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

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

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Lecture 27 This lecture will cover additional topics related to Chapter Rules of probability (Section ) 2.Expected value (Section 16.6)

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1. Rules of probability Last time, we covered four rules of probability 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

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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.

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Independent events Two events are independent if knowing whether one event has occurred does not change the probability that the other will occur. 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. For example, suppose a family has two children (not twins).

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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.

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Repeated addition The rule 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) = P(A) + P(B) 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

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Repeated multiplication The rule 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) = P(A) × P(B) 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

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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) (k-1) times once

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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 = The probability that you will have to roll twice is 5/6 × 1/6 = 5/36 = The probability that you will have to roll three times is 5/6 × 5/6 × 1/6 = 25/216 = …and so on.

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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) ] k times

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

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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

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Probability distribution The probability distribution for a random variable is 1,0001/5000 = /100 = –.01 –.0002 = outcomeprobability 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:

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Expected Value To find the expectation or expected value of a random variable, multiply each outcome by its probability, and add them up 1, outcomeProb. Expected value = =.21 = $ outcome × prob Example

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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.

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

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