More Probability LSP 121 LSP 121 Math and Tech Literacy II More Probability Greg Brewster DePaul University Topics More Probability Combining Probabilities AND – multiply probabilities OR – add probabilities Average trials to success “At Least Once” calculations Expected Values What’s the payoff? Should you ever play the lottery? Combining Events Sometimes you want to determine the probability of a combination of two event outcomes. Given outcomes A and B What is the probability that EITHER A OR B happen = Pr(A or B)? What is the probability that BOTH A AND B happen = Pr(A and B)? Greg Brewster, DePaul University Page 1 More Probability LSP 121 Either/Or Probabilities Mutually Exclusive Events If events A and B are mutually exclusive then P(A or B) = P(A) + P(B) Mutually exclusive means that these two event outcomes cannot both happen. Either/Or Probabilities Example: Suppose you roll a single die. What is the probability of rolling either a 2 or a 3? P(2 or 3) = P(2) + P(3) = 1/6 + 1/6 = 2/6 These events cannot both happen. A single roll can’t be both a 2 and a 3, so the equation works. Either/Or Probabilities Example: Suppose you draw a card from a deck of cards. What is the probability that it is either a King or a Queen? P(K or Q) = P(K) + P(Q) = 4/52 + 4/52 = 8/52 = 15.4% Greg Brewster, DePaul University Page 2 More Probability LSP 121 Either/Or Probabilities Example: Suppose you roll two dice. What is the probability that either the first roll is a 2 or the second roll is a 3? We cannot just add these probabilities together because it is possible that both of these events occur. We need the following more general equation: P(A or B) = P(A) + P(B) – P(both A and B) Either/Or Probabilities Example: Suppose you roll two dice. What is the probability that either the first roll is a 2 or the second roll is a 3? P((1st roll is 2) or (2nd roll is 3)) =P(1st roll is 2) + P(2nd roll is 3) – P(1st roll is 2 AND 2nd roll is 3) = 1/6 + 1/6 – 1/36 = 11/36 = 30.55% AND Probabilities for Independent Events If events A and B are independent then P(A and B) = P(A) x P(B) Independent means that the outcome of event A has no bearing on the outcome of the event B. Greg Brewster, DePaul University Page 3 More Probability LSP 121 Calculating Probabilities Independent Events For example, suppose you toss three coins. What is the theoretical probability of getting three tails? 1/2 x 1/2 x 1/2 = 1/8 = 12.5% Calculating Probabilities Independent Events A 100-year flood is a flood that occurs, on average, only once every 100 years. Find the theoretical probability that a 100year flood will strike a city in two consecutive years (1 in 100) x (1 in 100) = 0.01 x 0.01 = 0.0001 = 0.01% Calculating Probabilities Independent Events The Pick-Three game in the Illinois lottery is a game is a game where you must correctly match a 3-digit number to win. What is the probability that you win the Pick-Three game on a random try? P(win) = P((1st digit correct) AND (2nd digit correct) AND (3rd digit correct)) P(win) = P(1st digit correct) x P(2nd digit correct) x P(3rd digit correct) = 1/10 x 1/10 x 1/10 = 1/1000 = 0.001 = 0.1% Greg Brewster, DePaul University Page 4 More Probability LSP 121 Average Tries Before Success If you are repeatedly doing an activity (trials) and each repetition is independent and the probability of succeeding on each trial is p, then The average number of trials before you succeed is 1/p Average Tries Before Success On average, how many times do you need to play the Pick-Three game before you will win? Answer: the probability of success each time you play is p = 0.001. So the average number of times you need to play before you win is 1/p = 1000 times Average Tries Before Success On average, how many times do you need to roll a die before you get a 3? Answer: the probability of rolling a 3 on each try is p = 1/6. So the average number of times you need to roll before you roll a 3 is 1/p = 6 times Greg Brewster, DePaul University Page 5 More Probability LSP 121 Probability of At Least One Success in N Trials What is the theoretical probability of something happening at least once if you try it N times? P(at least one A in N trials) = 1 - [P(not A in one trial)]N Probability of At Least Once Example What is the probability that you will win the PickThree game at least once if you play 100 times? Probability of winning each time is 1/1000. Probability of not winning each time is (1 – 1/1000) = 999/1000. P(at least one win in 100 tries) = 1 - 0.999100 = 0.095 or 9.5% How about if you play 1000 times? P(at least one win in 1000 tries) = 1 - 0.9991000 = 0.632 or 63.2% Probability of At Least Once Example What is the theoretical probability that a region will experience at least one 100-year flood during the next 100 years? Probability of a flood is 1/100. Probability of no flood is 99/100. P(at least one flood in 100 years) = 1 - 0.99100 = 0.634 Greg Brewster, DePaul University Page 6 More Probability LSP 121 Probability of At Least Once Another Example You purchase 10 raffle tickets, for which the theoretical probability of winning some prize on a single ticket is 1 in 10. What is the theoretical probability that you will have at least one winning ticket? P(at least one winner in 10 tickets) = 1 - 0.910 = 0.651 Expected Value Now, what if we assign values to the outcomes we are working with? For example: Playing the Pick-Three game costs $1 Winning the Pick-Three game gives you $500 Then we can compute the expected value (also called the “expected winnings” or “expected losses”) of a game. Calculating Expected Value You compute the expected value by going through all the possible outcomes and adding up the (value x probability) for each outcome. Expected value = outcome 1 value x outcome 1 probability + outcome 2 value x outcome 2 probability +… Greg Brewster, DePaul University Page 7 More Probability LSP 121 Expected Value of Pick-Three Game The outcome values for the Pick-Three game are the following: Winning the game has value $499 (costs $1 to play and $500 winnings) and probability 0.001 Losing the game has value -$1 (costs $1 to play and no winnings) and probability 0.999 Expected value = $499 x 0.001 + (-$1 x 0.999) = -$0.50 = 50 cent loss per play Over the long run, on average, you expect to lose 50 cents per play. Play 100 times => lose $50 average Another Example Suppose that $1 lottery tickets have the following probabilities: 1 1 1 1 in in in in 5 win a free $1 ticket; 100 win $5; 100,000 to win $1000; 10 million to win $1 million. What is the expected value of a lottery ticket? Example I - Solution Ticket purchase: value -$1, prob 1.0 Win free ticket: value $1, prob 1/5 Win $5: value $5, prob 1/100 Win $1000: prob 1/100,000 Win $1million: prob 1/10,000,000 -$1 x 1= -1; $1 x 1/5 = $0.20; $5 x 1/100 = $0.05; $1000 x 1/100,000 = $0.01; $1,000,000 x 1/10,000,000 = $0.10 Greg Brewster, DePaul University Page 8 More Probability LSP 121 Example I - Solution Continued Now sum all the products: -$1 + 0.20 + 0.05 + 0.01 + 0.10 = -$0.64 Thus, averaged over many tickets, you should expect to lose $0.64 for each lottery ticket that you buy. If you buy 1000 tickets, you should expect to lose $640, on average. Example II The Insurance Business Suppose an insurance company sells policies for $500 each. The company knows that 10% of their clients will submit a successful claim that year and that successful claims average $1500 each. How much can the company expect to make per customer? Insurance Example Company makes $500 100% of the time (when a policy is sold) Company loses $1500 10% of the time ($500 x 1.0) + ( - $1500 x 0.1) = 500 – 150 = $350 Company gains $350 from each customer it sells a policy to, on average. Greg Brewster, DePaul University Page 9 More Probability LSP 121 Does it ever make sense to play Lotto? The Illinois Lottery “Lotto” game requires you to match six numbers. Odds of winning are 1 in 10,179,260 (according to their web site). So, probability of winning each time is p=1/10,179,261. The Jackpot you win changes all the time Expected value per play = -$1 + p * Jackpot So, if the Jackpot is larger than $10,179,261, then the expected value is greater than zero – all right! But, on average, you need to play 1/p = 10,179,261 times before you win. I wouldn’t recommend it. Greg Brewster, DePaul University Page 10
© Copyright 2026 Paperzz