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CHAPTER 3 DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS LEARNING OBJECTIVES • Understand random variables • Calculate means and variances • Determine probabilities from cumulative distribution functions • Understand the assumptions of the discrete probability distributions • Calculate probabilities, determine means and variances for each of the discrete probability distributions Concept of Random Variable • Summarize the outcome from a random experiment by a simple number • Used to describe the possible outcomes • Useful to associate a number with each outcome in the sample space • Variable that associates a number with the outcome of a random experiment is referred to as a random variable Definition of Random Variables • Function that assigns a real number to each outcome in the sample space • Denoted by an uppercase letter such as X • Value of the random variable is denoted by a lowercase letter such as x Two Types of Random Variables • Discrete and Continuous • Sometimes a measurement can assume any value in an interval of real numbers • Said to be a continuous random variable – Examples: Electrical current, length, pressure, temperature, time, voltage, weight • Sometimes the measurement is limited to integers • Said to be a discrete random variable – Examples: No. of scratches on a surface, No. of calls received per day, and No. of transmitted bits received in error Probability Distribution • Describes probabilities associated with the possible values of X • Discrete Case – Specified by just a list of the possible values along with the probability of each • Continuous Case – Used to describe the probability distribution • Convenient to express the probability in terms of a formula Discrete Probability Distribution • Definition – For a discrete random variable X with possible values x1, x2, x3, …, xn, the probability mass function (PMF) is • f(xi) = P(X = xi) • f(xi) 0 for all xi • f (x ) 1 x i Discrete Probability Distribution Examples • To better understand the PMF, consider • Example 1 – Tossing a 6-sided die – f(x)=1/6 for X=1,2,3,4,5,6 • Example 2 – Check whether the following can serve as probability distribution – 1. f(x)= (x-2)/2 for x=1,2,3,4; – 2. h(x)= x2/25 for x=0,1,2,3,4; Class Problem • • The sample space of a random experiment is {a,b,c,d,e,f}, and each outcome is equally likely. A random variable is defined as follows: Outcome a b x 0 0 c d 1.5 1.5 e f 2 3 Determine the probability mass function of X a) b) c) f(0)=P(X=0)=1/6+1/6=1/3 f(1.5)=P(X=1.5)=1/6+1/6=1/3 f(3)=P(X=3)=1/6 Cumulative Distribution Function • Useful to provide cumulative probabilities such as P(Xx) • Cumulative distribution function (CDF) of a discrete random variable X, denoted as F(x), is F ( x) P( X x) f ( x ) i x x i • F(x) has the following properties: – F(x)= P(X x)= f ( x ) xi x i – 0 F(x) 1 – If x y, then F(x) F(y) Cumulative Distribution FunctionExample • Determine the cumulative distribution of the random variable for which each outcome is equally likely in the following sample space: Outcome a x 0 b 0 c d e f 1.5 1.5 2 3 Solution Outcome a x f(x) b 0 0 1/6 1/6 c 1.5 1.5 1/6 0, x 0 1 , 0 x 1.5 3 2 F ( x) , 1.5 x 2 3 5 , 2 x 3 6 3 x 1, d 1/6 e f 2 3 1/6 1/6 Calculating Probabilities from PMF and CDF • Interested to determine – Probabilities from cumulative distribution functions – Cumulative distribution functions from probability mass functions – Vise versa • Consider the following example Example • Consider the following PMF f(y) y 1 2 3 4 f(y) .4 .3 .2 .1 0.4 y • Determine the CDF 0, 0.4, F ( y ) 0.7, 0.9 1 y 1 1 y 2 1 F(y) 2 y 3 3 y 4 4 y y 1 Example • Consider the following CDF 0, 0.4, F ( y ) 0.7, 0.9 1 y 1 1 y 2 2 y 3 3 y 4 4 y • Note that Y can take on 1, 2, 3, or 4. • Determine the PMF – P(Y=1)=0.4-0.0=0.4 – P(Y=2)=0.7-0.4=0.3, P(Y=3)=0.9-0.7=0.2, P(Y=4)=1.0-0.9=0.1 – Hence, y 1 2 3 4 f(y) .4 .3 .2 .1 Mean of a Discrete Random Variable • PMF provides complete information about the properties of a random variable • Useful to have some summary measures of these properties • Mean or expectation – Denoted by E(X) or represents an average value of the random variable • If a random variable takes on the values x1, x2,… , or xk with the probabilities f(x1), f(x2), …., and f(xk), its mean is = E(X)= x1.f(x1) + x2.f(x2)+ ...+ xk.f(xk) – Or = E(X)= xf ( x) x Examples • Find the mean of the probability distribution of the number of heads obtained in three flips of a coin – Probabilities • 1/8, 3/8, 3/8, and 1/8 • = (0)1/8 + (1)3/8 + (2)3/8 + (3) 1/8 = 3/2 • Suppose Y is the total showing on a pair of dice, find the mean of the probability distribution – Calculated as E[Y] 2( 1 ) 3( 2 ) 4( 3 ) 36 36 36 5( 4 ) 6( 5 ) 7( 6 ) 8( 5 ) 9( 4 ) 10( 3 ) 11( 2 ) 12( 1 ) 7 36 36 36 36 36 36 36 36 Variance of a Discrete Random Variable • Another important measure of the distribution of a random variable • Measures the spread or variability • Whereas mean measures central, the variance measures the deviation • Denoted as 2 or V(x) 2= V(X)=E(X-)2 = ( x )2 f ( x ) x • Standard deviation of X is =[V(X)]1/2 Examples • Find the variance of the probability distribution of the number of heads obtained in three flips of a coin – Variance 2= (0-3/2)2(1/8)+(1-3/2)2 (3/8)+(2-3/2)2(3/8)+(33/2)2(1/8)=0.75 • Find the variance of the total showing on a pair of dice – Variance 2 = V(X)= (2-7)2(1/36) + (3-7)2(2/36)+ … + (12-7)2(1/36) = Class Problem • If the range of X is the set {0, 1, 2, 3, 4} and P(X=x)= 0.2, determine the mean and variance of the random variable • Solution – X can take values of 0,1, 2, 3 or 4 – Summation of x f(x) – Hence, mean=2 – Variance can be calculated as – V(X)= (0-2)2 (0.2)+(1-2)2 (0.2)+…= Uniform Discrete Distribution • If X assumes the values x1,x2,….,xn with equal probability, then it has a discrete uniform distribution • Probability mass function f(xi)= 1/n for xi= x1,x2,….,xn n • Mean xi i 1 E[X]= n • Variance n 2 V(X)= (x i ) i 1 Examples • When a light bulb is selected randomly from a box containing a red, a blue, a white, and a yellow bulb – Sample space is {red, blue, white, yellow} with probability 1/4 – Hence, f(x)=1/4 • When a die is tossed – – – – Sample space is {1,2,3,4,5,6} with probability 1/6 f(x)=1/6 E[X]= (1*1/6+ 2*1/6+ 3*1/6+ 4*1/6+ 5*1/6+6*1/6)=3.5 V[X]=(1-3.5)2/6 + (2-3.5)2/6 + …+ (6-3.5)2/6 =35/12 Binomial Distribution • Two possible outcomes labeled success or failure • Referred to as a Bernoulli process • Conditions – Consists of n repeated trails – Results in two possible outcomes (Success or Failure) – Probability of success, denoted by p, remains constant – Trials are independent Examples • Tossing a coin 10 times 1. 2. 3. 4. • There are 10 trials, and they are identical Each trial has only two outcomes Probability of getting a H is 0.5 in each trial Trials (tosses) are independent In an operation, 5% of all machined parts are defective. 3 parts are randomly selected from the production line to determine if each of them is defective or good 1. 2. 3. 4. Three identical trials Each trial has two outcomes p=0.05 Independent trials PMF of the Binomial Distribution • Binomial distribution with parameters p and n=1,2,… • PMF n x – f(x)= p (1-p)n-x x – Where x =0,1, 2,…,n. n n! – x ( n x )! x! • E[X]= np • V (X)=np(1-p)=npq Example • In an operation, 5% of all parts machined by a firm are defective • If three parts are randomly selected from the production line, what is the probability that exactly one of them will be defective? • Solution – n=3, x=1, and p=0.05 – Substituting in the PMF 3 P(X=1)= f(1) = 1 0.051(1-0.05)3-1 P(X=1)=3*0.051(1-0.05)3-1 = 0.1354 Class Problem • The random variable X has a binomial distribution with n=10 and p=0.01, determine the following probabilities • P(X=5) 10 = (0.01)5(1-0.01)10-5) = 2.4 x10 8 5 • P(X2) =P(X=0)+P(X=1)+P(X=2)=0.904 +0.091+0.00045 =0.99 • P(3X<5) 4 =P(X=3)+ P(X=4)=1.14 x10 Geometric Distribution • • • • Closely related to the binomial experiment Trials are conducted until a success is obtained Let X denote the number of trials PMF – P(X = x) = f(x) = (1-p)x-1p x = 1,2,3,… • Mean and Variance – =E(x)=1/p and 2=V(x)=(1-p)/p2 Example • The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8 • Assume the trials are independent – What is the probability the first successful alignment requires exactly four trials? – What is the probability that the first successful alignment requires at most four trials? – What is the probability that the first successful alignment requires at least four trials? Example • Let X denote the number of trials to obtain in the first successful alignment. • Then X is a geometric random variable with p = 0.8 • Solution P(X = 4) = f(4) = 0.23(0.8) = 0.0064 P(X 4) = P(X=1) + P(X = 2) + P(X =3) + P(X = 4)=0.9984 P(X 4) = 1 P(X < 4) = 1 0.992 = 0.008 Class Problem • Assume that each of your calls to a popular radio station has a probability of 0.02 of connecting, that is, of not obtaining a busy signal (calls are independent) – What is the probability that your first call that connects is your tenth call? – What is the probability that it requires more than five calls for you to connect? – What is the mean number of calls needed to connect? Solutions • Let X denote the number of calls needed to obtain a connection • Then, X is a geometric random variable with p=0.02 • P(X = 10) = (1 0.02)9 0.02 0.9890.02 0.0167 • P(X>5) =1 P( X 4) 1 [ P( X 1) P( X 2) P( X 3) P( X 4)] 1 [0.02 0.98(0.02) .982 (0.02) 0.983 (0.02)] 1 0.0776 0.9224 • E(X) =1/0.02 = 50 Negative Binomial Distribution • Based on an experiment – Consists of a sequence of independent trials – Result in either a “S” or “F” – Probability of success is constant from trial to trial – Continues until a total of r success have been observed • PMF x 1 P(X = x) = f(x) = r 1 (1-p)x-rpr • Mean and Variance E[X] = r/p and V (X) = r(1-p)/p2 Example • Suppose that X is a negative binomial random variable with p=0.2 and r=4. Determine the following E(X), P(X = 20), P(X = 19), P(X = 21) • Solution • E(X) = r/p =4/0.2 = 20 19 x 1 16 4 x-r r ( 0 . 80 ) 0 . 2 0.0436 • P(X = 20) = r 1 (1-p) p = 3 18 15 4 ( 0 . 80 ) 0 . 2 0.0459 • P(X = 19) = 3 20 17 4 ( 0 . 80 ) 0 . 2 0.0411 • P(X = 21) = 3 Hypergeometric Distribution • Assumptions – Population consists of N objects (finite) – Each classified as a “S” or “F” and K success – n individuals is selected without replacement • Random variable of interest is X, the number of successes in the sample PMF of the Hypergeometric Distribution • X=number of successes in a sample of size n drawn from a population consisting of K successes and N-K failures • PMF is given K N K x n x f ( x) N n – for x satisfying max (0, n-N+K) x min (n,K) • Mean and Variance K and V (X) = n K N k N n E[X] = n N N N N 1 Example • A shipment of 20 machined parts contains 5 that are defective • If 10 of them are randomly chosen for inspection, what is the probability that 2 of the 10 will be defective? • Solution x = 2, n = 10, k = 5, and N=20 5 20 5 2 10 2 f(x=2)= = 0.348 20 10 Class Problem • A lot of 75 washers contains 5 in which the variability in thickness around the circumference of the washer is unacceptable • A sample of 10 washers is selected at random, without replacement. – What is the probability that none of the unacceptable washers is in the sample? – What is the probability that at least one unacceptable washer is in the sample? – What is the probability that exactly one unacceptable washer is in the sample? – What is the mean number of unacceptable washers in the sample? Solution • Let X denote the number of unacceptable washers in the sample of 10 • P(X = 0) K N K x n x N n 5 75 5 0 10 0 0.4786 75 10 • P(X 1) =1- P(X=0)=0.5214 • P(X = 1)= 5 75 5 1 10 1 0.3923 75 10 • E(X) =10*5/75=2/3 Poisson Distribution • Binomial, hypergeometric, and negative binomial distributions start with an experiment consisting of trials • Based on the number of outcomes occurring during a given time interval or in a specified regions • Examples – # of accidents that occur on a given highway during a 1week period – # of customers coming to a bank during a 1-hour interval – # of TVs sold at a department store during a given week – # of breakdowns of a washing machine per month Conditions • Consider the # of breakdowns of a washing machine per month example – Each breakdown is called an occurrence – Occurrences are random that they do not follow any pattern (unpredictable) – Occurrence is always considered with respect to an interval (one month) The Probability Mass Distribution • X = number of counts in the interval • Poisson random variable with > 0 x • PMF e f(x)= x=0,1,2, x! • Mean and Variance E[X] = , V (X) = Example • If a bank gets on average = 6 bad checks per day, what are the probabilities that it will receive four bad checks on any given day?10 bad checks on any two consecutive days? • Solution x = 4 and = 6, then f(4) = 6 4 e 6 = 0.135 4! e 12 1210 = 12 and x = 10, then f(10) = = 0.105 10! Class Problem • The number of failures of a testing instrument from contamination particle on the product is a Poisson random variable with a mean of 0.02 failure per hour. – What is the probability that the instrument does not fail in an 8-hour shift? – What is the probability of at least one failure in one 24hour day? Solution a) Let X denote the failure in 8 hours. Then, X has a Poisson distribution with =0.16 P(X=0)=0.8521 b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with =0.48 P(Y1) = 1-P(Y = 0) =0.3812 The Poisson Approximation To The Binomial Distribution • When n is large and p is small, binomial probabilities are often approximates by e x f(x)= x! for x = 0,1,2, … with = np Example • Assume 5% of the books at a certain bindery have defective bindings. Find the probability that 2 of 100 books bound by this bindery will have defective bindings, using – The binomial distribution – The Poisson distribution Solution • Using Binomial x = 2, n = 100, and p = 0.05 f(x) = 100 (0.05)2(0.95)98 = 0.081 2 • Using Poisson x = 2, = np = 100*0.05 = 5 e x f(2) = x! = 0.084 Next agenda • Continue our development of probability distributions with a discussion of several important continuous distributions