Review What is a sequenced experiment and how do you count the total outcomes? It is an experiment with a sequence of “k” trials, with n1 possible outcomes for the first trial, n2 for the second trial, etc. Total Outcomes = (n1)(n2) … (nk) What is a factorial experiment and how do you count the total outcomes? It is an experiment involving placing “n” objects in order. Total Outcomes = n! = (n)(n – 1)(n- 2) … (2)(1) What is the difference between a permutation and a combination? A permutation is an experiment INVOLVING PLACING A SUBSET OF “N” OBJECTS IN ORDER from a set of N objects; a combination is an experiment involving choosing a SUBSET of “n” objects from a set of N objects. How do you count the total outcome of each? NPn Total Outcomes = NPn = (N)! /(N – n)! NCn Total Outcomes = NCn = (N)! /[(n!)(N – n)!] Facts about Probabilities 0 P(Ei) 1 for any i. In other words, all probabilities for events or outcomes are between 0 and 1 inclusive. P(O1) + P(O2) + … + P(Ok) = 1. In other words, the sum of the probabilities of all possible basic outcomes of a random experiment is 1. Assigning Probabilities The Classical Approach The classical approach is appropriate for assigning probabilities when all outcomes are equally likely. Examples include draw a card from a fair deck, flipping a fair coin, rolling a fair die, etc. P (Oi) = 1/n, where n represents the total number of equally likely individual outcomes. P (rolling a two on a fair dice) P (pulling the ace of spades) P (flipping a head) The Relative Frequency Approach The relative frequency approach measures the proportion of the number of times an event [or outcome] occurs for the given number of times a random experiment is conducted. P (Ei) = Number of successes/number of trials As the number of trials approaches infinity, the P(Ei) approaches the true probability.
Class Discussion McCheesie fast food restaurant is trying to increase its share of the breakfast market by introducing breakfast burritos. Management would like to know the probability of 20 or fewer customers ordering the new product. [It is not worth’s management’s effort to introduce or carry a product line unless sales are 21 or more by 9 AM.] Management observes the number of days in the first month (30 days) in which 20 or less burritos were sold (Event A). This happened 25 days. P (less than 21 days) = Successes / Trials = 25 / 30 = .833 Subjective Probability Subjective Probability is based on perceived frequency, not observable trials. Many random experiments can only be performed once or twice. For example, Coke was replaced with New Coke. Management had some subjective beliefs about the effects on profits. They were proven wrong. [Bayesian probabilities which will be discussed in econometrics takes these into consideration.] Calculating the Probability of an Event The probability of an event equals the sum of the probabilities of each individual outcome in the event. P(E) = P(Oi); where Oi is an individual outcome of the event. Class Discussion Probability of rolling an odd number on a fair die P (an odd number) = P (1,3,5) = P (1) + P (3) + P (5) = 1/6 + 1/6 + 1/6 = .5 Probability of rolling a number greater than 4 on a fair die P (more than 4) = P (5,6) = P (5) + P (6) = 1/6 + 1/6 = 1/3 Probability of specific ranking of four commercial networks Count = n! = 4! = (4)(3)(2)(1) = 24 P (a specific outcome) = 1/n! = 1/24 .04167 Probability of a specific ranking of 2 of the four networks Count = NPn = N! /(N – n)! = 4! /(4 – 2)! = 12 P (a specific ranking of three out of five) = 1/12 .08333 Probability that the top two will be NBC and ABC Count = NCn = N! /n (N – n)! = 4! /[2! (4 – 2)!] = 6 P (the top two networks will be NBC and ABC) = 1/6 .16667
There are eight applicants for four job openings. Five are men and three are women. If all candidates are equally likely to be chosen, what is the probability that no women are chosen? [Hint: Choosing no women is the same as choosing all men.] P (A) = P (choosing four men from the five) = 5/70 = 1/14 P (Choosing any four) Count = NCn = N! /n (N – n)! = 5!/[4!(5 – 4)!] = 5 Count = NCn = N! /n (N – n)! = 8!/[4!(8 – 4)!] = 70 Relationships of Probabilities Complements P(A) + P(A) = 1; Why? A A = S, the sample space Addition Law P(A B) = P(A) + P(B) – P(A B); Why? ELEMENT IN THE INTERSECTION WOULD BE COUNTED UNDER P (A) AND AGAIN IN P (B) Class Discussion Hospital records show 12 % of patients are admitted for surgery, 16 % for obstetrics, and 2 % for both. If a new patient is admitted, what is the probability that s/he is admitted for either surgery or obstetrics, or both? Verbalize your answer. Records of auto accidents on a particular stretch of I81 show that 30% of accidents are caused by weather conditions, 20% result in bodily injury, and 42% are either weather related, result in injury, or both. You hear on the news of a wreck on I81 (on this particular stretch of road). What’s the probability the accident was caused by the weather and resulted in injury? Verbalize your answer.