negative

Positive

Negative

Disease incidence

Diagnosis sensitivity

Diagnosis specificity

False negative

False positive

0.0004

0.9996

0.8

0.2

0.1

0.9 Incidence of breast cancer among women ages 20–30 Mammography

performance

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Bayes’s Rule (Cont…)

If a sample space is decomposed in k disjoint events, A1, A2, … , Ak none with a null probability and P(A1) + P(A2) + … + P(Ak) = 1

And if C is any other event such that P(C) is not 0 or 1, then:

However, it is often intuitively much easier to work out answers with a probability tree than with these lengthy formulas.