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Expected Value and Binomial Distributions

Exploring how to find expected values and probabilities in discrete random variables, including examples of fair games and binomial experiments. Learn methods to calculate expected gains and losses in various scenarios.

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Expected Value and Binomial Distributions

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  1. Expectation

  2. OBJECTIVE Find Expected Value of a Distribution. Find the probability of a Binomial distribution.

  3. RELEVANCE Be able to find probabilities of discrete random variables.

  4. Definition…… • Expected value of a discrete random variable is equal to the mean of the random variable. • It plays a role in decision theory (games of chance). • Note: Although probabilities can never be negative, expected value CAN be negative.

  5. Formula…… • Note: It is the same as the formula for the mean.

  6. A Fair Game • In gambling games, an expected value of 0 implies that a game is a fair game (an unlikely occurrence!) • In a profit and loss analysis, an expected value of 0 represents the break-even point. A game is fair if the expectation = 0

  7. 1000 tickets are sold at $1.00 each for a color TV valued at $350. What is the expected value of the gain if a person buys one ticket? Set these up as gains minus losses. Example……

  8. 1000 tickets are sold at $1.00 each for a color TV valued at $350. What is the expected value of the gain if a person buys one ticket? • Answer: • Note: -$0.65 does not mean you lose 65 cents since the person can only win a TV valued at $350. • It means the average of the losses is $0.65 for each of the 1000 ticket holders.

  9. Example • A ski resort loses $70,000 per season when it does NOT snow heavily and makes $250,000 profit when it DOES snow heavily. The probability of having a good season is 40%. Find the expectation of the profit.

  10. A ski resort loses $70,000 per season when it does NOT snow heavily and makes $250,000 profit when it DOES snow heavily. The probability of having a good season is 40%. Find the expectation of the profit. • Answer:

  11. Example • 1000 tickets are sold at $1 each for 4 prizes of $100, $50, $25, and $10. What is the expected value if a person buys 2 tickets?

  12. 1000 tickets are sold at $1 each for 4 prizes of $100, $50, $25, and $10. What is the expected value if a person buys 2 tickets? • Answer:

  13. Example • At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy 1 ticket. What is the expected value of your gain?

  14. At a raffle, 1500 tickets are sold at $2 each for four prizes of $500, $250, $150, and $75. You buy 1 ticket. What is the expected value of your gain? • Answer: • Because the expected value is negative, you can expect to lose an average of $1.35 for each ticket that you buy.

  15. Binomial Distributions

  16. Many types of probabilities have 2 outcomes. a. A coin: heads or tails b. Baby Born: Male or Female c. T/F Test: True or False • Some situations can be modified or reduced to 2 outcomes. a. A medical treatment: effective or ineffective b. Multiple choice test: correct or incorrect

  17. Binomial Experiment satisfies 4 requirements….. • 1. Has 2 outcomes or reduces to 2. • 2. Fixed # of Trials • 3. Outcomes for each trial must be independent • 4. The probability of success must remain the same for each trial This leads to…

  18. Binomial Distribution….. • Definition – The outcomes of a binomial experiment and their corresponding probabilities.

  19. Notation for the Binomial…… • “n” – number of trials • “p” – the numerical probability of success • “q” – the numerical probability of failure; q = 1 - p • “x” – the # of successes in “n” trials x will always be a whole number – no decimals!

  20. There are many ways to find the probability of a binomial. • One way is to use the formula below:

  21. Example: A coin is tossed 3 times. Find the probability of getting exactly 2 heads. Notations Needed: n = 3 p = ½ q = 1 – ½ = ½ x = 2 Let’s solve the following example using several different methods…..

  22. 1st Way: Using a tree diagram, find the sample space for 3 coins: HHH THH HHT THT HTH TTH HTT TTT There are 3 out of 8 possibilities of getting exactly 2 heads. Probability = 3/8 or 0.375 A coin is tossed 3 times. Find the probability of getting exactly 2 heads.

  23. 2nd way : Using the binomial formula. Remember….. n = 3 p = ½ q = ½ x = 2 A coin is tossed 3 times. Find the probability of getting exactly 2 heads.

  24. 3rd Way: Chart in Book P. 711 in book (or look on copy) Answer: 0.375 A coin is tossed 3 times. Find the probability of getting exactly 2 heads.

  25. 4th and BEST way! – Graphing Calculator Calculator Keys: 2nd Vars 0 or A: binompdf (n,p,x) You will enter binompdf(3, ½, 2) Enter Answer: 0.375 A coin is tossed 3 times. Find the probability of getting exactly 2 heads.

  26. Public Opinion Reported that 5% of Americans are afraid of being alone in the house at night. If a random sample of 20 Americans is selected, find the probability that there are exactly 5 people who are afraid of being alone in the house at night. Answer: n = 20 p = .05 x = 5 Binompdf(20, .05, 5)= 0.002 Example

  27. A burglar alarm system has 6 fail-safe components. The probability of each failing is 0.05. Find the probability that exactly 3 will fail. Answer: n = 6 p = 0.05 x = 3 Binompdf(6, 0.05, 3)= 0.002 Example

  28. A student takes a random guess at 5 multiple choice questions. Find the probability that the student gets exactly 3 correct. Each question has 4 possible choices. Answer: n = 5 p = ¼ x = 3 Binompdf(5,1/4,3)= 0.088 Example

  29. Binomial Distributions…Continued At Least And At Most

  30. Public Opinion reported that 5% of Americans are afraid of the dark. If a random sample of 20 is selected, find the probability that A. At most 3 are afraid of the dark. B. At least 3 are afraid of the dark. Example

  31. Public Opinion reported that 5% of Americans are afraid of the dark. If a random sample of 20 is selected, find the probability that…..at most 3 are afraid of the dark. • This is the same as finding the probabilities for x = 0, 1, 2, and 3 • Add the probabilities together. • n = 20 • p = .05 • x = 0, 1, 2, 3

  32. The Long Way……. • Using the calculator: P(at most 3) = bpdf(20, .05, 0) + bpdf(20, .05, 1) + bpdf(20, .05, 2) + bpdf(20, .05, 3) P(at most 3) = 0.358 + 0.377 + 0.189 + 0.060 = 0.984.

  33. Using your graphing calculator: Put x’s in L1 Set a formula for L2. bpdf(20, .05, L1) The sum of L2 is your answer. Is there a shorter way?.......

  34. Public Opinion reported that 5% of Americans are afraid of the dark. If a random sample of 20 is selected, find the probability that…..at least 3 are afraid of the dark. • n = 20 • p = .05 • x = 3, 4, 5, ………, 20

  35. The answer is 0.075.

  36. You Try….. • A burglar alarm system has 6 fail-safe components. The probability of each failing is 0.05. Find these probabilities. a. Fewer than 3 will fail b. None will fail c. More than 3 will fail

  37. n = 6 p = .05 x = 0, 1, 2 The answer is 0.998 A burglar alarm system has 6 fail-safe components. The probability of each failing is 0.05. Find these probabilities.A. Fewer than 3 will fail

  38. n = 6 p = .05 x = 0 The answer is 0.735 You DO NOT need the lists for this one because there is only one x. A burglar alarm system has 6 fail-safe components. The probability of each failing is 0.05. Find these probabilities.B. None will fail

  39. n= 6 p = .05 x = 4, 5, 6 The answer is 0.00008642 A burglar alarm system has 6 fail-safe components. The probability of each failing is 0.05. Find these probabilities.C. More than 3 will fail

  40. Assignment…… • Worksheet

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