Chapter 3: Random Variables and Probability Distributions Definition and nomenclature A random variable is a function that associates a real number with each element in the sample space. We use an uppercasel letter such as X to denote the random variable. We use a lowercase letter such as x for one of its values. Example: Consider a random variable Y which takes on all values y for which y > 5. JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Defining Probabilities: Random Variables Examples: Out of 100 heart catheterization procedures performed at a local hospital each year, the probability that more than five of them will result in complications is P(X > 5) Drywall anchors are sold in packs of 50 at the local hardware store. The probability that no more than 3 will be defective is P(Y < 3) JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Discrete Random Variables Exercise 2.51 P.59 (Modified) A box contains 500 envelopes (75 have $100, 150 have $25, 275 have $10). Someone spends $75 to buy 3 envelopes. The individual opens each envelope to see if the envelope contains a $10 bill. The sample space describing the presence of $10 bills (H) vs. bills that are not $10 (N) is: S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH} The random variable associated with this situation, X, reflects the outcome of the experiment. X is the number of envelopes that contain $10 X = {0, 1, 2, 3} Why no more than 3? Why 0? Note: if the number of possible solutions is countable, the variable is discrete S = {NNN, NNH, NHN, HNN, NHH, HNH, HHN, HHH} X = {0, 1, 2, 3} the probability distribution function (x, f(x)) – see definition 3.4, pg. 84 JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Discrete Probability Distributions 1 The probability that an envelope contains a $10 bill is 275/500 or .55 What is the probability that there are no $10 bills in the group of three envelopes? P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125 What is the probability that only one envelope contains a $10 bill? P(X = 1) = 3 * (0.55)*(1-0.55)* (1-0.55) = 0.334125 Why 3 for the X = 1 case? Three items in the sample space for X = 1 NNH NHN HNN P(X=0) = P(not in the 1st envelope ∩ not in the 2nd ∩ not in the 3rd) = (1-275/500)3 = (0.45)3 = 0.09112 P(0) =(1-0.55)^3 = 0.091125 P(1) =3*((0.55)*(1-0.55)^2) = 0.334125 P(2) =3*(0.55^2*(1-0.55)) = 0.408375 P(3) = 0.55^3 = 0.166375 JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Discrete Probability Distributions 2 Summary of probabilities for all possible outcomes P(X = 0) =(1-0.55) * (1-0.55) *(1-0.55) = 0.091125 P(X = 1) = 3*(0.55)*(1-0.55)* (1-0.55) = 0.334125 P(X = 2) = 3*(0.55^2*(1-0.55)) = 0.408375 P(X = 3) = 0.55^3 = 0.166375 The probability distribution associated with the number of $10 bills is given by: P(X=0) = P(not in the 1st envelope ∩ not in the 2nd ∩ not in the 3rd) = (1-275/500)3 = (0.45)3 = 0.09112 P(0) =(1-0.55)^3 = 0.091125 P(1) =3*((0.55)*(1-0.55)^2) = 0.334125 P(2) =3*(0.55^2*(1-0.55)) = 0.408375 P(3) = 0.55^3 = 0.166375 x 1 2 3 P(X = x) JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Another View The probability histogram JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Another Discrete Probability Example Given: A shipment consists of 8 computers 3 of the 8 are defective Experiment: Randomly select 2 computers Definition: random variable X = # of defective computers selected What is the probability distribution for X? Possible values for X: X = 0 X =1 X = 2 Let’s start with P(X=0) [0 defectives and 2 nondefectives are selected] Recall that P = specified target / all possible (all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers) P(X = 0) = P(0 defectives and 2 nondefective) = (all ways to get 0 out of 3 defectives) ∩ (all ways to get 2 out of 5 nondefective) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers) JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Discrete Probability Example What is the probability distribution for X? Possible values for X: X = 0 X =1 X = 2 Let’s calculate P(X=1) [1 defective and 1 nondefective are selected] (all ways to get 1 out of 3 defectives) ∩ (all ways to get 1 out of 5 nondefectives) (all ways to choose 2 out of 8 computers) (all ways to choose 2 out of 8 computers) x 1 2 P(X = x) JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Discrete Probability Distributions The discrete probability distribution function (pdf) f(x) = P(X = x) ≥ 0 Σx f(x) = 1 The cumulative distribution, F(x) F(x) = P(X ≤ x) = Σt ≤ x f(t) Note the importance of case: F not same as f JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Probability Distributions From our example, the probability that no more than 2 of the envelopes contain $10 bills is P(X ≤ 2) = F (2) = _________________ F(2) = f(0) + f(1) + f(2) = .833625 Another way to calculate F(2)  (1 - f(3)) The probability that no fewer than 2 envelopes contain $10 bills is P(X ≥ 2) = 1 - P(X ≤ 1) = 1 – F (1) = ________ 1 – F(1) = 1 – (f(0) + f(1)) = 1 - .425 = .575 Another way to calculate P(X ≥ 2) is f(2) + f(3) F(2) = f(0) + f(1) + f(2) = .833625  (OR 1 - f(3)) 1 – F(1) = 1 – (f(0) + f(1)) = 1 - .425 = .575  (OR f(2) + f(3)) JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Your Turn … The output of the same type of circuit board from two assembly lines is mixed into one storage tray. In a tray of 10 circuit boards, 6 are from line A and 4 from line B. If the inspector chooses 2 boards from the tray, show the probability distribution function for the number of selected boards coming from line A. x P(x) 1 2 P(x = 0) = JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Continuous Probability Distributions In general, The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is The probability that a given part will fail before 1000 hours of use is Probability density function f(x) JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Visualizing Continuous Distributions The probability that the average daily temperature in Georgia during the month of August falls between 90 and 95 degrees is The probability that a given part will fail before 1000 hours of use is JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Continuous Probability Calculations The continuous probability density function (pdf) f(x) ≥ 0, for all x ∈ R The cumulative distribution, F(x) Example: the uniform distribution (i.e., f(x) = 1, 1 < x < 2) 1. what is the area of the rectangle? (1) The total area under the curve is P(S) and so will always be 1. JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019

Example: Problem 3.7, pg. 92 The total number of hours, measured in units of 100 hours x, 0 < x < 1 f(x) = 2-x, 1 ≤ x < 2 0, elsewhere P(X < 120 hours) = P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) NOTE: You will need to integrate two different functions over two different ranges. b) P(50 hours < X < 100 hours) = Which function(s) will be used? { P(X < 1.2) = P(X < 1) + P (1 < X < 1.2) = ∫01xdx + ∫11.2 (2-x)dx = (x2/2)|01 + (2x- x2/2)|11.2 =0.68 P(.5 < X < 1) = 0.375 JMB Chapter 3 Lecture 1 9th ed EGR 252 JMB 2019