Chapter 17 (Specific Probability Models)

Transcription

1 Chapter 17 (Specific Probability Models) We are going to look at some specific probabilities that have the following attributes... a.) There are only 2 possible outcomes (success or failure). 2.) The probability of success is the same for every trial (we will call this "p") The probability of failure is the same for every trial ("q") p + q = 1 III.) The trials are independent. First trial result has no effect on subsequent trials. Situation like this occure often and are called Bernoulli Trials. (Common examples, flipping a coin, shooting a free throw,...) Geometric Probability Model A Geometric Probability Model is made up of Bernoulli trials and answers the question "What is the probability of the first success occurring on the n th trial?" Rolling a die...what is the probablity that the first time I will roll a "1" is on the 4th roll? It is fairly simple to come up with the following formulas...

2 Formulas for Geometric Probability Model: P(X) = q x 1 p where X is number of trials until first success, q is probability of failure, p is probability of success. E(X) = µ = 1 p SD(X) = σ = q p 2 91% of all e mail is SPAM. What is the probability that the first e mail you see in the morning is an actual e mail and not spam? What is the probability that the first e mail you see that is not spam is the 4th one you look at? What is the probability that the first e mail you see that is not spam is the 10th one you look at? How many e mails should you expect to have to look at until you see one that is not spam?

3 Just a little bit more about independence? Sometimes we think trials are independent when they actually are not quite so. Example: A cereal company has decided to put cards of famous athletes in each box of cereal (We actually talked about this earlier). 20% of the boxes contain "Aaron Rodgers" cards. If you open a box a week, how many boxes should you expect to open until you get an "Aaron Rodgers" card? Are trials independent? Does opening the first box change the probability of the next box having an "Aaron Rodgers" card? What is the company is really small and only made 10 boxes? Well, isn't this true of any finite number of boxes? The 10% Condition: If Bernoulli trials are not truly independent it is okay to proceed as long as the sample is smaller than 10% of the population. One last Geometric Example 6% of blood donors are type O negative (Universal donors). If donors line up in line, what is the probability that the first O negative donor is one of the first 4 people in line? Is this Geometric (3 requirements)? 10% condition met? Now let's answer the question... How about using a calculator? (see Tech Sheet)

4 Binomial Probability Model A Binomial Probability Model uses Bernoulli trials and answers the question "what is the probability of k successes in n trials?" Let's take a little SURPRISE quiz!!!! 5 Question Multiple Choice Quiz 1.) The answer to this question is A.) AB.) BC.) C D.) D 2.) Which of the following made up words has the made up meaning "to contemplate" A.) brork B.) strompt C.) hogit D.) prescht 3.) What number is Mr. Rumpel thinking of right now (as he writes the quiz)? A.) 3 B.) 7 C.) 9 D.) 13 4.) What is the first number in the sequence,,,? A.) 8 B.) 9 C.) 5 D.) 7 5.) What letter fills in the blank to make the word Mr. Rumpel wants it to make "h t" A.) a B.) i C.) o D.) u

5 What is the probability that you got all 5 right? What is the probability that you got 1 right? What is the probability that you got 2 right? What is the probability that you got 3 right? What is the probability that you got 4 right? What is the probability that you got 0 right? Lets figure it out another way. Let's look at the probability of getting one right Let's call getting the answer right a success (S) and getting it wrong a failure (F) we have 5 ways to get 1 right and 4 wrong SFFFF FSFFF FFSFF FFFSF FFFFS looking at the first set the probability of getting the first question right is 1/4 and the probability of getting the next wrong wrong are each 3/4

6 so the probability of getting SFFFF is (1/4) x (3/4) x (3/4) x (3/4) x (3/4) = 81/1024 the probability of getting FSFFF is (3/4) x (1/4) x (3/4) x (3/4) x (3/4) = 81/1024 FFSFF and FFFSF and FFFFS will all be the same so really the probability of getting exactly one question right is 5 x 81/1024 or 405/1024 What did we really just do? Well we just took the probability of getting 1 question right times the probability of getting 4 questions wrong time the number of ways to arrange one right out of 5 questions... Mathemaical Side Trip

7 When you are looking at how many different orders you can pick out of, this is called a Combination. If I want to figure out how many ways I can pick 3 out of 10 people to be on a committee, this would be a combination of 10 choosing 3 and it has the following formula... Combination of choosing k from n = n! k!(n k)!! is the symbol for "Factorial" and means the product of that number and every integer down to 1. 7! = 7 x 6 x 5 x 4 x3 x 2 x 1 = 5040 If I want to figure out how many ways I can pick 3 out of 10 people to be on a committee, this would be a combination of 10 choosing 3 and it has the following formula... Combination of choosing k from n = n! So n = 10 and k = 3 k!(n k)! 10 or 10 C 3 = 10! = = = ! 7! 6 x

8 So the number of ways of getting one question correct out of 5 is 5C 1 = = 5 The calculators can figure out COMBINATIONS for you Math PRB n C k Where n is the total number you are selecting from and k is the number you are selecting. You actually type in the number of n then hit Math PRB n C k then the number for k then enter. Back to regularly scheduled discussion What did we really just do? Well we just took the probability of getting 1 question right times the probability of getting 4 questions wrong time the number of ways to arrange one right out of 5 questions... So how many ways can we pick one question to get right out of 5? 5C 1 = 5 So the answer is 5 x (1/4) 1 x (3/4) 3 =.527 Be honest...so you know where all of those numbers come from?

9 What about the probability of getting exactly 2 questions right? Well how many arrangements can I have with 2 S's and 3 F's SSFFF SFSFF SFFSF SFFFS FSSFF FSFSF FSFFS FFSSF FFSFS FFFSS which is much easier to get by going 5 C 2 For all ten of those arrangements the probability of success is 1/4 so two S's = (1/4) 2 and the probability of three F's is = (3/4) 3 So generally speaking the probability of getting some number of questions right is equal to a combination of the total number of questions picking that number times the probability of getting a question right raised to that number times the probability of getting a question wrong raised to 5 that number. lets call the total number of questions n lets call the number right k lets call the probability of getting a question right p lets call the probability of getting a question wrong q

10 We get: probability of r successes = n C k p k q n k This is the formula for a BINOMIAL PROBABILITY If I roll a die 10 times what is the probability of getting a "1" on exactly 2 of those rolls? Well n = 10, k = 2, p = 1/6 and q = 5/6 So 10 C 2 (1/6) 2 (5/6) 8 = Official Formula Page: n = number of trials p = probability of success q = probability of failure X = number of successes in n trials P(X) = n C x p x q n x, where n C x = n! x!(n x)! E(X) = µ = np SD(X) = σ = npq

11 65% of all college freshmen who eventually get their degree change their major at least once. There are 10 former B W students attending the UW EC as freshmen this year. What is the probability that 3 of them will change their major by the time they get their degree? What is the probability that less than 4 of them will change their major by the time they get their degree? What is the probability that at least 5 of them will change their major by the time they get their degree? See tech sheet on calculator use Assignment Growification Page ,18,19,25,26