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Chapter 11 Probability Copyright © 2015, 2011, and 2007 Pearson Education, Inc. 1 Chapter 11: Probability 11.1 11.2 11.3 11.4 11.5 Basic Concepts Events Involving “Not” and “Or” Conditional Probability and Events Involving “And” Binomial Probability Expected Value and Simulation Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 2 Section 11-5 Expected Value and Simulation Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 3 Expected Value • Determine expected value of a random variable and expected net winnings in a game of chance. • Determine if a game of chance is a fair game. • Use expected value to make business and insurance decisions. • Use simulation in genetic processes such as flower color and birth gender. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 4 Expected Value Children in third grade were surveyed and told to pick the number of hours that they play electronic games each day. The probability distribution is given below. # of Hours x Probability P(x) 0 0.3 1 0.4 2 0.2 3 0.1 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 5 Expected Value Compute a “weighted average” by multiplying each possible time value by its probability and then adding the products. 0(0.3) 1(0.4) 2(0.2) 3(0.1) 1.1 1.1 hours is the expected value (or the mathematical expectation) of the quantity of time spent playing electronic games. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 6 Expected Value If a random variable x can have any of the values x1, x2 , x3 ,…, xn, and the corresponding probabilities of these values occurring are P(x1), P(x2), P(x3), …, P(xn), then the expected value of x is given by E ( x) x1 P( x1 ) x2 P( x2 ) xn P( xn ). Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 7 Example: Finding the Expected Number of Boys Find the expected number of boys for a three-child family. Assume girls and boys are equally likely. Solution S = {ggg, ggb, gbg, gbb, bgg, bgb, bbg, bbb} The probability distribution is on the right. # Boys x 0 1 2 3 Probability P(x) 1/8 3/8 3/8 1/8 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. Product x P( x) 0 3/8 6/8 3/8 8 Example: Finding the Expected Number of Boys Solution (continued) The expected value is the sum of the third column: 3 6 3 12 0 8 8 8 8 3 1.5. 2 So the expected number of boys is 1.5. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 9 Example: Finding Expected Winnings A player pays $3 to play the following game: He rolls a die and receives $7 if he tosses a 6 and $1 for anything else. Find the player’s expected net winnings for the game. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 10 Example: Finding Expected Winnings Solution The information for the game is displayed below. Die Outcome Payoff 1, 2, 3, 4, or 5 $1 Net P(x) x –$2 5/6 6 $7 $4 1/6 x P( x) –$10/6 $4/6 Expected value: E(x) = –$6/6 = –$1.00 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 11 Games and Gambling A game in which the expected net winnings are zero is called a fair game. A game with negative expected winnings is unfair against the player. A game with positive expected net winnings is unfair in favor of the player. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 12 Example: Finding the Cost for a Fair Game What should the game in the previous example cost so that it is a fair game? Solution Because the cost of $3 resulted in a net loss of $1, we can conclude that the $3 cost was $1 too high. A fair cost to play the game would be $3 – $1 = $2. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 13 Example: Finding Expected Winnings in Roulette One simple type of roulette is played with an ivory ball and a wheel set in motion. The wheel contains thirty-eight compartments. Eighteen of the compartments are black, eighteen are red, one is labeled “zero,” and one is labeled “double zero.” (These last two are neither black nor red.) In this case, assume the player places $1 on either red or black. If the player picks the correct color of the compartment in which the ball finally lands, the payoff is $2. Otherwise, the payoff is zero. Find the expected net winnings. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 14 Example: Finding Expected Winnings in Roulette Solution By the expected value formula, expected net winnings are 18 20 1 E (net winnings) ($1) ($1) $ 38 38 19 The expected net loss here is $1/19, or about 5.3¢, per play. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 15 Investments Expected value can be a useful tool for evaluating investment opportunities. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 16 Example: Finding Expected Investment Profits Mark is going to invest in the stock of one of the two companies below. Based on his research, a $6000 investment could give the following returns. Company ABC Profit or Probability Loss x P(x) –$400 0.2 $800 0.5 $1300 0.3 Company PDQ Profit or Probability Loss x P(x) $600 0.8 1000 0.2 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 17 Example: Finding Expected Investment Profits Find the expected profit (or loss) for each of the two stocks. Solution ABC: –$400(0.2) + $800(0.5) + $1300(0.3) = $710 PDQ: $600(0.8) + $1000(0.2) = $680 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 18 Business and Insurance Expected value can be used to help make decisions in various areas of business, including insurance. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 19 Example: Expected Lumber Revenue A lumber wholesaler is planning on purchasing a load of lumber. He calculates that the probabilities of reselling the load for $9500, $9000, or $8500 are 0.25, 0.60, and 0.15, respectively. In order to ensure an expected profit of at least $2500, how much can he afford to pay for the load? Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 20 Example: Expected Lumber Revenue Solution The expected revenue from sales can be found below. Income x P(x) x P( x) $9500 .25 $2375 $9000 .60 $5400 $8500 .15 $1275 Expected revenue: E(x) = $9050 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 21 Example: Expected Lumber Revenue Solution (continued) expected profit = expected revenue – cost cost = expected revenue – expected profit To have an expected profit of $2500, he can pay up to $9050 – $2500 = $6550. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 22 Simulation An important area within probability theory is the process called simulation. It is possible to study a complicated, or unclear, phenomenon by simulating, or imitating, it with a simpler phenomenon involving the same basic probabilities. Simulation methods (also called Monte Carlo methods) require huge numbers of random digits, so computers are used to produce them. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 23 Example: Simulating Genetic Processes Toss two coins 50 times and use the results to approximate the probability that the crossing of Rr pea plants will produce three successive redflowered offspring. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 24 Example: Simulating Genetic Processes Solution We actually tossed two coins 50 times and got the following sequence. th, hh, th, tt, th, hh, ht, th, ht, th, hh, hh, tt, th, hh, ht, ht, ht, ht, th, hh, hh, hh, tt, ht, tt, hh, ht, ht, hh, tt, tt, tt, th, tt, tt, hh, ht, ht, ht, hh, tt, th, hh, tt, hh, ht, tt, tt, tt By the color interpretation described in the text, this gives the following sequence of flower colors in the offspring. (Only “both tails” gives white.) Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 25 Example: Simulating Genetic Processes Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 26 Example: Simulating Genetic Processes We now have an experimental list of 48 sets of three successive plants, the 1st, 2nd, and 3rd entries, then the 2nd, 3rd, and 4th entries, and so on. Do you see why there are 48 in all? Now we just count up the number of these sets of three that are “red-red-red.” Since there are 20 of those, our empirical probability of three successive red offspring, obtained through simulation, is 20/48 = 5/12, or about 0.417. Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 27 Example: Simulating Births with Random Numbers A couple plans to have five children. Use random number simulation to estimate the probability that they will have more than three boys. Solution Let each sequence of five digits, as they appear in Table 20 (page 632 from text), represent a family with five children, and (arbitrarily) associate odd digits with boys, even digits with girls. (Recall that 0 is even.) Verify that, of the fifty families simulated, only the ten marked with arrows have more than 3 boys (4 boys or 5 boys). P(more than 3 boys) 10 / 50 0.20 Copyright © 2016, 2012, and 2008 Pearson Education, Inc. 28