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) P(X > 5) P(Y < 3) a random variable, X, is a function that associates a real number with each element in the sample space and x is one of the values that X can take. JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

Discrete Random Variables Problem 2.53 Page 55 Modified Assume someone spends $75 to buy 3 envelopes. 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} 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. 66 JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

Discrete Probability Distributions 1 The probability that the envelope contains a $10 bill is 275/500 or .55 What is the probability that there are no $10 bills in the group? 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 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 (students fill in the table) JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

Discrete Probability Distributions 2 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 (students fill in the table) x 1 2 3 P(X = x) JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

Example 3.8, pg 80 Shipment: 8 computers of which 3 are defective Random purchase of 2 computers What is the probability distribution for the random variable X = defective computers purchased? Possibilities: X = 0 X =1 X = 2 Let’s start with P(X=0) P = specified target / all possible (0 defectives and 2 nondefectives are selected) (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 EGR 252.001 Spring 2009

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) JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

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  (OR 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  (OR 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 EGR 252.001 Spring 2009

Another View The probability histogram JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

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 associated with the selected boards being from line A. x P(x) 1 2 P(x = 0) = JMB Chapter 3 Lecture 1 EGR 252.001 Spring 2009

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 EGR 252.001 Spring 2009

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 EGR 252.001 Spring 2009

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 EGR 252.001 Spring 2009

Example: Problem 3.7, pg. 88 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 EGR 252.001 Spring 2009