Chapter 17 Probability Models Copyright © 2010 Pearson Education, Inc.

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.

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).

The Geometric Model (cont.) Geometric probability model for Bernoulli trials: Geom(p) p = probability of success q = 1 – p = probability of failure X = number 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 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.

Groovy M&M’s A new sales gimmick has 30% of the M&M’s covered in speckles. These ‘groovy’ candies are mixed randomly with normal candies as they are bagged. You buy a bag and remove candies one at a time looking for speckles. Does this represent a Bernoulli trial? Explain. What is the probability that the first speckled one we see is the fourth candy we pick? What is P(X=3) What is P(X=5) How many candies on average would we need to look at to find a speckled one?

A new sales gimmick has 30% of the M&M’s covered in speckles A new sales gimmick has 30% of the M&M’s covered in speckles. These ‘groovy’ candies are mixed randomly with normal candies as they are bagged. You buy a bag and remove candies one at a time looking for speckles. Does this represent Bernoulli trials? Explain. What is the probability that the first speckled one we see is the fourth candy we pick? What is the probability that the first speckled candy is one of the first three we look at? What is the probability we end up with two speckled candies in a handful of five?

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).

The Binomial Model (cont.) In n trials, there are ways to have k successes. Read nCk as “n choose k.” Note: n! = n  (n – 1)  …  2  1, and n! is read as “n factorial.”

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) = nCx px qn–x

Geometpdf: first success on trial x Geometcdf: first success on or before trial x Binompdf: x successes in n trials Binomcdf: x or fewer successes in n trials

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…

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

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.

Spam A company that monitors email claims that out of the over 1 billion emails sent every day 91% of them are spam. When you first check your email in the morning, about how many emails should you have to open before you find the first real email? What is the probability that the 4th message in your inbox is the first one that isn’t spam?

Spam 91% of email messages are spam. Suppose there are 25 messages in your inbox. What are the mean and standard deviation of the real messages in your inbox? What is the probability you will find only 1 or 2 real messages?

Spam 91% of email messages are spam. Over the course of a week your spam filter let 151 out of 1422 messages reach your inbox. Should you be concerned? In other words, what is the probability that only 151 of 1422 messages are real?

Arrows An archer is able to hit the bull’s eye 80% of the time. If she shoots 6 arrows, what is the probability of each of the following? Her first bull’s eye is the third arrow she misses the bull’s eye at least once first bull’s eye is on the 4th or 5th arrow she gets exactly 4 bull’s eyes she gets at least 4 bull’s eyes she gets at most 4 bull’s eyes

Arrows An archer is able to hit the bull’s eye 80% of the time. If she shoots 6 arrows, what is the probability of each of the following? How many bull’s eyes do you expect her to get? With what standard deviation? If she keeps shooting until she hits the bull’s eye, how long do you expect it to take?

Arrows An archer is able to hit the bull’s eye 80% of the time. Say she shoots 10 arrows. Find the mean and standard deviation of the number of bull’s eyes she may get. What is the probability: she never misses? there are no more than 8 bull’s eyes? there are exactly 8 bulls eyes? she hits the bull’s eye more often than she misses?

Arrows An archer is able to hit the bull’s eye 80% of the time. She will be shooting 200 arrows in a competition. What are the mean and standard deviation of the number of bull’s eyes she might get? Is a normal model appropriate here? Use the 68-95-99.7 rule to describe the distribution of the number of bull’s eyes she might get. Would you be surprised if she made only 140 bull’s eyes?

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.

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

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.