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CHAPTER 4Probability - Sections: 4.1 to 4.4

-- Introduction

Probability theory is the study of randomness. The outcomes of a random process depend solelyon chance. For example, choosing a person at random from a large population and recording theperson's height is a random process. One possible outcome of this random process is whether theperson is more than 72 inches tall. Probability theory makes it possible to answer the question:What is the likelihood that the next person we are going to choose, is going to be more than 72inches tall?

-- Definitions

Empirical concept of probability

The empirical concept of probability is that ofrelative frequency, the ratio of the total numberof occurrences of an event to the total number of times the experiment is repeated.

When the number of trials is large, the relative frequency provides a satisfactory measure of theprobability associated with the specific event. This is one of the so-called laws of large numbersof probability theory.

Sample space

In probability and statistics the term experiment is used in a very wide sense and refers to anyprocedure that yields a collection of outcomes. A random experiment is one whose outcomesdepend only on chance and it can be repeated under identical conditions.

The set whose elements are all the possible outcomes of an experiment is called the samplespace of the experiment. The concept of the sample space plays a fundamental role indetermining probabilities for individual outcomes of a random process. Your understanding ofthe sample space should be that ofa complete and as detailed as possible listing of all thebasic (elementary) outcomes of a random experiment.

Events

In most cases we are not interested in just one single outcome of a random phenomenon, insteadwe are interested in one or more of them.

An event is any sub-collection of elementary outcomes from the sample space.

We say that an event has occurred, if the outcome of the experiment is one of the outcomes thatare included in the event. For example, when you roll a die all the possible outcomes (the sample

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space) are the numbers 1, 2, 3, 4, 5, or 6. An event, in this case, is any combination of thesenumbers, for example that the number that shows up on the die is a multiple of 3. For thisexample, the event occurs if a 3 or 6 shows up.

Probabilities

A probability is a numerical value between 0 and 1 (inclusive) assigned to elementary outcomesof the sample space. The probability of an event is found by adding up the probabilities assignedto the simple outcomes that make up the event. For example, the probability of the event "thenumber is a multiple of 3" is 1/6 + 1/6 or 1/3, which is the sum of the probabilities of rolling a 3,or a 6.

The following two general rules apply to the numbers that we call 'probabilities'

1. The value of a 'probability' can never be negative or greater than 1.2. The sum of the probabilities of all the outcomes that are included in the sample

space should always equal 1.

Equally likely assumption

We say that the outcomes of an experiment are equally likely to happen, if they all have the sameprobability of occurring. For example, the outcomes of tossing a die.

-- LAWS OF PROBABILITY

Disjoint (or mutually exclusive) events

Two or more events are said to be disjoint, or mutually exclusive, if they do not have anyoutcomes in common. Consequently, disjoint events cannot occur simultaneously.

-- the addition l aw of probabil ity for disjoin t events

For any two disjoint events A, and B:

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

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-- law of the complement

Assume the event A. The event whose outcomes are all the outcomes in the sample space exceptfor the outcomes that make up A, is called the complement of A and is denoted by:

not A orA

c

.

The law of the complement states that for any event A:

P(not A) = 1 - P(A)

Independent events

We say that two events are dependent if the occurrence of one event causes a change in theprobability of occurrence (or nonoccurrence) of another event.

In some situations the fact that an event A has already occurred, does not cause a change in theprobability of occurrence associated with anotherevent B. In this case we say that the two events(A and B) are independent.

-- the multi pli cation law for independent events

The multiplication law of probability for independent events states that:

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

i.e. the probability of the simultaneous occurrence of two independent events A and B equals the

product of the probability of A and the probability of B

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-- RANDOM VARIABLES

A random variable X, Y, or Z, ... , assigns of a numerical value (for example the height of aperson) to each outcome of a random process.

Example 1

N = The number of people in l ine at a tell er machine.Possible Values of N = { 0, 1, 2, 3, 4, 5, . . . }

Here we say that N is a discrete random variable. The values of N are finite, or in some casescountable infinite.

Example 2

T = The time one has to wait in li ne at a tell er machine.Possible Values of T = { any positive interval of time }

Here we say that T is a continuous random variable. Every number greater than zero is apossible value, with no exception. There are no gaps in between successive values that therandom variable T can take.

-- Probability Distributions

Fordiscrete random variables, the probability distribution of a random variable X is theassignment (mapping) of probabilities to the values of X.

Any assignment of probabilities to the values of a discrete random variable (r.v.) shouldcomply with the following two probability axioms:

1. The probability of each value of the r.v. must be less than or equal to 1 and greater thanor equal to 0.

2. The sum of the probabilities of all the values of the r.v. must equal 1.-- The Mean of a Discrete Random Variable X

Example 1

N = The number of people in l ine at a tell er machine.Possible Values of N = { 0, 1, 2, 3, 4, 5, . . . }

The number of people that you found in line the last 10 times that you went to an automatic tellermachine (ATM), are shown in the table below.

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The weighted average of the number of people that you found in line each time that you went tothe ATM for these 10 times, can be computed as follows:

Weighted Average

= SUM of {(Number of People in Line)*(Relative Frequency)}

= (0)(.10) + (1)(.40) + (2)(.30) + (3)(.10) + (4)(.10)

= 1.7

Example 2

The number of heads (X) that show up when a fair coin is tossed 4 times, is a random variable.The probability distribution function (pdf) of the random variable X (i.e., the mapping ofprobability values to each of the possible outcomes of X), is shown below.

Probability Distribution Function (pdf)

of the Number of Heads X in 4 Tosses of a Fair Coin

HTTH

HHTT

HTTT HTHT HHHT

THTT THHT HHTH

TTHT THTH HTHH

TTTT TTTH TTHH THHH HHHH

X = 0 X = 1 X = 2 X = 3 X = 4P(X = 0) = 1/16 =

.0625P(X = 1) = 4/16 =

.2500P(X = 2) = 6/16 =

.3750P(X = 3) = 4/16 =

.2500P(X = 0) = 1/16 =

.0625

The mean (weighted average) of the random variable X is computed as follows:

People in Line Frequency Relative Frequency

0 1 .101 4 .402 3 .303 1 .10

4 1 .10Total 10 1.00

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Mean of X ()= SUM of { (x)* P(X =x)}

= (0)(.0625) + (1)(.2500) + (2)(.3750) + (3)(.2500) + (4)(.0625)

= 2

-- The Variance of a Discrete Random Variable X

As expected, the mean of a random variable X tells us where the center of the distribution is.Variability in the values of X is measured by using either the variance () or the standarddeviation () of the random variable X.Example 1

Compute the variance and standard deviation of the random variable X, the number of heads thatcan show up in 4 tosses of a fair coin.

Variance of X (VarX or 2)= SUM of {( - x)2 P(X = x)}= (2 - 0) (.0625) + (2 - 1) (.2500) + (2 - 2) (.3750) + (2 - 3) (.2500) + (2 - 4) (.0625)

= 1

Standard Deviation of X ()= SQRT (VarX) = SQRT (1) = 1

EXERCISES

1. A large US bank wants to find out the mean and standard deviation of the number ofpeople in line at one of its ATM locations. The probability distribution of the random variable N,( N = the number of people in line) based on the banks records, is given in the following table.

People in Line (N) 0 1 2 3 4 5 6 >6

P(N = n) .25 .42 .15 .10 .05 .02 .01 .00

Compute the mean and the standard deviation of the r. v. N.Answers: = 1.38, 2 = 1.6756, or = SQRT(1.6756) = 1.29442. The probability distribution of the number of raisins N in a cookie is as follows:

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Number of raisins (N) 0 1 2 3 4 5 >5

P(N = n) .05 .10 .20 .40 .15 .10 .00

Find the mean , variance, and standard deviation for the r. v. N, the number of raisins in a

cookie.

Answer: = 2.8, 2 = 1.56, = 1.2493. A fair die is rolled. If an even number shows up, you win as many dollars as the number thatshows up on the die. If an odd number shows up, you lose $2.20. Let X = what you win or loseeach time you play the game.

1. What is the probability distribution function of X?2. Compute the mean and the standard deviation of X.

Answers:

(1)

Possible values of X $2 $4 $6 - $2.20

P(X = x) 1/6 1/6 1/6 3/6

(2) mean = $0.90, standard deviation = $3.31

4. Each day a bakery bakes four cakes at the cost of $8 each, and prices them at $25 dollars each.

(Any cake not sold at the end of the day is discarded.) The demand X for cakes on any day hasthe following probability distribution.

Daily demand for cakes (X) 0 1 2 3 4

P(X = x) .10 .20 .50 .10 .10

Let Y be the profit from cakes sold on any day.

Describe the probability distribution of the Y. (Hint: Profit can also be negative. For example ifthe bakery does not sell any cakes, then the profit for that day will be - $32)

1. Find the expected (mean) profit of the bakery, if each day the bakery bakes 4 cakes.2. The bakery's management wonders whether baking 3, or 2 cakes will be more profitable

for the bakery in the long run. How many cakes should the bakery bake each day (4, 3, or2) in order to make it most profitable in the long run?

-- Continuous Random Variables

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The probability distribution of a continuous random variable X is described by a density curve,much like normal curves.

Here events are expressed as: { X > a }, { X < b }, or { a < X < b }.

Probabilities for these events are found from the areas that correspond to them, under theparticular density curve.

Example 1

Let T = time in minutes that one has to wait in line at a teller machine.

The probability distribution of the r. v. T could be described by a curve such as the one in thefigure below.

EXERCISE

The weight of food packed in certain containers is a normally distributed random variable with amean weight of 500 lb and a standard deviation of 5 lb. Suppose a container is picked at random.Find the probability that it contains:

1. more than 510 lb2. less than 498 lb3. between 491 and 503 lb

Answers:

1.

0.02282. 0.34463. 0.6898