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As a measure of the chance, or probability, with which we can expect the event to occur, it is convenient to assign a number between 0 and 1. There are two important procedures by means of which we can estimate the probability of an event. If an event can occur in h different ways out of a total number of n possible ways, all of which are equally likely, then the probability of the event is h n. Since there are two equally likely ways in which the coin can come up-namely, heads and tails assuming it does not roll away or stand on its edge -and of these two ways a head can arise in only one way, we reason that the required probability is 1 2.
In arriving at this, we assume that the coin is fair, i. If after n repetitions of an experiment, where n is very large, an event is observed to occur in h of these, then the probability of the event is h n.
This is also called the empirical probability of the event. Both the classical and frequency approaches have serious drawbacks, the first because the words "equally likely" are vague and the second because the "large number" involved is vague. Because of these difficulties, mathematicians have been led to an axiomatic approach to probability. The Axioms of ProbabilitySuppose we have a sample space S. If S is discrete, all subsets correspond to events and conversely, but if S is nondiscrete, only special subsets called measurable correspond to events.
To each event A in the class C of events, we associate a real number P A. Then P is called a probability function, and P A the probability of the event A, if the following axioms are satisfied. Theorem For every event A,i. It follows that we can arbitrarily choose any nonnegative numbers for the probabilities of these simple events as long as 14 is satisfied. In particular, if we assume equal probabilities for all simple events, then 15 and if A is any event made up of h such simple events, we have 16 This is equivalent to the classical approach to probability given on page 5.
We could of course use other procedures for assigning probabilities, such as the frequency approach of page 5. Assigning probabilities provides a mathematical model, the success of which must be tested by experiment in much the same manner that theories in physics or other sciences must be tested by experiment.
Find the probability of a 2 or 5 turning up. If we assign equal probabilities to the sample points, i. Denote by P the probability of B given that A has occurred.
Since A is known to have occurred, it becomes the new sample space replacing the original S. In words, 18 says that the probability that both A and B occur is equal to the probability that A occurs times the probability that B occurs given that A has occurred. We call P the conditional probability of B given A, i. It is easy to show that conditional probability satisfies the axioms on page 5.
Since B is the union of the events 1, 2, or 3 turning up, we see by Theorem that assuming equal probabilities for the sample points. The result is easily generalized to n events. Conversely, if 21 holds, then A and B are independent.
Independence of more than three events is easily defined. Then if A is any event, we have the following important theorem: Theorem Bayes' Rule : 24 This enables us to find the probabilities of the various events A 1 , A 2 , ,A n that can cause A to occur. For this reason Bayes' theorem is often referred to as a theorem on the probability of causes. Combinatorial AnalysisIn many cases the number of sample points in a sample space is not very large, and so direct enumeration or counting of sample points needed to obtain probabilities is not difficult.
However, problems arise where direct counting becomes a practical impossibility. In such cases use is made of combinatorial analysis, which could also be called a sophisticated way of counting.
Fundamental Principle of Counting: Tree DiagramsIf one thing can be accomplished in n 1 different ways and after this a second thing can be accomplished in n 2 different ways,. A diagram, called a tree diagram because of its appearance Fig.
PermutationsSuppose that we are given n distinct objects and wish to arrange r of these objects in a line. We call n P r the number of permutations of n objects taken r at a time. CombinationsIn a permutation we are interested in the order of arrangement of the objects.
For example, abc is a different permutation from bca. In many problems, however, we are interested only in selecting or choosing objects without regard to order. Such selections are called combinations. For example, abc and bca are the same combination. The total number of combinations of r objects selected from n also called the combinations of n things taken r at a time is denoted by n C r or We have see Problem 1.
When n is large, a direct evaluation of n! The symbol in 33 means that the ratio of the left side to the right side approaches 1 as n.
Computing technology has largely eclipsed the value of Stirling's formula for numerical computations, but the approximation remains valuable for theoretical estimates see Appendix A. A card is drawn at random from an ordinary deck of 52 playing cards. Describe the sample space if consideration of suits a is not, b is, taken into account. Denoting hearts, spades, diamonds, and clubs, respectively, by 1, 2, 3, 4, for example, we can indicate a jack of spades by 11,2.
The sample space then consists of the 52 points shown in Fig. Find the probability that it is a an ace, b a jack of hearts, c a three of clubs or a six of diamonds, d a heart, e any suit except hearts, f a ten or a spade, g neither a four nor a club.
Let us use for brevity H, S, D, C to indicate heart, spade, diamond, club, respectively, and 1, 2 13 for ace, two, , king. Then 3 H means three of hearts, while 3 H means three or heart. Let us use the sample space of Problem 1.
It could also have been arrived at by simply reasoning that there are 13 numbers and so each has probability 1 13 of being drawn. Method 2by part c. Method 3Since events R and W are mutually exclusive, it follows from 4 , page 5, that a The first card can be drawn in any one of 52 ways, and since there is replacement, the second card can also be drawn in any one of 52 ways.
Then both cards can be drawn in 52 52 ways, all equally likely. In such a case there are 4 ways of choosing an ace on the first draw and 4 ways of choosing an ace on the second draw so that the number of ways of choosing aces on the first and second draws is 4 4.
Then the required probability is b The first card can be drawn in any one of 52 ways, and since there is no replacement, the second card can be drawn in any one of 51 ways.
Then both cards can be drawn in 52 51 ways, all equally likely. In such a case there are 4 ways of choosing an ace on the first draw and 3 ways of choosing an ace on the second draw so that the number of ways of choosing aces on the first and second draws is 4 3.
Then the required probability is 1. Three balls are drawn successively from the box of Problem 1. Find the probability that they are drawn in the order red, white, and blue if each ball is a replaced, b not replaced. Find the probability of a 4 turning up at least once in two tosses of a fair die. Method 1Events A 1 and A 2 are not mutually exclusive, but they are independent.
One bag contains 4 white balls and 2 black balls; another contains 3 white balls and 5 black balls. If one ball is drawn from each bag, find the probability that a both are white, b both are black, c one is white and one is black. Prove Theorem , page 7. Extensions to larger values of n are easily made.
Box I contains 3 red and 2 blue marbles while Box II contains 2 red and 8 blue marbles. A fair coin is tossed. If the coin turns up heads, a marble is chosen from Box I; if it turns up tails, a marble is chosen from Box II.
Find the probability that a red marble is chosen. Therefore, the probability of choosing a red marble isBayes' theorem 1. Prove Bayes' theorem Theorem , page 8. Suppose in Problem 1. What is the probability that Box I was chosen i. Let us use the same terminology as in Problem 1. We seek the probability that Box I was chosen given that a red marble is known to have been chosen. A committee of 3 members is to be formed consisting of one representative each from labor, management, and the public.
If there are 3 possible representatives from labor, 2 from management, and 4 from the public, determine how many different committees can be formed using a the fundamental principle of counting and b a tree diagram. With each of these ways we can choose a public representative in 4 different ways. Then the tree diagram of Fig. From this tree diagram we can list all these different committees, e.
This is also called the number of permutations of n different objects taken n at a time and is denoted by n P n. In how many ways can 10 people be seated on a bench if only 4 seats are available?
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Skip to search form Skip to main content You are currently offline. Some features of the site may not work correctly. Spiegel Published Mathematics. Part I: Probability. Chapter 1: Basic Probability.
The tutorial approach i. The prerequisite is one year of calculus. Schaum's outline of theory and problems of probability and statistics Schaum's outline series Murray R Spiegel. Therein, pp. Create a free account to download.
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A short summary of this paper. The prerequisite is one year of calculus. Probability, Statistics and Random Processes. Get Book. It should also be of considerable value to those taking courses in mathematics, management, commerce, or any of the numerous other fields in which Probabili…, This book is designed for use as a supplement to all current standard texts or as a textbook for a formal course on Statistics.
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The first edition of Schaum's Probability and Statistics by Murray R. Spiegel century that a rigorous mathematical theory based on axioms, definitions, and In many problems, however, we are interested only in selecting or choosing objects.Reply
Schaum's outline of theory and problems of statistics Murray R. elementary probability theory and applications, which paves the way for a study of sampling.Reply