4594827 Probability Concepts and Applications

Lirenso: RES 341

1 Probability: Concepts and Applications

Objectives • Define terminology related to probability. • Differentiate between probability and non-probability sampling designs. • Describe the probability distribution of a random variable. • Explain the concept of a normal probability distribution. • Explain the standard normal distribution. 1. Probability Probability theory is an important part of statistical theory that bridges descriptive and inferential statistics. It is the science of uncertainty or chance, or likelihood. A probability value ranges between 0 and 1 inclusive and represents the likelihood that a particular event will happen. A probability value of 0 means there is no chance that an will happen and a value of 1 means there is 100 percent chance that the event will happen. Understanding probability is helpful for decision-making. Conducting an experiment or sample test provides an outcome that can be used to compute the chance of events occurring in the future. An experiment is the observation of some activity or the act of taking some measurement. Whereas, an outcome is a particular result of an experiment. The collection of one or more outcomes of an experiment is known as an event. For example, a market testing of a sample of new breakfast cereal, new beer, new wine, new magazine, etc. gives the Director of Production or Director of Marketing a company a preliminary idea (outcome) whether consumers would like the product if it is produced and distributed in bulk. There are three definitions of probability. The first one is known as classical probability. The classical definition applies when there are n equally likely outcomes to an experiment. It is obtained by dividing the number of favorable outcomes by the total number of possible outcomes. The probability of certain events is already known or the resulting probabilities are definitive. For example: (1) The chance that a woman gives birth to a male or female baby (p = 0.50 or ½), (2) The chance that tail or head appears in a toss of coin (p = 0.50 or ½), and (3) The chance that one spot will appear in die-rolling (p = 0.16 or 1/6). The second one is empirical probability that is based on past experience. This is determined dividing the number of times an event happens by the total number of observations. For example:

Lirenso: RES 341

2

(1) 383 of 751 business graduates were employed in the past. The probability that a particular graduate will be employed in his or her major area is 383/751 = 0.51 or 51%. (2) The probability that your income tax return will be audited if there are two million mailed to your district office and 2,400 are to be audited is 2,400/2,000,000 = 0.0012 or 0.12%. The third is a subjective probability. Subjective probability is a probability assigned to an event based on whatever evidence is available. It is an educated guess. Unlike empirical probability, it is not based on past experience. Subjective probability is obtained by evaluating the available options and by assigning the probability. Examples of events that require computing subjective probability: (1) Estimating the probability that a person wins a jackpot lottery. (2) Estimating the probability that the GM will lose its first ranking in car sales. 2. Events Events can be classified as mutually exclusive, joint, independent, conditional or complement. Combining probabilities of events requires using the rules of addition and multiplication (see table below). A) Two events are mutually exclusive if by virtue of one event happening the other cannot happen. For example, A business can’t be bankrupt, break-even and profitable at the same. It can only be one of the three. Similarly, being a male or female are mutually exclusive and collectively exhaustive events. None one is both and everyone is one or the other. Example: The two most common primary causes of death in the US are heart attack and cancer. Heart attack is the cause for one-third (0.33) of the Americans who die each year and cancer is the cause for one-fifth (0.20) of the deaths each year. If 2003 is like 2002, the probability that a randomly selected American will die of either a heart attack or cancer is the sum of these two probabilities. 0.33 + 0.20 = 0.53. [Special Rule of Addition].

B) Events are joint if two or more events happen at the same time. For example, driving autos is one event and talking on the cell phone is another event. When you see someone talking on the cell phone while driving an automobile it is a joint event. Example: If 90% of the Citibank customers have a saving account, 40% have a market rate account, and 60% have both, the probability that a randomly selected Citibank customer will have either saving or market rate account will be computed as: 0.90 + 0.40 – 0.60 = 0.70 [General Rule of Addition].

C) Events are independent if the occurrence of one event does not affect the occurrence of another event.

Lirenso: RES 341

3

Example: Suppose the probability that a student gets an “A” grade is 0.50 in statistics and 0.60 in History. Assume that the grade received in statistics is independent of the grade received in History. The probability that the student will receive an “A” grade in both subjects (statistics and History) is computed as: (0.50) x (0.60) = 0.30 [Special Rule of Multiplication]. (1) Two traffic lights on Broadway Road operate independently. Your probability of being stopped at the first one is 0.4 and your probability of being stopped at the second one is 0.7. The probability of being stopped at: A) both lights = 0.4 x 0.7 = 0.28 B) neither light = 0.6 x 0.3 = 0.18 C) the first but not the second = 0.4 x 0.3 = 0.12 D) the second but not the first = 0.7 x 0.6 = 0.42

BASIC RULES OF PROBABILITY No.

EVENTS

1

Mutually exclusive

2

Not mutually exclusive (Joint/ Compound)

3

Independent

4

Conditional

5

Complement

FEATURES No overlapping events - if one event happens the other one can’t occur at the same time Overlapping/ Concurrent events – two or more events happen at the same time The occurrence of event A has no effect on the occurrence of another event B Dependent events – the probability of a particular event occurring given that another event has occurred All events in the sample space that are not part of the specified event – determined by subtracting the probability of an event not happening from the probability of happening

KEY CONNECTING WORDS The probability of A occurring or the probability of B occurring

APPLICABLE RULE

FORMULA

Special Rule of Addition

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

The probability that either A may occur or B may occur followed by the possibility that both A and B may occur. The probability that A and B will occur

General Rule of Addition

P (A or B) = P(A)+ P(B) – P(A and B)

Special Rule of Multiplication

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

P(B|A) – probability that event B will occur given that event A has already occurred

General Rule of Multiplication

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

Event Not occurring – or Neither/nor will happen

Complement Rule

P(A) = 1 – P (~ A)

Lirenso: RES 341

4

D) Events are conditional if a particular event occurs given that another event has occurred. Bayesian theorem can be used to revise prior probabilities and validate earlier decisions. For example: Three defective electric toothbrushes were accidentally shipped to a drugstore by Clean-brush products along with 17 non-defective ones. A) What is the probability that the first two electric toothbrushes sold will be returned to the drugstore because they are defective? (3/20) + (2/19) = 6/380 = 0.01579 [General Rule of Multiplication]. B) What is the probability that the first two electric toothbrushes sold will not be defective? (17/20) + (16/19) = 272/380 = 0.7158 [General Rule of Multiplication].

E) The set of all possible outcomes of an experiment is known as sample space. The sample space for die-rolling is {1,2,3,4,5,6}. An event in the sample space that is not part of the specified event is known as the complement. Under conditions where one even occurs as a subset of another event, the probability of the first event cannot be higher than the probability of the one for which it is a subset. Example: Suppose the probability that mortgage loans are approved by lenders in your state is 0.70 (70%). By the complement rule, the probability that loans may not be approved for a randomly selected applicant is 0.30 (30%). 1 - .70 = .30

A contingency table with two variables, values of one variable placed in rows and the other in columns, can be used to estimate simple & joint probabilities of the two events. This can be demonstrated by using the following example. An airline company completed an on-board passenger survey of 400 customers in an attempt to measure the number of bags checked by those traveling on business or for pleasure. Results are shown in the table below: Passenger Business Pleasure Total

0 Checked 50 20 70

1 Checked 60 80 140

2 Checked 30 90 120

3 or more 10 60 70

Total 150 250 400

1) The probability that a customer checks exactly 2 bags 120/400 = 0.30 2) The probability that a passenger who responded to the survey was a business customer is 150/400 = 0.375 3) The probability that a customer checked fewer than 2 bags 70/400 + 140/400 = 210/400 = 0.525 4) The probability that a customer checked more than one bag is 120/400 + 70/400 = 190/400 = 0.47

Lirenso: RES 341

5

The breakdown of the simple and joint probabilities can be displayed graphically using the tree diagram.

Tree Diagram for Airlines Passengers 50 60 30 10

150

400

60 250

90 80 20

3. Discrete Probability Distributions Probabilities values for experiments whose outcomes are numerical are known as random variables. Random variables can be discrete (have a finite number of sample space) or continuous (have an infinite number of sample space). An example of discrete probability distributions is binomial distribution [See Table A2 in the Appendix]. Properties of Binomial distribution: • Each trial (X) may be selected from infinite population without replacement or from a finite population (N) with replacement. • Each trial (X) is mutually exclusive and collectively exhaustive • Each trial (X) has two possible outcomes, success or failure. • Each trial has a fixed probability of success or failure (π). Binomial probability formula for any trial is: P(x) = ___n! πx(1-π)n-x x!(n-x)! Where: n is the number of trials. x is the number of success or failures π is the probability of success or failure on each trial. The above formula can also be written as follows: P(x) = nCr π x (1−π)n-x Where nCr stands for combination.

Lirenso: RES 341

6

The probability notation summary for different values of P (x) is given in the table below. It is important to understand the meaning of terms such as “at least”, “more than”, “at most”, “less than”, “between X1 and X2 inclusive”, and “between X1 and X2.” They indicate cumulative binomial probabilities that involve adding the values up or down. MS Excel can be used to determine the binomial probabilities of an event. Find the probability that x takes on a value that is … At least x More than x At most x Less than x Between x1 and x2 inclusive Between x1 and x2

Probability Notation

P(X ≥ x) P(X > x) P(X ≤ x) P(X < x) P(x1 ≤ x ≤ x2) P(x1 < x < x2)

Example: Suppose AAA road emergency service has been 60% successful in the past reaching its customers within 30 minutes. What is the probability of obtaining three successful calls within an acceptable time in a sample of five monitored service calls in this particular sequence? This problem can be solved in thre different ways. 1) It can be computed using the binomial probability distribution formula. In the above problem: π = 0.60, n = 5, X = 3. P(x=3) = 5! 0.603(1-0.60)5-3 3!(5-3)! = 0.346 2) This value can also be read directly from the Binomial Distribution table. In the table where n = 5, the intersection of row x = 3 and column π = 0.60 would give a probability of 0.346. 3) We can use Excel, probability distributions function. Other types of discrete probability distribution are Hypergeometric, Poisson and Exponential. Hypergeometric is different from binomial and its probability changes from one trial (experiment) to another. Poisson is known as a probability of zero and works with averages, instead of π. 4. Normal Probability Distributions One particular continuous probability distribution is normal distribution. It is a symmetric, bell-shaped distribution that takes a normal random variable, with μ being the mean of the normal random variable, X being any value below or above μ, and ơ being the standard deviation from the center of distribution.

Lirenso: RES 341

7

Properties of the normal probability distribution: • The normal curve is bell shaped and has a single peak at the exact center of the distribution. • The mean, median, and mode of distribution are equal and located at the peak. • Half of the area under the curve is above the peak, and other half is below it. There are several families of the normal distribution curve. The standard one has a random variable Z, μ = 0 and ơ = 1. Based on the Empirical Rule, there are 3ơ on each side of the curve. Probability value is zero at 4ơ and beyond from the center. The probability values for each Z-value are listed in Table A19 for Z value located to the left of μ and in the same table for the Z-value located to the right of μ). As explained in Week Three Lecture, the Z-value measures the number of standard deviations that a data value is from the population mean. The value of Z is obtained using the formula: Z = X – μ (‘s’ can be substituted for ơ if the population parameter is unknown) ơ A negative Z value shows that the random variable is located to the left of μ and a positive Z value means the random variable is located to the right of the center. The rules for solving different normal probability problems are given below. Rules for Solving Normal Probability Problems Manually . Rule 1 2

To Find… P(Z < z) P(Z > z)

Area under the curve… below a value of Z above a value of Z

3

P(z1 < Z < z2)

Between two values of Z

4

P(Z = z)

exactly the value of Z

Look Up… the Z value and use the table directly The Z value and subtract the value in the table from 1 OR the negative of the Z value and use the table directly Both Z values and subtract the lower value from the higher value the Z value in table directly (left side of the curve if Z-value is negative (right side of the curve if Z-value is positive) in Table A19.

Example: The amount of money requested in home loan applications at Dawn River Federal Savings are approximately normally distributed with μ = $70,000 and ơ = $20,00. A loan application is received this morning. What is the probability that: A) the amount requested is $80,000 or more? The answer is 0.3085. Using Z = X – μ/ ơ Z = 80,000 – 70,000/20000 = 0.5. When Z = 0.50, the probability area is 0.3085 in standard normal table or 1.00 – 0.6915 = 0.3085. B) The amount requested is less than $65,000? The answer is 0.4012. Using Z = X – μ/ ơ, Z = 65,000 – 70,000/20000 = -0.25. This can be directly obtained from standard normal table, using row Z = -0.2 and column 0.05.

Lirenso: RES 341

8

5. The Principles of Counting If the number of possible outcomes in an experiment is small, it is relatively easy to count them. If, however, there are a large number of possible outcomes, it would be tedious to count all the possibilities. To facilitate counting, three counting formulas will be examined: the multiplication formula, the permutation formula, and the combination formula. A) The Multiplication Formula is: Total number of arrangements = (m)(n). Example: If a salesperson has seven shirts and 5 ties to display, 35 outfits are possible. This means he or she has 35 different ways displaying the outfits. This can be extended to more than two events. For three events m, n, and o: Total number of arrangements _ (m)(n)(o). B) The Permutation Formula The permutation formula is applied to find the possible number of arrangements when there is only one group of objects. Order is very important in permutation. Note the arrangements a b c and b a c are different permutations. The formula to count the total number of different permutations is:

where: n is the total number of objects (pool). r is the number of objects selected at a time. Before we solve the two problems illustrated, note that permutations and combinations use a notation called n factorial. It is written n! and means the product of n(n-1), (n-2), (n-3), (n-4), (n-5), etc. For instance, 5! Means 5 x 4 x 3 x 2 x 1 = 120.

By definition, zero factorial, written 0!, is 1. That is, 0! = 1. Example: A flag with three stripes of 3 colors can use any of six colors. How many flags are possible. This is a permutation problem and the order of arranging the flags is important.

Lirenso: RES 341

n

Pr = 6P3

=

9

6! = 120 (6-3)!

C) The Combination Formula If the order of the selected objects is not important, any selection is called a combination. The formula to count the number of r object combinations from a set of n objects is:

Example: You have 6 colors to choose from and you wish to choose 3 for a flag. How many choices are possible? This is a combination problem the order of arranging colors in not important. n

Cr = 6C3

=

6! = 20 3!(6-3)!

Discussion Questions 1. Give three real-life or business examples for each of the following: A) Mutually exclusive events B) Joint events C) Independent events D) Conditional events 2. Chances are 50-50 that a newborn baby will be a girl. For families with five children, what is the probability that all the children are girls? [Hint: use binomial probability distribution]

3. Assume a restaurant business succeeds 60% of the time. Suppose that there are three such restaurants open in your city. When they don’t compete with one another, it is reasonable to believe that their relative success would be independent. [Hint: use binomial probability distribution] A) What is the probability that all the three businesses succeed? B) What is the probability that all the three businesses fail? C) What is the probability that at least one business succeeds? 4. A recent study of the hourly wages of maintenance crews for major airlines showed that the mean hourly salary was $16.50, with a standard deviation of $3.50. If we select a crew member at random, what is the probability the crew member earns:

Lirenso: RES 341

a. b. c. d.

10

Exactly $18.00 per hour? Between $16.50 and $20.00 per hour? More than $20.00 per hour? Less than $15.00 per hour?

5. How do we know whether a given counting problem takes the permutation or combination method of selecting or arranging objects?