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Chapter 17 Probability Models math2200. I don’t care about my [free throw shooting] percentages. I keep telling everyone that I make them when they count.

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Presentation on theme: "Chapter 17 Probability Models math2200. I don’t care about my [free throw shooting] percentages. I keep telling everyone that I make them when they count."— Presentation transcript:

1 Chapter 17 Probability Models math2200

2 I don’t care about my [free throw shooting] percentages. I keep telling everyone that I make them when they count. -- Shaquille O'Neal, in post-game interviews recorded by WOAI-TV on November 7, 2003.

3 O’Neal’s free throws Suppose Shaq shoots 45.1% on average. Let X be the number of free throws Shaq needs to shoot until he makes one. Pr (X=2)=? Pr (X=5)=? E(X)=?

4 Bernoulli trials Only two possible outcomes –Success or failure Probability of success, denoted by p the same for every trial The trials are independent Examples –tossing a coin –Free throw in a basketball game assuming every time the player starts all over.

5 Can we model drawing without replacement by Bernoulli trials? The draws are not independent when sampling without replacement in finite population. But they are be treated as independent if the population is large –Rule of thumb: the sample size is smaller than 10% of the population

6 Geometric model How long does it take to achieve a success in Bernoulli trials? A Geometric probability model tells us the probability for a random variable that counts the number of Bernoulli trials until the first success Geom(p) –p = probability of success –q = 1-p = probability of failure –X : number of trials until the first success occurs –P (X=x) = q x-1 p –E (X) = 1/p –Var (X) = q/p 2

7 What is the probability that Shaq makes his first free throw in the first four attempts? 1-P(NNNN) = 1-(1-0.451) 4 = 0.9092 or P(X=1)+P(X=2)+P(X=3)+P(X=4)

8 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.

9 The Binomial Model (cont.) There are ways to have k successes in n trials. –Read n C k as “n choose k.”

10 The Binomial Model (cont.) Binom (n, p) n = number of trials p = probability of success q = 1 – p = probability of failure X = number of successes in n trials

11 How do we find E(X) and Var(X)? Find P(X=x) directly Binomial random variable can be viewed as the sum of the outcome of n Bernoulli trials –Let Y1,…, Yn be the outcomes of n Bernoulli trials –E (Y1) =…= E (Yn) = p*1+q*0=p, E(X) = np –Var (Y1) =…= Var (Yn) = (1-p) 2 *p+(0-p) 2 *q = pq, Var (X) = npq. In general if Y1,…,Yn are independent and have the same mean µ and variance σ 2 and X = Y1+…+Yn, then E(X) = E(Y1)+…+E(Yn)=nµ and Var(X) = Var(Y1)+…+Var(Yn)=nσ 2.

12 Mean and variance of sum Suppose Y1,…,Yn are independent and have the same mean µ and variance σ 2 Let X = Y1+…+Yn E(X) = E(Y1)+…+E(Yn)=nµ Var(X) = Var(Y1)+…+Var(Yn)=nσ 2

13 If Shaq shoots 20 free throws, what is the probability that he makes no more than two? Binom(n, p), p=0.451, n=20 P(X=0 or 1 or 2) = P(X=0) + P(X=1) + P(X=2) = 0.0009

14 Normal approximation to Binomial If X ~ Binomial (n, p), n =10000, p =0.451, P(X<2000)=? When Success/failure condition (np >= 10 and nq>=10) is satisfied, Binomial (n,p) can be approximated by Normal with mean np and variance npq. P (X<2000)=P ( Z< (2000-np) / sqrt (npq)) = P(Z< -50.4428) =normalcdf (-1E99, -50.4428,0,1) = 0

15 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. 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.

16 Poisson model Binomial(n,p) is approximated by Poisson(np) if np<10. Let λ=np, we can use Poisson model to approximate the probability. Poisson(λ) –λ : mean number of occurrences –X: number of occurrences

17 Poisson Model (cont.) The Poisson model is also used directly to model the probability of the occurrence of events. It scales to the sample size –The average occurrence in a sample of size 35,000 is 3.85 –The average occurrence in a sample of size 3,500 is 0.385 Occurrence of the past events doesn’t change the probability of future events.

18 An application of Poisson model In 1946, the British statistician R.D. Clarke studied the distribution of hits of flying bombs in London during World War II. Were targeted or due to chance.

19 Flying bomb (cont’) The average number of hits per square is then 537/576=.9323 hits per square. Given the number of hits following a Poisson Model P (X=0) = [e^(-0.923)*(-0.923)^0] / 0! = 0.393647 0.393647* 576 = 226.7 No need to move people from one sector to another, even after several hits! # of hits012345 # of cells with # of hits above229211933571 Poisson Fit226.7211.498.530.67.11.6

20 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.

21 What have we learned? –Geometric model When we’re interested in the number of Bernoulli trials until the first success. –Binomial model When we’re interested in the number of successes in a fixed number of Bernoulli trials. –Normal model To approximate a Binomial model when we expect at least 10 successes and 10 failures. –Poisson model To approximate a Binomial model when the probability of success, p, is very small and the number of trials, n, is very large.

22 TI-83 2 nd + VARS (DISTR) pdf: P(X=x) when X is a discrete r.v. –geometpdf(prob,x) –binompdf(n,prob,x) –poissonpdf(mean,x) cdf: P(X<=x) –geometcdf(prob,x) –binomcdf(n,prob,x) –poissoncdf(mean,x)


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