Presentation on theme: "Chapter 3. Discrete Probability Distributions"— Presentation transcript:
1Chapter 3. Discrete Probability Distributions 3.1 The Binomial Distribution3.2 The Geometric and Negative Binomial Distributions3.3 The Hypergeometric Distribution3.4 The Poisson Distribution3.5 The Multinomial Distribution
23.1 The Binomial Distribution 3.1.1 Bernoulli Random Variables(1/2) To model- the outcome of a coin toss,- whether a valve is open or shut,- whether an item is defective or not,- any other process that has only two possible outcomes.The outcomes are labeled 0 and 1The random variable is defined by the parameter p, 0 p 1, which is the probability that the outcome is 1.
33.1.1 Bernoulli Random Variables(2/2) Expectations
43.1.2 Definition of the Binomial Distribution(1/5) Consider an experiment consisting ofn Bernoulli trials (X1, ……, Xn)that are independent andthat each have a constant probability p of success.Then the total number of successes X, that is X=X1+…+Xn, is a random variable that has a binomial distribution with parameters n and p, which is writtenX~B(n,p)
53.1.2 Definition of the Binomial Distribution(2/5) The probability mass function of a B(n,p) random variable isfor , with
63.1.2 Definition of the Binomial Distribution(3/5) Ex) X~B(8,0.5)
73.1.2 Definition of the Binomial Distribution(4/5) Ex) X~B(8,0.5)0.0040.2730.2190.1090.03143215678xProbability432156780.0040.6360.8550.9650.9960.10000.3630.1440.035
83.1.1 Definition of the Binomial Distribution(5/5) Symmetric Binomial Distributions :A B(n,0.5) distribution is a symmetric probability distribution for any value of the parameter n. The distribution is symmetric about the expected value n/2.
9Example 24 : Air Force Scrambles(1/3) 16 planes.A probability of 0.25 that the engines of a particular plane does not start at a given attempt.Then, the number of planes successfully launched has a binomial distribution with parameters n = 16 and p = 0.75The expected number of plane launched is
10Example 24 : Air Force Scrambles(2/3) The variance isThe probability that exactly 12 planes scramble successfully isThe probability that at least 14 planes scramble successfully is
11Example 24 : Air Force Scrambles(3/3) 0.0000.0520.1800.2080.0540.00697513111315xProbability1086241214160.0010.0200.1100.1340.0100.225975131113151086241214160.0000.0790.3690.8020.9900.0070.0010.0270.1890.9360.1000.594
12Proportion of successes in Bernoulli Trials LetThen, if
133. 2 The Geometric and Negative Binomial Distributions 3. 2 3.2 The Geometric and Negative Binomial Distributions Definition of the Geometric Distribution(1/2)The number of trials up to and including the first success in a sequence of independent Bernoulli trials with a constant success probability p has a geometric distribution with parameter p.The probability mass function isfor
143.2.1 Definition of the Geometric Distribution(2/2) The cumulative distribution function isThe expectations
16Example 24 : Air Force Scrambles(1/2) If the mechanics are unsuccessful in starting the engines, then they must wait 5 minutes before trying again.The distribution of the number of attempts needed to start a plane’s engine geometric distribution with p = 0.75.The probability that the engines start on the third attempt is
17Example 24 : Air Force Scrambles(2/2) The probability that the plane is launched within 10 minutes of the first attempt to start the engines isThe expected number of attempts required to start the engines is
183.2.2 Definition of the Negative Binomial Distribution(1/2) The number of trials up to and including the r th success in a sequence of independent Bernoulli trials with a constant success probability p has a negative binomial distribution with parameter p and rThe probability mass function isfor
193.2.2 Definition of the Negative Binomial Distribution(2/2) The expectations
20Example 12 : Personnel Recruitment(1/2) Suppose that a company wishes to hire three new workers and each applicant interviewed has a probability of 0.6 of being found acceptable.The distribution of the total number of applicants that the company needs to interview Negative Binomial distribution with parameter p = 0.6 and r = 3.The probability that exactly six applicants need to be interviewed is
21Example 12 : Personnel Recruitment(2/2) If the company has a budget that allows up to six applicants to be interviewed, then the probability that the budget is sufficient isThe expected number of interviews required is
223. 3 The Hypergeometric Distribution 3. 3 3.3 The Hypergeometric Distribution Definition of the Hypergeometric Distribution(1/3)Consider a collection of N items of which r are of a certain kind.If one of the items is chosen at random, the probability that it is of the special kind is clearlyConsequently, if n items are chosen at random with replacement, then clearly the distribution of X, the number of defective items chosen, is
233.3.1 Definition of the Hypergeometric Distribution(2/3) However, if n items are chosen at random without replacement, then the distribution of X is the hypergeometric distribution.The hypergeometric distribution has a probability mass function given byfor
243.3.1 Definition of the Hypergeometric Distribution(3/3) The expectationsIt represents the distribution of the number of items of a certain kind in a random sample of size n drawn without replacement from a population of size N that contains r items of this kind.
25Let X and Y be independent binomial random variables such that Let us consider the conditional probability mass function of X given that X+Y=n.That is, the conditional distribution of Xgiven the value of X+Y is hypergeometric!
26Expectation of X:Let Xi be the following random variable:Xi=1 if the ith selection is acceptable; 0 otherwise.Then,Let X be the hypergeometric random variable with parameters (r,N.n). Then,This givesVariance of X:
27Since Xi is a Bernoulli random variable, Also for i<j,Hence,Cf. Let Then,andAs N goes to infinity,
28Example 17 : Milk Container Contents(1/3) Suppose that milk is shipped to retail outlets in boxes that hold 16 milk containers. One particular box, which happens to contain six under weight containers, is opened for inspection, and five containers are chosen at random.The distribution of the number of underweight milk containers in the sample chosen by inspector Hypergeometric distribution with N=16, r=6, and n=5.
29Example 17 : Milk Container Contents(2/3) The probability that the inspector chooses exactly two underweight containers isThe expected number of underweight containers chosen by the inspector is
30Example 17 : Milk Container Contents(3/3) 0.0580.0340.0020.2060.4120.28843215Number of underweight milk containers found by inspectorProbability
313. 4 The Poisson Distribution 3. 4 3.4 The Poisson Distribution Definition of the Poisson Distribution(1/3)The distribution ofThe number of defects in an itemThe number of radioactive particles emitted by a substanceThe number of telephone calls received by an operator with a certain time limitThat is, the number of “events” that occur within certain specified boundaries.
323.4.1 Definition of the Poisson Distribution(2/3) A random variable X distributed as a Poisson random variable with parameter λ, which is writtenhas a probability mass functionfor
333.4.1 Definition of the Poisson Distribution(3/3) The Poisson distribution is often useful to model the number of times that a certain event occurs per unit of time, distance, or volume, and it has a mean and variance both equal to the parameter value λ.The expectations
34The Poisson random variable can be used as an approximation for a binomial random variables with parameters (n,p) when n is large and p is small:Let X be a binomial random variable with parameters (n,p) andThen,That is,For large n and small p,Hence,
35Example 3 : Software Errors(1/3) Suppose that the number of errors in a piece of software has a Poisson distribution with parameter λ=3.[The expected number of errors]= [The variance in the number of errors] = 3.The probability that a piece of software has no error is
36Example 3 : Software Errors(2/3) The probability that there are three or more errors in a piece of software is
37Example 3 : Software Errors(3/3) 0.1010.0500.1680.0010.008975131113Number of software errorsProbability108624120.0220.0030.1490.224975131113108624120.1010.0500.1680.0010.0080.0220.0030.1490.2240.000
383. 5 The Multinomial Distribution 3. 5 3.5 The Multinomial Distribution Definition of the Multinomial Distribution(1/2)Consider a sequence of n independent trials where each individual trial can have k outcomes that occur with constant probability value p1,…, pk with p1+···+pk = 1. The random variable X1,…, Xk that count the number of occurrences of each outcome are said to have a multinomial distribution.The joint probability mass function isfor nonnegative integer values of the satisfying
393.5.1 Definition of the Multinomial Distribution(2/2) The random variables X1,…, Xk have expectation and variances given bybut they are not independent. (why?)
40Example 1 : Machine Breakdowns(1/4) Suppose that the machine breakdowns are attributable to electrical faults, mechanical faults, and operator misuse, and these causes occur with probabilities of 0.2, 0.5, and 0.3, respectively.The engineer is interested in predicting the causes of the next ten breakdowns.X1: the number of breakdowns due to electrical reasons.X2: the number of breakdowns due to mechanical reasons.X3: the number of breakdowns due to operator misuse.
41Example 1 : Machine Breakdowns(2/4) X1+X2+X3=10If the breakdown causes can be assumed to be independent of one another, then the probability mass function isThe probability that there will be three electrical breakdowns, five mechanical breakdowns, and two misuse breakdowns is
42Example 1 : Machine Breakdowns(3/4) The expected number of electrical breakdowns isThe expected number of mechanical breakdowns isThe expected number of misuse breakdowns is
43Example 1 : Machine Breakdowns(4/4) If the engineer is interested in the probability of there being no more two electrical breakdowns, then this can be calculated by nothing that X1~B(10, 0.2), so that