Probability Concepts: Conditional Probability & Random Variables

Topics covered 2.4 Conditional probability 2.5 Independence 3.1 Introduction to Random Variables 3.2 Probability distributions 4.1 Probability density functions 4.2 Cumulative distribution functions. 1. 3.1 Problem 10 : The number of pumps in use at a 6 pump and a 4 pump station will be observed tomorrow at noon. Give the possible values for each random variable a. T, the total pumps in use T = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10} b. X the difference in the number of pumps in station 1 and station 2. X = {-4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6} c. U the maximum number of pumps in use at either station U = {0, 1, 2, 3, 4, 5, 6} d. Z, the number of stations having exactly two pumps in use. Z = {0, 1, 2} 2. 3.2 Problem 19 : The textbook is posted P(Y=0) = P(Both arrive on Wed) = .3 * .3 = .09 P(Y=1) = P(W,Th) OR (Th,W) OR P(Th,Th) = .3*.4 + .4*.3 + .4*.4 = .4 P(Y=2) = (W,F) OR (Th,F) OR (F,W) OR (F,Th) OR (F,F) = .32 P(Y=3) = 1 - (.09+.4+.32) = .19

3. You have three uniform random variables U(0,1), X,Y,Z. Either using calculus or Monte Carlo simulations (i.e. a ton of random samples on the computer). This is much easier with simulations than with calculus a. What is the probability density function of X + Y + Z? b. What is the cumulative probability function of X + Y + Z? Graph it. 4. An urn contains n balls numbered 1,2, ... ,n. We select at random r balls. (a) with replacement, (b) without replacement. What is the probability that the largest selected number is m? a) (m/n) r - (m-1 / n) r b) [ (M-1)C(R-1) ] / [ (N)C(R) ]

5. The population of Nicosia is 75% Greek and 25% Turkish. 20% of the Greeks and 10% of the Turks speak English. A visitor to the town meets someone who speaks English. What is the probability that she is Greek? GIVEN: P(Greek) = 0.75; P(Turk) = 0.25; P(English|Greek) = 0.2; P(English|Turk) = 0.10 DERIVE: P(English) = 0.2*0.75 + 0.10*0.25 = 0.175 BAYES: P(Greek|English) = [ P(English|Greek) * P(Greek) ] / P(English) = [ 0.2 * 0.75 ] / 0.175 = 0.857, or 85.7% 6. Huygens's problem. A and B throw alternatively pairs of dice in that order. A wins if she scores 6 points before B gets 7 points, in which case B wins. If A starts the game what is the probability that A wins? Ways for A to get 6: (1,5), (5,1), (2,4), (4,2), (3,3) Prob(A win) = 5/6 2 = 5/36; Prob(A not win) = 31/36 Ways for B to get 7: (1,6), (6,1), (2,5), (5,2), (3,4), (4,3) Prob: 6/36 = 6/6 2 = 1/6; P(B not win) = 5/6 For A to win given A starts… 5/36 + (31/36 * 5/6 * 5/36) + (31/36 * 5/6 * 31/36 * 5/6 * 5/6) = 30/61 = 0.49, or 49%

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