# Chap 5 probability, Lecture notes for Probability and Statistics. Indian Institute of Management (IIM)

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Chapter 5

Copyright © 2015 McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

A Survey of Probability Concepts

Chapter 5

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Learning Objectives  LO5-1 Define the terms probability, experiment, event, and

outcome.  LO5-2 Assign probabilities using a classical, empirical, or subjective approach.  LO5-3 Calculate probabilities using the rules of addition.  LO5-4 Calculate probabilities using the rules of multiplication.  LO5-5 Compute probabilities using a contingency table.  LO5-6 Calculate probabilities using Bayes’ theorem.  LO5-7 Determine the number of outcomes using principles of counting.

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Probability

PROBABILITY A value between zero and one, inclusive, describing the relative possibility (chance or likelihood) an event will occur.

LO5-1 Define the terms probability, experiment, event, and outcome.

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Probability

LO5-1 Define the terms probability, experiment, event, and outcome.

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 Frequently expressed as a decimal, such as .70, .27, or .50.  It may be given as a fraction such as 7/10, 27/100, or 1/2.   The closer a probability is to 0, the more improbable it is the

event will happen. The closer the probability is to 1, the more sure we are it will happen

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Experiment, Outcome, and Event  An experiment is a process that leads to the occurrence of one and only

one of several possible results.  An outcome is the particular result of an experiment.  An event is the collection/set of one or more outcomes of an experiment.

LO5-1

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Experiment, Outcome, and Event Example

 In the die-rolling experiment, there are six possible outcomes, but there are many possible events.

 Counting the number of members of the board of directors for Fortune 500 companies over 60 years of age, the number of possible outcomes can be anywhere from 0 to the total number of members.

LO5-1

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Experiment, Outcome, and Event Example LO5-1

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Ways to Assign Probabilities There are three ways to assign probabilities:

1. CLASSICAL PROBABILITY Based on the assumption that the outcomes of an experiment are equally likely.

2. EMPIRICAL PROBABILITY The probability of an event happening is the fraction of the time similar events happened in the past.

3. SUBJECTIVE PROBABILITY The likelihood (probability) of a particular event happening that is assigned by an individual based on whatever information is available.

LO5-2 Assign probabilities using a classical, empirical, or subjective approach.

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Classical Probability

Consider an experiment of rolling a six-sided die. What is the probability of the event: “an even number of spots appear face up”? The possible outcomes are:

There are three “favorable” outcomes (a two, a four, and a six) in the collection of six equally likely possible outcomes.

LO5-2

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Empirical Probability

The formula to determine an empirical probability is:

Empirical approach to probability is based on what is called the Law of Large Numbers.

LO5-2

The key to establishing probabilities empirically: a larger number of observations provides a more accurate estimate of the probability.

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Empirical Probability - Example On February 1, 2003, the Space Shuttle Columbia exploded. This was the second disaster in 123 space missions for NASA. On the basis of this information, what is the probability that a future mission is successfully completed?

98.0 123 121 

flights of number Total flights l successfuof Numberflight l successfua ofy Probabilit

LO5-2

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Subjective Probability

 If there is little or no data or information to calculate a probability, it may be arrived at subjectively.

 Example : Illustrations of subjective probability are: 1. Estimating the likelihood a person will be married

before the age of 30. 2. Estimating the likelihood the U.S. budget deficit

will be reduced by half in the next 10 years.

LO5-2

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Summarizing Probability

LO5-2

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Rules of Addition for Computing Probabilities  Special Rule of Addition : If two events A and B are mutually exclusive, the probability of one or

the other event occurring equals the sum of their probabilities.

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

 For three mutually exclusive events designated A, B, and C, the rule is written:

LO5-3 Calculate probabilities using rules of addition.

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

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Special Rule of Addition  Events are mutually exclusive if the occurrence of any one event means that none of the others can occur at the same time.

 An example of mutually exclusive is the die-tossing experiment of the events “a number 4 or larger” and “a number 2 or smaller.” If the outcome is in the first group {4, 5, and 6}, then it cannot also be in the second group {1 and 2}.

LO5-3 Calculate probabilities using rules of addition.

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Special Rule of Addition - Example Mutually Exclusive Events

A machine fills plastic bags with a mixture of beans, broccoli, and other vegetables. Most of the bags contain the correct weight, but because of the variation in the size of the beans and other vegetables, a package might be underweight or overweight. If a check of 4,000 packages filled in the past month revealed as follows:

LO5-3

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Special Rule of Addition - Example Mutually Exclusive Events

What is the probability that a particular package will be either underweight or overweight?

P(A or C) = P(A) + P(C) = 0.025 + 0.075 = 0.10

The complement rule can also be used: Note that P(A or C) = P(~B), so P(~B) = 1 – P(B) = 1 - 0.900 = 0.10

LO5-3

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The Complement Rule  The complement rule is used to determine the probability of an event occurring by

subtracting the probability of the event not occurring from 1.

P(A) + P(~A) = 1

or P(A) = 1 - P(~A).

LO5-1

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The Complement Rule - Example An experiment has two mutually exclusive outcomes. Based on the rules of probability, the sum of the probabilities must be one.

If the probability of the first outcome is 0.61, then logically, AND by the complement rule, the probability of the other outcome is (1.0 - 0.61) = 0.39.

P(B) = 1 - P(~B) = 1 – 0.61 = 0.39

LO5-1

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Joint Probability

LO5-3

 Sometimes, the outcomes of an experiment may not be mutually exclusive.

 For example, the survey of Florida Tourist Commission reveal that from 200 selecting tourist samples, there were 120 tourists went to Disney World and 100 tourists went to Busch Gardens. What is the probability that a person selected visited either Disney World or Busch Gardens?

To solve an above problem, we use the rule to find a joint probability.

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Rules of Addition for Computing Probabilities

LO5-3

The General Rule of Addition - If A and B are two events that are not mutually exclusive, then P(A or B) is given by the following formula:

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

P( A and B) is called a joint probability.

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The General Rule of Addition The diagram shows the results of a survey of 200 tourists who visited Florida during the year. The results revealed that 120 went to Disney World, 100 went to Busch Gardens, and 60 visited both.

What is the probability a selected person visited either Disney World or Busch Gardens?

To answer the question, 1. Add the probability that a tourist visited Disney

World and the probability he or she visited Busch Gardens

2. Subtract the probability of visiting both

LO5-3

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The General Rule of Addition Example Continued:

P(Disney or Busch) = P(Disney) + P(Busch) - P(both Disney and Busch) = 120/200 + 100/200 – 60/200 = 0.60 + 0.50 – 0.30 = 0.80

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LO5-3

The probability a selected person visited either Disney World or Busch Gardens is 0.80

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General Rule of Addition – Example What is the probability that a card chosen at random from a standard deck of cards will be either a king or a heart?

Solution 1.Add the probability of a king (there are 4 in a deck of 52 cards) to the probability of a heart (there are 13 in a deck of 52 cards) 2.17 out of 52 cards meet the requirement, we have counted the king of hearts twice. Then, we need to subtract 1 card from the 17 so the king of hearts is counted only once.

LO5-3

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General Rule of Addition – Example

P(A or B) = P(A) + P(B) - P(A and B) = 4/52 + 13/52 - 1/52 = 16/52, or 0.3077

Thus, there are 16 cards that are either hearts or kings. So the probability is 16/52 = 0.3077.

LO5-3