# Chapter 4 Using Probability and Probability Distributions

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Chapter 4 Using Probability and Probability Distributions
Business Statistics: A Decision-Making Approach 7th Edition Chapter 4 Using Probability and Probability Distributions

Important Terms Random Variable - Represents a possible numerical value from a random event and it can vary from trial to trial Probability – the chance that an uncertain event will occur Experiment – a process that produces outcomes for uncertain events Sample Space (or event) – the collection of all possible experimental outcomes

Basic Rule of Probability
Individual Values Sum of All Values 0 ≤ P(Ei) ≤ 1 For any event Ei where: k = Number of individual outcomes in the sample space ei = ith individual outcome The value of a probability is between 0 and 1 0 = no chance of occurring 1 = 100% change of occurring The sum of the probabilities of all the outcomes in a sample space must = 1 or 100%

Simple probability The probability of an event is the number of favorable outcomes divided by the total number of possible equally-likely outcomes. This assumes the outcomes are all equally weighted. Textbook: classical probability

Visualizing Events (or sample space)
Contingency Tables Tree Diagrams Ace Not Ace Total Black Red Total Sample Space 2 24 Ace Sample Space Black Card Not an Ace Full Deck of 52 Cards Ace Red Card Not an Ace

Experimental Outcome Example
A automobile consultant records fuel type and vehicle type for a sample of vehicles 2 Fuel types: Gasoline, Diesel 3 Vehicle types: Truck, Car, SUV 6 possible experimental outcomes: e1 Gasoline, Truck e2 Gasoline, Car e3 Gasoline, SUV e4 Diesel, Truck e5 Diesel, Car e6 Diesel, SUV Truck e1 e2 e3 e4 e5 e6 Car Gasoline SUV Truck Diesel Car SUV

Simple probability What is the probability that a card drawn at random from a deck of cards will be an ace? 52 cards in the deck and 4 are aces The probability is ???? Each card represents a possible outcome  52 possible outcomes.

Simple probability There are 36 possible outcomes when a pair of dice is thrown.

Simple probability Calculate the probability that the sum of the two dice will be equal to 5? Four of the outcomes have a total of 5: (1,4; 2,3; 3,2; 4,1) Probability of the two dice adding up to 5 is ????.

Simple probability Calculate the probability that the sum of the two dice will be equal to 12? Only one (6,6) ????

Simple Probability The probability of event A is denoted by P(A)
Example: Suppose a coin is flipped 3 times. What is the probability of getting two tails and one head? The sample space consists of 8 possible outcomes. S = {TTT, TTH, THT, THH, HTT, HTH, HHT, HHH} the probability of getting any particular outcome is 1/8. Getting two tails and one head: A = {TTH, THT, HTT} P(A) = ????

Essential Concepts and Rules of Probability
Independent event Dependent event Joint and Conditional Probabilities Mutually Exclusive Event Addition Rule Complement Rule Multiplication Rule

Independent vs. Dependent Events
Independent: Occurrence of one does not influence the probability of occurrence of the other Dependent: Occurrence of one affects the probability of the other

Examples Independent Events A = heads on one flip of fair coin
B = heads on second flip of same coin Result of second flip does not depend on the result of the first flip. Dependent Events X = rain forecasted on the news Y = take umbrella to work Probability of the second event is affected by the occurrence of the first event

Probability Types Simple (Marginal) probability Joint probability
Involves only a single random variable, the outcome of which is uncertain Joint probability Involves two or more random variables, in which the outcome of all is uncertain Conditional probability (will be discussed soon!) Download “Probability Type Example ” Excel file

Exercise 4-22-B (page 158)

Practice Exercise 4-23: try questions. Electrical Mechanical Total
Electrical Mechanical Total Lincoln 28 39 67 Tyler 64 69 133 92 108 200

Conditional Probability: ex 1
A conditional probability is the probability of an event given that another event has occurred. Involves two or more random variables, in which the outcome of at least one is known Conditional probability for any two events A , B: Notation

Conditional Probability: ex 1
Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. What is the probability that a car has a CD player, given that it has AC (use the table on the next slide)? i.e., we want to find P(CD | AC)

Conditional Probability: ex 1
What is the probability that a car has a CD player, given that it has AC ? CD No CD Total AC .2 .5 .7 No AC .2 .1 .3 Total .4 .6 1.0

Conditional Probability: ex 2
What is the probability that the total of two dice will be greater than 8 given that the first die is a 6? This can be computed by considering only outcomes for which the first die is a 6. Then, determine the proportion of these outcomes that total more than 8.

Conditional Probability: ex 2
There are 6 outcomes for which the first die is a 6. And of these, there are four that total more than 8. (6,3; 6,4; 6,5; 6,6)

Conditional Probability: ex 2
6 outcomes for which the first die is a 6 There are four that total more than 8 The probability of a total greater than 8 given that the first die is 6 is 4/6 = 2/3. More formally, this probability can be written as: p(total>8 | Die 1 = 6) = 2/3(6,3; 6,4; 6,5; 6,6). Conditional probability using “Probability Type Example” Excel file

Mutually Exclusive Events
If A occurs, then B cannot occur A and B have no common elements B A card cannot be Black and Red at the same time. A Red Cards Black Cards

Mutually Exclusive Events
If events A and B are mutually exclusive, then the probability of A or B is p(A or B) = p(A) + p(B) What is the probability of rolling a die and getting either a 1 or a 6? impossible to get both a 1 and a 6 p(1 or 6) = p(1) + p(6) = 1/6 + 1/6 = 1/3

Not Mutually Exclusive Events
If the events are not mutually exclusive, P(A or B) = P(A)+ P(B) – P(A and B) + = A B A B P(A or B) = P(A) + P(B) - P(A and B) Overlap: joint probability

Not Mutually Exclusive Events
What is the probability that a card will be either an ace or a spade? p(ace) = 4/52 and p(spade) = 13/52 The only way both can be drawn is to draw the ace of spades. There is only one ace of spades: p(ace and spade) = 1/52 . The probability of an ace or a spade can be: p(ace) + p(spade) - p(ace and spade) = 4/ /52 - 1/52 = 16/52 = 4/13

Rule of Addition Suppose we have two events and want to know the probability that either event occurs. Mutually exclusive: P(A or B) = P(A) + P(B) Not Mutually exclusive: P(A or B) = P(A)+ P(B) – P(A and B)

Complement Rule The complement of an event E is the collection of all possible elementary events not contained in event E. The complement of event E is represented by E. Complement Rule: Or, E If there is a 40% ( E ) chance it will rain, there is a 60% ( E ) chance it won’t rain! E

Rule of Multiplication
The rule of multiplication applies to the situation when we want to know the probability that both events (event A and event B) occur. Addition: either events Independent: P(A ∩ B) = P(A) P(B) Dependent: P(A ∩ B) = P(A) P(B|A)

Independent Events If A and B are independent, then the probability that events A and B both occur is: p(A and B) = p(A) x p(B) What is the probability that a coin will come up with heads twice in a row? Two events must occur: a head on the first toss and a head on the second toss. The probability of each event is 1/2 the probability of both events is: 1/2 x 1/2 = 1/4.

Independent Events What is the probability that the first card is the ace of clubs (put it back: replace) and the second card is a club (any club)? The probability of the first event is 1/52 (only one ace of club) The probability of the second event is 13/52 = 1/4 (composed of clubs) Answer is 1/52 x 1/4 = 1/208

Dependent Events If A and B are not independent, then the probability of A and B both occur is: p(A and B) = p(A) x p(B|A) where p(B|A) is the conditional probability of B given A.

Dependent Events If someone draws a card at random from a deck and then, without replacing the first card, draws a second card, what is the probability that both cards will be aces?  4 of the 52 cards are aces First: p(A) = 4/52 = 1/13. Of the 51 remaining cards, 3 are aces. So, p(B|A) = 3/51 = 1/17 Probability of A and B is: 1/13 x 1/17 = 1/221.

Practice Page 180 Exercise 4-26 Exercise 4-30

Summary Probability Rules Addition Rule for Two Events
Addition Rule for Mutually Exclusive Events Conditional Probability Multiplication Rules

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