Please turn off cell phones, pagers, etc. The lecture will begin shortly.

1. Please turn off cell phones, pagers, etc. The lecture will begin shortly.

2. Announcements • Exam 3 will be held next Wednesday, April 5. • It will cover material from Lectures 20-27, • or chapters 12, 13 and 16. • Each student may bring one sheet of notes • (8.5 × 11 inches, both sides) to use during the • exam • Monday’s class (Lecture 28) will be a review and • cover example questions.

3. Lecture 27 This lecture will cover additional topics related to Chapter 16. • Rules of probability (Section 16.4-16.5) • Expected value (Section 16.6)

4. Rule 0: 0 ≤ P(A) ≤ 1 Rule 1: P(not A) = 1 – P(A) Rule 2: P(A or B) = P(A) + P(B) if the events A and B are mutually exclusive Rule 3: P(A and B) = P(A) × P(B) if the events A and B are independent 1. Rules of probability Last time, we covered four rules of probability

5. Mutually exclusive Two events are mutually exclusive if they cannot both happen For example, the events A: it will rain tomorrow B: it will snow tomorrow are not mutually exclusive, because it is possible to have rain and snow on the same day. On the other hand, the events A: a person has no siblings B: a person has three or more siblings are mutually exclusive, because both cannot be true.

6. Independent events Two events are independent if knowing whether one event has occurred does not change the probability that the other will occur. For example, suppose a family has two children (not twins). The events A: first child is a girl B: second child is a boy are independent, because the sex of the first child has no influence on the sex of the second child. The events A: first child is a girl B: the family has at least one girl are not independent, because knowing A has occurred increases the probability of B.

7. Comments • If two events are mutually exclusive, they cannot be • independent. • If two events are independent, they cannot be • mutually exclusive. • Don’t fall prey to the gambler’s fallacy. The gambler’s fallacy arises when a person erroneously believes that independent events influence each other. For example, the person may believe that he’s “due” to win the lottery soon, because he hasn’t won in a long time.

8. Repeated addition The rule P(A or B) = P(A) + P(B) can be extended to three or more events, provided that they are all mutually exclusive (i.e., that no two of them can occur at the same time). P(A or B or C or…) = P(A) + P(B) + P(C) + … For example, if you roll a fair die, the probability of getting an even number is P(even number) = P(2 or 4 or 6) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 1/2

9. Repeated multiplication The rule P(A and B) = P(A) × P(B) can be extended to three or more events, provided that they are all mutually independent (i.e., that no event has any influence on any other) P(A and B and C and…) = P(A) × P(B) × P(C) × … For example, if you toss a coin three times, the probability of getting “HHH” is P(H and H and H) = P(H) × P(H) × P(H) = 1/2 × 1/2 × 1/2 = 1/8

10. (k-1) times once When will it happen? The rule of repeated multiplication allows us to figure out the probability that an event occurs for the first time on the kth trial, for any value of k=1,2,3,… The probability that the event A occurs for the first time on trial k is P(doesn’t occur) × P(doesn’t occur) × … × P(doesn’t occur) × P(does occur)

11. Example If you roll a pair of dice once, the probability of getting a 7 is 6/36 = 1/6. Suppose you decide to keep rolling until you get a 7. The probability that you will only have to roll once is 1/6 = 0.167. The probability that you will have to roll twice is 5/6 × 1/6 = 5/36 = 0.139. The probability that you will have to roll three times is 5/6 × 5/6 ×1/6 = 25/216 = 0.116. …and so on.

12. k times Probability of happening by a certain time The rule of repeated multiplication also allows us to figure out the probability that an event occurs sometime within the first k trials. The probability that the event occurs sometime within the first k trials is 1 – [ P(doesn’t) × P(doesn’t) × … × P(doesn’t) ]

13. 100 1000 1 – (.99999) ≈ .0001 1 – (.99999) ≈ .001 10000 1 – (.99999) ≈ .095 Example Suppose that, in any year, the earth will be hit by an asteroid with probability 1/100,000 = .00001. The probability that it will be hit in the next century is The probability that it will be hit in the next millenium is The probability that it will be hit in the next 10,000 years is

14. 2. Expected value Often the outcome of an experiment is a number. If the outcome of an experiment is a number, that number is called a random variable. Examples of random variables: • How much you win in a lottery • The blood pressure of a subject chosen at random • from a population • The exam score of a student chosen at random • from a class

15. outcome probability 1 1/100 = 0.01 1,000 1/5000 = 0.0002 0 1 – .01 – .0002 = 0.9898 Probability distribution The probability distribution for a random variable is • a list of all possible outcomes, and • the probabilities of those outcomes • (must add up to one) For example, suppose that you have a ticket for a lottery in which you win $1000 with probability 1/5,000 and $1 with probability 1/100. The distribution is:

16. outcome outcome × prob 1 0.01 1,000 0.0002 Prob. 0 0.9898 Expected Value To find the expectation or expected value of a random variable, • multiply each outcome by its probability, and • add them up Example .01 0.2 0 Expected value = .01 + 0.2 + 0 = .21 = $0.21

17. Interpretation of Expected Value The expected value is the average value of the outcome over the long run, if the experiment were repeated many times. (For this reason, it is sometimes called the mean. But don’t confuse it with a sample mean, which we learned about in Lecture 11.) In the lottery example, it would be the amount that you would win on average if you played the lottery over and over. It is not the amount that you win if you play just once. (If you play just once, it’s impossible to win $.21.

18. outcome prob. outcome × prob. 1 1/6 1/6 2 1/6 2/6 3 1/6 3/6 4 1/6 4/6 5 1/6 5/6 6 1/6 6/6 Another example The experiment is to roll a single die. What is the expected value? Expected value = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 21/6 = 3.5