# Binomial and Poisson Probability Distributions

## Presentation on theme: "Binomial and Poisson Probability Distributions"— Presentation transcript:

Binomial and Poisson Probability Distributions
Binomial Probability Distribution l Consider a situation where there are only two possible outcomes (a “Bernoulli trial”) Example: u flipping a coin The outcome is either a head or tail u rolling a dice For example, 6 or not 6 (i.e. 1, 2, 3, 4, 5) Label the possible outcomes by the variable k We want to find the probability P(k) for event k to occur Since k can take on only 2 values we define those values as: k = 0 or k = 1 u Let the probability for outcome “k” to occur be: P(k = 0) = q (remember 0 ≤ q ≤ 1) u something must happen so P(k = 0) + P(k = 1) = (mutually exclusive events) P(k = 1) = p = 1 - q u We can write the probability distribution P(k) as: P(k) = pkq1-k (Bernoulli distribution) u coin toss: define probability for a head as P(1) P(k=1) = 0.5 and P(0=tail) = 0.5 too! u dice rolling: define probability for a six to be rolled from a six sided dice as P(k=1) P(k=1) = 1/6 and P(k=0=not a six) = 5/6. James Bernoulli (Jacob I) born in Basel, Switzerland Dec. 27, 1654-Aug. 16, 1705 He is one 8 mathematicians in the Bernoulli family. (from Wikipedia) R.Kass/Sp12 P416 Lecture 2

Mean and Variance of a discrete distribution
l What is the mean () of P(k)? l What is the Variance (2) of P(k)? l Let’s do something more complicated: Suppose we have N trials (e.g. we flip a coin N times) what is the probability to get m successes (= heads)? l Consider tossing a coin twice. The possible outcomes are: no heads: P(m = 0) = q2 one head: P(m = 1) = qp + pq (toss 1 is a tail, toss 2 is a head or toss 1 is head, toss 2 is a tail) = 2pq two heads: P(m = 2) = p2 Note: P(m=0)+P(m=1)+P(m=2)=q2+ qp + pq +p2= (p+q)2 = 1 (as it should!) l We want the probability distribution P(m, N, p) where: m = number of success (e.g. number of heads in a coin toss) N = number of trials (e.g. number of coin tosses) p = probability for a success (e.g. 0.5 for a head) Mean and Variance of a discrete distribution (remember p+q=1) we don't care which of the tosses is a head so there are two outcomes that give one head R.Kass/Sp12 P416 Lecture 2

number of “two color” combinations
l If we look at the three choices for the coin flip example, each term is of the form: CmpmqN-m m = 0, 1, 2, N = 2 for our example, q = 1 - p always! coefficient Cm takes into account the number of ways an outcome can occur without regard to order. for m = 0 or 2 there is only one way for the outcome (both tosses give heads or tails): C0 = C2 = 1 for m = 1 (one head, two tosses) there are two ways that this can occur: C1 = 2. l Binomial coefficients: number of ways of taking N things m at time 0! = 1! = 1, 2! = 1·2 = 2, 3! = 1·2·3 = 6, m! = 1·2·3···m Order of occurrence is not important u e.g. 2 tosses, one head case (m = 1) n we don't care if toss 1 produced the head or if toss 2 produced the head Unordered groups such as our example are called combinations Ordered arrangements are called permutations For N distinguishable objects, if we want to group them m at a time, the number of permutations: u example: If we tossed a coin twice (N = 2), there are two ways for getting one head (m = 1) u example: Suppose we have 3 balls, one white, one red, and one blue. n Number of possible pairs we could have, keeping track of order is 6 (rw, wr, rb, br, wb, bw): n If order is not important (rw = wr), then the binomial formula gives number of “two color” combinations R.Kass/Sp12 P416 Lecture 2

l Binomial distribution: the probability of m success out of N trials:
p is probability of a success and q = 1 - p is probability of a failure l To show that the binomial distribution is properly normalized, use Binomial Theorem: Þ binomial distribution is properly normalized R.Kass/Sp12 P416 Lecture 2

Mean of binomial distribution:
A cute way of evaluating the above sum is to take the derivative: Variance of binomial distribution (obtained using similar trick): R.Kass/Sp12 P416 Lecture 2

Example: Suppose you observed m special events (success) in a sample of N events
u The measured probability (“efficiency”) for a special event to occur is: What is the error (standard deviation) on the probability ("error on the efficiency"): The sample size (N) should be as large as possible to reduce the uncertainty in the probability measurement. Let’s relate the above result to Lab 2 where we throw darts to measure the value of p. If we inscribe a circle inside a square with side=s then the ratio of the area of the circle to the rectangle is: we will derive this later in the course r s So, if we throw darts at random at our rectangle then the probability (ºe) of a dart landing inside the circle is just the ratio of the two areas, p/4. The we can determine p using: p= 4e. The error in p is related to the error in e by: We can estimate how well we can measure p by this method by assuming that e=p/4= ( …)/4: This formula “says” that to improve our estimate of p by a factor of 10 we have to throw 100 (N) times as many darts! Clearly, this is an inefficient way to determine p. R.Kass/Sp12 P416 Lecture 2

Pete Rose’s lifetime batting average: 0.303
Example: Suppose a baseball player's batting average is (3 for 10 on average). u Consider the case where the player either gets a hit or makes an out (forget about walks here!). prob. for a hit: p = 0.30 prob. for "no hit”: q = 1 - p = 0.7 u On average how many hits does the player get in 100 at bats? = Np = 100·0.30 = 30 hits u What's the standard deviation for the number of hits in 100 at bats?  = (Npq)1/2 = (100·0.30·0.7)1/2 ≈ 4.6 hits we expect ≈ 30 ± 5 hits per 100 at bats u Consider a game where the player bats 4 times: probability of 0/4 = (0.7)4 = 24% probability of 1/4 = [4!/(3!1!)](0.3)1(0.7)3 = 41% probability of 2/4 = [4!/(2!2!)](0.3)2(0.7)2 = 26% probability of 3/4 = [4!/(1!3!)](0.3)3(0.7)1 = 8% probability of 4/4 = [4!/(0!4!)](0.3)4(0.7)0 = 1% probability of getting at least one hit = 1 - P(0) = =76% Pete Rose’s lifetime batting average: 0.303 R.Kass/Sp12 P416 Lecture 2

Poisson Probability Distribution
l The Poisson distribution is a widely used discrete probability distribution. l Consider the following conditions: p is very small and approaches 0 u example: a 100 sided dice instead of a 6 sided dice, p = 1/100 instead of 1/6 u example: a 1000 sided dice, p = 1/1000 N is very large and approaches ∞ example: throwing 100 or 1000 dice instead of 2 dice The product Np is finite l Example: radioactive decay Suppose we have 25 mg of an element very large number of atoms: N ≈ 1020 (avogadro’s number is large!) Suppose the lifetime of this element t = 1012 years ≈ 5x1019 seconds probability of a given nucleus to decay in one second is very small: p = 1/t = 2x10-20/sec BUT Np = 2/sec finite! The number of decays in a time interval is a Poisson process. l Poisson distribution can be derived by taking the appropriate limits of the binomial distribution Siméon Denis Poisson June 21, 1781-April 25, 1840 uradioactive decay unumber of Prussian soldiers kicked to death by horses per year ! uquality control, failure rate predictions N>>m R.Kass/Sp12 P416 Lecture 2

The mean and variance of a Poisson distribution are the same number!
u m is always an integer ≥ 0 u does not have to be an integer It is easy to show that:  = Np = mean of a Poisson distribution 2 = Np =  = variance of a Poisson distribution l Radioactivity example with an average of 2 decays/sec: i) What’s the probability of zero decays in one second? ii) What’s the probability of more than one decay in one second? iii) Estimate the most probable number of decays/sec? u To solve this problem its convenient to maximize lnP(m, ) instead of P(m, ). The mean and variance of a Poisson distribution are the same number! R.Kass/Sp12 P416 Lecture 2

If you observed m events in a “counting” experiment, the error on m is
u In order to handle the factorial when take the derivative we use Stirling's Approximation: The most probable value for m is just the average of the distribution u This is only approximate since Stirlings Approximation is only valid for large m. u Strictly speaking m can only take on integer values while  is not restricted to be an integer. If you observed m events in a “counting” experiment, the error on m is ln10!= ln10-10=13.03 ®14% ln50!= ln50-50=145.60®1.9% R.Kass/Sp12 P416 Lecture 2

Comparison of Binomial and Poisson distributions with mean m = 1
Not much difference between them! N N For N large and m fixed: Binomial Þ Poisson R.Kass/Sp12 P416 Lecture 2

Uniform distribution and Random Numbers
What is a uniform probability distribution: p(x)? p(x)=constant (c) for a £ x £b p(x)=zero everywhere else Therefore p(x1)dx1= p(x2)dx2 if dx1=dx2 Þ equal intervals give equal probabilities For a uniform distribution with a=0, b=1 we have p(x)=1 1 p(x) What is a random number generator ? A number picked at random from a uniform distribution with limits [0,1] All major computer languages (C, C++) come with a random number generator. In C++ rand() gives a random number. The following C++ program generates 5 random numbers: #include<iostream> #include<math.h> #include<stdlib.h> // compile and link using: c++ -o abc abc.C -lm with source code named abc.C int main() { // program to calculate and print out some random numbers int K, points; srand(1); //initial random number generator points=4; for(K=0;K<points;K++) std::cout<<" K "<<K<<" random number "<<rand()/(1.0*RAND_MAX)<<"\n"; } return 0; x 1 If we generate “a lot” of random numbers all equal intervals should contain the same amount of numbers. For example: generate: 106 random numbers expect: 105 numbers [0.0, 0.1] 105 numbers [0.45, 0.55] K= 0 random number K= 1 random number K= 2 random number K= 3 random number K= 4 random number R.Kass/Sp12 P416 Lecture 2

A C++ program to throw dice
#include<iostream> #include<cstdio> #include<math.h> #include<stdlib.h> // compile and link on unix using: cxx -o abc abc.C -lm // where abc.C is the file that contains the source code int main() { // program to roll dice int numroll;// number of rolls int K,dice; float roll[7]; //keeps track of how often a 1,2..6 occurs float prob; // probability of rolling a 1,2,3..6 // srand(1); //initialize random number generator std::cout<<"give number of rolls of the dice "<<"\n"; std::cin>>numroll; std::cout<<"number of rolls of the dice "<< numroll<<"\n"; for(K=0;K<=6;K++) { roll[K]=0; } for(K=0;K<numroll;K++) // % is the modulus operator dice =1+rand()%6; roll[dice]++; // cout<<"dice "<<dice1<<"\n"; } for(K=1;K<=6;K++) prob=roll[K]/numroll; std::cout<<"roll of dice "<<K<<" number "<<roll[K]<<" prob "<<prob<<"\n"; return 0; R.Kass/Sp12 P416 Lecture 2

Download ppt "Binomial and Poisson Probability Distributions"

Similar presentations