1. Chapter 5, continued...

2. IV. Binomial Probability Distribution The binomial is used to calculate the probability of observing x successes in n trials. Ex. What’s the probability of getting 5 “heads” when you toss a coin 7 times?

3. A. Binomial Experiment Properties 1. The experiment consists of a sequence of n identical trials. 2. Two outcomes are possible for each trial: “success” and “failure”. 3. The probability of a success, denoted by p, doesn’t change from trial to trial. 4. The trials are independent.

4. Example A car saleswoman approaches 3 prospective buyers. Past experience tells her that a shopper will buy 1/10 times. 1. n=3 identical trials 2. S=buys a car; F=doesn’t buy a car 3. p=.10, (1-p)=.90 4. Trials are independent since customers don’t know each other. The action of one customer can’t affect the probability the next buys a car.

5. The problem What is the probability that exactly 2 of the 3 customers buy a car? S F S S S S S F F F F F F S SSS (x=3) SSF (2) SFS (2) SFF (1) FSS (2) FSF (1) FFS (1) FFF (0)

6. Using the counting rule You can see from the probability tree that there are 3 possible outcomes that have exactly 2 successes. But you can use the counting rule from chapter 4. These 3 outcomes are (SSF), (SFS), and (FSS).

7. Calculating probabilities Using the multiplication law, you can find the probability of observing each outcome: P(SSF)=p*p*(1-p)=.1(.1)(.9)=.009 P(SFS)=p*(1-p)*p=.1(.9)(.1)=.009 P(FSS)=(1-p)*p*p=.9(.1)(.1)=.009

8. General formulas In general, the probability of x successes in a particular sequence of n trial outcomes: Finally we need to combine the number of outcomes with the probability of each to give us a Binomial Probability Function.

9. C. Binomial Probability Function Where f(x) is the probability of x successes in n trials. In our example, what is the probability that exactly 2 of the 3 customers buy a car? f(2)=3*(.1) 2 (1-.1) (3-2) =.027 Not good odds!!!!

10. D. Binomial E(x) and Var(x) The expected value: E(x)=  =np Variance: Var(x)=  2 =np(1-p)

11. V. Poisson Probability Distribution Siméon Denis PoissonSiméon Denis Poisson (1781-1840), a French mathematician, developed the Poisson to estimate the probability of a number of occurrences over a specified interval of time or space. It’s said that the French Army used Poisson’s methods to predict the incidence of soldiers’ death by mule kicks. Tres debonair, no?

12. A. Properties of a Poisson Experiment 1. The probability of an occurrence is the same for any two intervals of equal length. 2. The occurrence or nonoccurrence in any interval is independent of the same in any other interval. Example: Experiment is the number of cars through a drive-thru window in 1 hour. Can you see how this experiment satisfies the conditions?

13. B. Poisson Probability Function f(x) is the probability of exactly x occurrences in an interval.  is the expected value or mean # of occurrences in an interval e is the ln(1) or approximately 2.71828.

14. C. Example with time intervals At a drive up window at a local bank, past experience tells us that the mean number of cars in a 15 minute period is  =10 cars. x is the random variable, # of cars in a 15 minute span. What does this have to do with mules?

15. Problems What is the probability of exactly 7 cars in any 15 minute span? How would this be useful?

16. Problems Find the probability of 1 car in 5 minutes. First we need to convert  =10 from a 15 minute span to a 5 minute span. If  =10 cars in 15 minutes,  =10/3=3.33 cars in 5 minutes.

17. D. Example with distance intervals Suppose a pipeline needs 1 repair every 100 miles, per year (  =1 per 100 miles). If 1000 miles of pipe are built, what is the probability that there will need to be 5 repairs in a year? If  =1 per 100 miles,  =10 per 1000 miles.