1. Chapter 23C: Expected Values of Discrete Random Variables The mean, or expected value, of a discrete random variable is 1

2. Center The mean of the probability distribution is the expected value of X, denoted E(X) E(X) is also denoted by the Greek letter µ (mu)

3. Interpretation E(x) is a “long run” average; if you perform the experiment many times and observe the random variable x each time, then the average x of these observed x-values will get closer to E(x) as you observe more and more values of the random variable x.

4. Interpretation E(x) is not the value of the random variable x that you “expect” to observe if you perform the experiment once

5. Expected Value I flip a coin. If it lands on heads you win $5. If it lands on tails, you win nothing. What is the expected value?

6. Expected Value I flip the same coin. This time if it lands on heads you win $5, but if it lands on tails, you win $3. What is the expected value?

7. This time, you win $5 if the coin lands on heads, but if it lands on tails, you owe me $6. Calculate the expected value. Expected Value If you play ten times, you Can expect to lose $5.

8. Example: coin tossing Suppose a fair coin is tossed 3 times and we let x = the number of heads. Find m = E(x). First we must find the probability distribution of x.

9. Example (cont.) Possible values of x: 0, 1, 2, 3. p(1)? An outcome where x = 1: THT P(THT)? (½)(½)(½)=1/8 How many ways can we get 1 head in 3 tosses? 3 C 1 =3

10. Example (cont.)

11. So the probability distribution of x is: x 0 1 2 3 p(x) 1/83/83/81/8

12. Example zSo the probability distribution of x is: x0123 p(x)1/83/83/81/8

13. Roulette The roulette wheel has alternating black and red slots numbered 1 through 36. There are also 2 green slots numbered 0 and 00. A bet on any one of the 38 numbers (1-36, 0, or 00) pays odds of 35:1; that is... If you bet $1 on the winning number, you receive $36, so your winnings are $35 American Roulette 0 - 00 (The European version has only one 0.)

14. Roulette Wheel: Expected Value of a $1 bet on a single number Let x be your winnings resulting from a $1 bet on a single number; x has 2 possible values x-135 p(x)37/381/38 E(x)= -1(37/38) + 35(1/38) = -.05 So on average the house wins 5 cents on every such bet. A “fair” game would have E(x)=0. The roulette wheels are spinning 24/7, winning big $$ for the house, resulting in …

16. Expected Values of Discrete Random Variables 16 Given: E(x)= -1(37/38) + 35(1/38) = -.05 How can you make this a “fair” game?

17. Homework Page 616 (1 – 14) 17

18. Chapter 23D The Binomial Distribution 18

19. Binomial example Take 5 coins and toss them once. What’s the probability that you flip exactly 3 heads and 2 tails?

20. Binomial example Solution: One way to get exactly 3 heads: HHHTT What’s the probability of this exact arrangement? P(heads) x P(heads) x P(heads) x P(tails) x P(tails) = (½) 3 x ( ½) 2 Another way to get exactly 3 heads: THHHT Probability of this exact outcome = = (½) 1 x ( ½) 3 x (½) 1 = (½) 3 x ( ½) 2

21. Binomial example Solution: Another way to get exactly 3 heads: THHHT Probability of this exact outcome = = (½) 1 x ( ½) 3 x (½) 1 = (½) 3 x ( ½) 2

22. Binomial distribution In fact, (½) 3 x ( ½) 2 is the probability of each unique outcome that has exactly 3 heads and 2 tails. The overall probability of 3 heads and 2 tails is: [(½) 3 x ( ½) 2 ]+ [(½) 3 x ( ½) 2 ] + [(½) 3 x ( ½) 2 ] +….. for as many unique arrangements as there are— but how many are there?

23. OutcomeProbability THHHT(1/2) 3 x (1/2) 2 HHHTT (1/2) 3 x (1/2) 2 TTHHH (1/2) 3 x (1/2) 2 HTTHH(1/2) 3 x (1/2) 2 HHTTH(1/2) 3 x (1/2) 2 THTHH(1/2) 3 x (1/2) 2 HTHTH(1/2) 3 x (1/2) 2 HHTHT(1/2) 3 x (1/2) 2 THHTH(1/2) 3 x (1/2) 2 HTHHT(1/2) 3 x (1/2) 2 10 arrangements x (1/2) 3 x (1/2) 2 Probability of each unique outcome (note: they are all equal) ways to arrange 3 heads 5 C 3 = 5!/(3!2!) = 10

24.  P(3 heads and 2 tails) = x P(heads) 3 x P(tails) 2 = 10 x (½) 5 = 31.25%

25. x p(x) 03451 2 Binomial distribution function: X= the number of heads tossed number of heads p(x) number of heads 10 ways 31.25%

26. Example 2 As voters exit the polls, you ask a representative random sample of 6 voters if they voted for proposition 100. If the true percentage of voters who vote for the proposition is 55.1%, what is the probability that, in your sample, exactly 2 voted for the proposition and 4 did not?

27. Solution: Outcome Probability YYNNNN = (.551) 2 x (.449) 4 NYYNNN = (.551) 2 x (.449) 4 NNYYNN = (.551) 2 x (.449) 4 NNNYYN = (.551) 2 x (.449) 4 NNNNYY = (.551) 2 x (.449) 4. ways to arrange 2 Yes votes among 6 voters 15 arrangements x (.551) 2 x (.449) 4  P(2 yes votes exactly) = x (.551) 2 x (.449) 4 = 18.5%

28. Binomial distribution, generally 1-p = probability of failure p = probability of success X = # successes out of n trials n = number of trials Note the general pattern emerging  if you have only two possible outcomes (call them 1/0 or yes/no or success/failure) in n independent trials, then the probability of exactly X “successes”=

29. Binomial distribution: example If I toss a coin 20 times, what’s the probability of getting exactly 10 heads?

30. Homework Page 620 (3-5) Page 622 (Odds) 30