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

Bernoulli Trials The basis for the probability models we will examine in this chapter is the We have Bernoulli trials if: there are possible outcomes ( and ). the probability of success, p, is the trials are Bernoulli trial. two success failure constant. independent.

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 Geometric models are completely specified by one parameter, , and are denoted Geom( ). first success. p, the probability of success p

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) = qx-1p

Independence One of the important requirements for Bernoulli trials is that the trials be When we don’t have an infinite population, the trials are But, there is a rule that allows us to pretend we have trials: The % 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 % of the population. independent. not independent. independent 10 10

TI Tips Under Distributions you will find both geometpdf and geometcdf Use geometpdf(p,x) to find the probability of any outcome. p is the individual probability of success and x is the number of trials until you get a success. Ex. geometpdf(.2,5) ≈ Use geometcdf(p, x) to find the probability of success the xth trial. Ex. Geometcdf(.2, 4) ≈ individual 0.082 on or before 0.590

Ex. Approximately 3.6% of all (untreated) Jonathan apples had bitter pit in a study conducted by the botanists Ratkowsky and Martin (Australian Journal of Agricultural Research, Vol. 23, pp.783-790). Bitter pit is a disease of apples resulting in a soggy core, which can be caused either by over watering the apple tree or by a calcium deficiency in the soil. Let n be a random variable that represents the first Jonathan apple chosen at random that has bitter pit. a) Find the probability that n = 3, n = 5, n= 12. b) Find the probability that n < 5. c) Find the probability that n > 5. d) State the mean and standard deviation. geometpdf(.036,3) ≈ 0.033 geometpdf(.036,5) ≈ 0.031 geometpdf(.036,12) ≈ 0.024 geometcdf(.036,5) ≈ 0.167 P(5 - ∞) = 1 – P(1-4) = 1 - geometcdf(.036,4) ≈ 0.864 μ = ≈ 27.78 apples σ = ≈ 27.27 apples

Assignment P. 398 #7-12

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

The Binomial Model A Binomial model tells us the probability for a random variable that counts the number of in a fixed number of Two parameters define the Binomial model: ; and, . We denote this Binom(n, p). successes Bernoulli trials. n, the number of trials p, the probability of success

The Binomial Model (cont.) In n trials, there are ways to have k successes. Read nCk as Note: n! = n x (n-1) x … x 2 x 1, and n! is read as Can also be shown as “n choose k.” “n factorial.” Find the following. a) b) 10C4 = 210 = 10

The Binomial Model (cont.) Binomial probability model for Bernoulli trials: Binom(n,p) n = p = q = X = P(X = x) = number of trials probability of success 1 – p = probability of failure # of successes in n trials nCx px qn-x

TI tips Under Distributions you will find both binompdf and binomcdf Use binompdf(n,p,x) to find the probability of any outcome. n is the number of trials, p is the individual probability of success and x is the number of successes. Ex. binompdf(5,.2,2) Use binomcdf(n,p, x) to find the probability of getting successes in n trials. Ex. binomcdf(5,.2, 2) individual ≈ 0.205 x or fewer ≈ 0.942

Ex. A basketball player makes 70% of his freethrows Ex. A basketball player makes 70% of his freethrows. Assuming the shots are independent, what is the probability of each of the following? a) He makes exactly 3 out of the next 5 attempts. b) He makes at most 3 out of the next 5. c) He makes at least 3 out of the next 5. binompdf(5,.7,3) ≈ 0.309 P(0-3) = binomcdf(5,.7,3) ≈ 0.472 P(3-5) = 1 – P(0-2) = 1 – binomcdf(5,.7,2) ≈ 0.837

The Normal Model to the Rescue As long as the 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 successes and failures: np ≥ 10 and nq ≥ 10 Success/Failure 10 10

Continuous Random Variables When we use the Normal model to approximate the Binomial model, we are using a random variable to approximate a random variable. So, when we use the Normal model, we no longer calculate the probability that the random variable equals a value, but only that it lies two values. continuous discrete particular between

Ex. An orchard owner knows that he’ll have to use about 6% of the apples he harvests for cider because they will have bruises or blemishes. He expects a tree to produce about 300 apples. a) Verify that you can use the normal model to approximate the distribution. b) Find the probability there will be no more than a dozen cider apples. c) Is it likely there will be more than 50 cider apples? Explain. np = 18 > 10 and nq = 282 > 10, therefore N(18, 4.113) is appropriate. .5 – N(12,18,18,4.113) ≈ 0.072 P(x < 12) = P(0-12) = binomcdf(300,.06,12) ≈ .085 No, 50 is ≈ 7.8 SD above the mean.

What Can Go Wrong? Be sure you have Bernoulli trials. You need outcomes per trial, a constant , and Remember that the provides a reasonable substitute for Don’t confuse models. Don’t use the Normal approximation with You need at least successes and failures to use the Normal approximation. two probability of success independence. 10% Condition independence. Geometric and Binomial small n. 10 10

Assignment P. 399 #13-15, 17, 19-21, 26, 29