1. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 1

2. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 17 Probability Models

3. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 3 Bernoulli Trials The basis for the probability models we will examine in this chapter is the Bernoulli trial. We have Bernoulli trials if: there are two possible outcomes (success and failure). the probability of success, p, is constant. the trials are independent.

4. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 4 The Geometric Model A single Bernoulli trial is usually not all that interesting. A Geometric probability model tells us the probability for a random variable that counts the number of Bernoulli trials until the first success. Geometric models are completely specified by one parameter, p, the probability of success, and are denoted Geom(p).

5. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 5 The Geometric Model (cont.) Geometric probability model for Bernoulli trials: Geom(p) p = probability of success q = 1 – p = probability of failure X = # of trials until the first success occurs P(X = x) = q x-1 p

6. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 6 Independence One of the important requirements for Bernoulli trials is that the trials be independent. When we don’t have an infinite population, the trials are not independent. But, there is a rule that allows us to pretend we have independent trials: The 10% condition: Bernoulli trials must be independent. If that assumption is violated, it is still okay to proceed as long as the sample is smaller than 10% of the population.

7. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 7 Example: Speckled Skittles A new gimmick has 30% of Skittles candies covered with speckles. These “groovy” candies are mixed randomly with the normal candies as they are put into the bags for distribution and sale. You buy a bag and remove candies one at a time looking for speckles. a. What is the probability that the first speckled one we see is the fourth candy we get? b. What’s the probability that the first speckled one is the tenth one? c. Write the general formula. d. How many do we expect to have to check, on average, to find a speckled one? Check 10% condition!!

8. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 8 Example continued…. a. P(first speckled is fourth candy)=(.7) 3 (.3)=.1029 b. P(first speckled is tenth candy)=(.7) 10 (.3)=.0085 c. General Formula: P(X=x)=q x-1 p d. μ= 1/p = 1/.3 = 3 1/3 candies

9. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 9 Example 2: Jumper Cables Suppose that 40% of the students who drive to BHS carry jumper cables in their car. Your car has a dead battery and you don’t have jumper cables, so you decide to stop students who are headed to the parking lot and ask them whether they have a pair. a. What is the probability that the third student you ask will have jumper cables? b. What is the probability that first student you ask will have jumper cables? c. What is the probability that three or fewer students must be stopped to find cables? d. What is the average number of students you would have to stop to find jumper cables?

10. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 10 Example 2 continued…. a. P(first person with cables is third one asked)=(.6) 2 (.4)=.144 b. P(first person with cables is first one asked)=(.6) 0 (.4)=.4 c. P(x≤3)=P(1 or 2 or 3)=P(1) +P(2)+P(3) = (.6) 0 (.4)+ (.6) 1 (.4)+ (.6) 2 (.4) =.4 +.24 +.144 =.784 d. μ = 1/p = 1/.4 = 2.5 students

11. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 11 Geometric Model with Calculator geometpdf(p,x) p is the probability of success x is the number you are “waiting” for Based on the last example, what is the probability of the 4 th student have the jumper cables you need. geometpdf(.4,4) =.0864 What is the probability that you will find jumper cables in 4 or fewer students asked? geometcdf(.4,4) =.8704 …….or……… geometpdf(.4,1)+geometpdf(.4,2)+ geometpdf(.4,3)+ geometpdf(.4,4) =.8704

12. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 12 Example: Skittles again! Back to the speckled skittles....what is the probability that we’ll find two speckled ones in a handful of five candies? This is a little bit different…notice we are not waiting until something “special” happens? We are looking for a number of successes in a group.

13. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 13 How do you find the answer? Would this work? P(2 speckled out of 5) = (.3) 2 (.7) 3 NO Why not? This method says that the first 2 would be speckled and the last three are not. Can you get 2 speckled in another way? We need to count the number of ways that getting 2 speckled out of 5 can happen. Do you remember anything from Algebra II that might help?

14. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 14 The Binomial Model A Binomial model tells us the probability for a random variable that counts the number of successes in a fixed number of Bernoulli trials. Two parameters define the Binomial model: n, the number of trials; and, p, the probability of success. We denote this Binom(n, p).

15. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 15 The Binomial Model (cont.) In n trials, there are ways to have k successes. Read n C k as “n choose k.” Note: n! = n x (n-1) x … x 2 x 1, and n! is read as “n factorial.”

16. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 16 The Binomial Model (cont.) Binomial probability model for Bernoulli trials: Binom(n,p) n = number of trials p = probability of success q = 1 – p = probability of failure X = # of successes in n trials P(X = x) = n C x p x q n-x

17. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 17 Back to the Skittles…. Now lets solve it: P(2 speckled out of 5) = 5 C 2 (.3) 2 (.7) 3 = 10(.3) 2 (.7) 3 =

18. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 18 Calculator for Binomial Models binompdf(n,p,x) n is the number in the group p is the probability of success x is the number of successes you want Based on the last example, what is the probability of 2 out 5 being speckled? P(X=2) = binompdf(5,.3,2) = What is the probability that you will get 2 or fewer speckled Skittles out of 5? P(X<=2) = binomcdf(5,.3,2) = What is the probability that you will get at least 2 speckled Skittles out of 5? P(X>=2) = 1-binomcdf(5,.3,1) =

19. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 19 Before you do a problem: Check to see if there are Bernoulli trials. 1. 2. 3. If you can’t say “independent”, you must show the 10% condition to be true. State the model that you are using, with correct symbolic notation.(Geom(p) or Binom(n,p)) Set up your probability: P(of what?)= answer

20. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 20 Example: Blood Donors People with O-negative blood are called “universal” donors because O-negative blood can be given to anyone else, regardless of the recipient’s blood type. Only about 6% of people have O-negative blood. a. Suppose 20 donors come to the blood drive. What are the mean and standard deviation of the number of universal donors among them? b. What is the probability that there are 4 universal donors in a group of 100 donors? c. What is the probability that there are 2 or 3 universal donors? d. What is the probability that there are at most 3 universal donors in 55 donations? e. What is the probability that there are at least 4 universal donors in a 150 donations? Don’t forget…… a. If donors line up at random for a blood drive, how many do you expect to examine before you find someone who has O-negative blood? b. What is the probability that the first O-neg donor is the 9 th person in line? c. What’s the probability that the first O-negative donor found is the one of the first four people in line?

21. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 21 Blood Example What do you have to do first? Lot’s of stuff. E(X)=16.7 P(X<=4)=.2193 E(X)=1.2 SD(X)=1.06 P(X=2 or X=3)=.3106

22. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 22 Example – More Blood Suppose the Red Cross anticipates the need for at least 1850 units of O-negative blood this year. It estimates that it will collect blood from 32,000 donors. How great is the risk that they will fall short of meeting its need? We need to calculate the probability of getting exactly 1850 units of O-neg blood from 32,000 donors. What would that be? That’s just the beginning – they need at least 1850 units – so we will need to do it again for 1851 and 1852 and so on…..

23. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 23 The Normal Model to the Rescue! When dealing with a large number of trials in a Binomial situation, making direct calculations of the probabilities becomes tedious (or outright impossible). Fortunately, the Normal model comes to the rescue…

24. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 24 The Normal Model to the Rescue (cont.) As long as the Success/Failure Condition holds, we can use the Normal model to approximate Binomial probabilities. Success/failure condition: A Binomial model is approximately Normal if we expect at least 10 successes and 10 failures: np ≥ 10 and nq ≥ 10

25. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 25 Blood and more blood…. Check Bernoulli and success failure condition: np ≥ 10 and nq ≥ 10 32000(.06) ≥ 10 and 32000(.94) ≥ 10 ≥ 10 ≥ 10 So, we can use a Normal model, using the mean and standard deviation from the Binomial model. μ = np = 32000(.06) = 1920 and SD= N(1920,42.48) So we can find the probability P(X>1850)=P(z>-1.65)=.95

26. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 26 Continuous Random Variables When we use the Normal model to approximate the Binomial model, we are using a continuous random variable to approximate a discrete random variable. So, when we use the Normal model, we no longer calculate the probability that the random variable equals a particular value, but only that it lies between two values.

27. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 27 What Can Go Wrong? Be sure you have Bernoulli trials. You need two outcomes per trial, a constant probability of success, and independence. Remember that the 10% Condition provides a reasonable substitute for independence. Don’t confuse Geometric and Binomial models. Don’t use the Normal approximation with small n. You need at least 10 successes and 10 failures to use the Normal approximation.

28. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 28 What have we learned? Bernoulli trials show up in lots of places. Depending on the random variable of interest, we might be dealing with a Geometric model Binomial model Normal model

29. Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 17- 29 What have we learned? (cont.) Geometric model When we’re interested in the number of Bernoulli trials until the next success. Binomial model When we’re interested in the number of successes in a certain number of Bernoulli trials. Normal model To approximate a Binomial model when we expect at least 10 successes and 10 failures.