essential statistics chapter 111 general rules of probability

Essential Statistics Chapter 11 1

Chapter 11

General Rules of Probability


Probability Rules from Chapter 9


Venn Diagrams

Two disjoint events:

Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common:


If two events A and B do not influence each other, and if knowledge about one does not change the probability of the other, the events are said to be independent of each other.

If two events are independent, the probability that they both happen is found by multiplying their individual probabilities:

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

Multiplication Rulefor Independent Events


Multiplication Rule for Independent Events

Example Suppose that about 20% of incoming male freshmen smoke.

Suppose these freshmen are randomly assigned in pairs to dorm rooms (assignments are independent).

The probability of a match (both smokers or both non-smokers):– both are smokers: 0.04 = (0.20)(0.20)– neither is a smoker: 0.64 = (0.80)(0.80)– only one is a smoker: ?

} 68%

32% (100% 68%) What if pairs are self-selected?


Addition Rule: for Disjoint Events

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


General Addition Rule

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


Case Study

Student DemographicsAt a certain university, 80% of the students were in-state students (event A), 30% of the students were part-time students (event B), and 20% of the students were both in-state and part-time students (event {A and B}). So we have that P(A) = 0.80, P(B) = 0.30, and P(A and B) = 0.20.

What is the probability that a student is either an in-state student or a part-time student?


Other Students

P(A or B) = P(A) + P(B) P(A and B) = 0.80 + 0.30 0.20 = 0.90

All Students

Part-time (B)

0.30{A and B}

0.20

Case Study

In-state (A)

0.80


Other Students

All Students

Part-time (B)

0.30

Case Study

{A and B}

0.20In-state (A)

0.80In-state, but not

part-time (A but not B):0.80 0.20 = 0.60


The probability of one event occurring, given that another event has occurred is called a conditional probability.

The conditional probability of B given A is denoted by P(B|A)– the proportion of all occurrences of A for

which B also occurs

Conditional Probability


Conditional Probability

P(A)

B) andP(A A)|P(B

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


Case Study

Student DemographicsIn-state (event A): P(A) = 0.80Part-time (event B): P(B) = 0.30Both in-state and part-time: P(A and B) = 0.20.

Given that a student is in-state (A), what is the probability that the student is part-time (B)?

0.25 0.80

0.20

P(A)

B) andP(A A)|P(B


General Multiplication Rule

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

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

For ANY two events, the probability that they both happen is found by multiplying the probability of one of the events by the conditional probability of the remaining event given that the other occurs:


Case Study

Student Demographics

At a certain university, 20% of freshmen smoke, and 25% of all students are freshmen.Let A be the event that a student is a freshman, and let B be the event that a student smokes.So we have that P(A) = 0.25, and P(B|A) = 0.20.

What is the probability that a student smokes and is a freshman?


Case Study

Student Demographics

P(A) = 0.25 , P(B|A) = 0.20

5% of all students are freshmen smokers.

P(A and B) = P(A) P(B|A)= 0.25 0.20= 0.05


Independent Events

Two events A and B that both have

positive probability are independent if

P(B|A) = P(B)

– General Multiplication Rule:P(A and B) = P(A) P(B|A)

– Multiplication Rule for independent events:P(A and B) = P(A) P(B)


Tree Diagrams Useful for solving probability problems that

involve several stages Often combine several of the basic probability

rules to solve a more complex problem– probability of reaching the end of any complete

“branch” is the product of the probabilities on the segments of the branch (multiplication rule)

– probability of an event is found by adding the probabilities of all branches that are part of the event (addition rule)


Case StudyBinge Drinking and Accidents

At a certain college, 30% of the students engage in binge drinking. Among college-aged binge drinkers, 18% have been involved in an alcohol-related automobile accident, while only 9% of non-binge drinkers of the same age have been involved in such accidents.Let event A = {accident related to alcohol}.Let event B = {binge drinker}.So we have P(A|B)=0.18, P(A|’not B’)=0.09, & P(B)=0.30 .

What is the probability that a randomly selected student has been involved in an alcohol-related automobile accident?


Case StudyBinge Drinking and Accidents

P(Accident) = P(A) = 0.054 + 0.063 = 0.117

0.054

0.637

0.063

0.246

P(A and B)

= P(B)P(A|B)

= (0.30)(0.18)

Accident

No accident

Accident

No accident

0.09

0.82

0.18

0.91

Bingedrinker

Non-binge drinker

0.30

0.70

essential statistics chapter 111 general rules of probability

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