Chapter 15. Probability, continued. The sample space of a random experiment is the set of all possible outcomes of the experiment.

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1 Chapter 15. Probability, continued Review: The sample space of a random experiment is the set of all possible outcomes of the experiment. We have to count outcomes; a useful tool is the multiplication rule: count outcomes in two or more stages; suppose in each stage the number of choices doesn't depend on choices made in other stages; then the number of outcomes is the product of the number of choices in each stage. Example: five candidates in an election, the one with the most votes is President, 2 nd most is VP, 3 rd most is Secretary. Outcomes: Section 3. Permutations and combinations A standard deck of cards has four suits (Hearts, Diamonds, Spades and Clubs) and 13 values (Ace, King, Queen, Jack, 10, 9, 8, 7, 6, 5, 4, 3, 2), making 4 * 13 = 52 cards. A poker hand consists of five cards. How many different poker hands are there? There are 52 * 51 * 50 * 49 * 48 = 311,875,200 hands, if we pay attention to the order in which the cards are dealt. 1

2 In general, if you have n items, and you choose r of them, and you care about the order in which you choose them, then the number of choices is called the number of permutations on n objects taken r at a time = np r = n * (n 1) * (n 2) *... * (n r + 1) (notice that there are r numbers being multiplied). So the number of 4 member committees (in which order matters) from 35 people = 35 P 4 = 35 * 34 * 33 * 32. The number of 5 card hands (in which order matters) from 52 cards = 52 P 5 = 52 * 51 * 50 * 49 * 48. In draw poker, the order we receive the cards doesn t matter. When we counted the ordered hands above, we counted each unordered hand many times. How many ordered hands does each unordered hand give rise to? There are 5 * 4 * 3 * 2 * 1 = 5! = 120 orderings of a five card hand. So the number of unordered hands is 52 * 51 * 50 * 49 * 48)/( 5 * 4 * 3 * 2 * 1) = 311,875,200/120 = 2,598,960 In general, if you have n items, and you choose r of them, and you don t care about the order in which you choose them, then the number of choices is called the number of combinations of n objects taken r at a time = nc r = n P r / r! = [n * (n 1) * (n 2) *... * (n r + 1)]/[r * (r 1) * (r 2) *...*1] 2

3 Suppose there are 10 horses entered in a race. In how many ways can one pick a) the top three finishers regardless of order? b) the first-, second-, and third-place finishers in the race? b) asks for the number of permutations = 10 P 3 = 10 * 9 * 8 = 720. a) asks for the number of combinations = 10 C 3 = 720/3! = 720/6 = 120. Look at book's problems and determine if you are asked for permutations or combinations 3

4 Sections 4. Probability spaces The probability of something happening is a number between 0 and 1; if a random experiment is carried out repeatedly, the probability of an outcome is supposed to represent the fraction of the experiments for which that outcome is expected to occur. The book gives the example of a basketball player who, over a long period of time, shoots 2135/2362 =.904 of his free throws. So one might conclude that if he has a free throw attempt (that s the random experiment in this case),.904 is the probability of success and =.096 is the probability of failure. In general, for some random experiment, you must first determine the sample space S = {o 1,...,o N }. Next you must assign a probability Pr (o i ) to each outcome o i. Each probability is between 0 and 1, and Pr (o 1 ) + + Pr (o N ) = 1. This is called a probability assignment. If we toss a fair coin two times, and note on each toss whether it is heads or tails, then the sample space S = {HH, HT, TH, TT}. Since each of these outcomes is equally likely, Pr (HH) = Pr (HT) = Pr (TH) = Pr (TT) = ¼ =.25. If we toss a fair coin two times, and note the number of heads, then the sample space S = {2, 1, 0}. But these outcomes are not equally likely, since HT and TH both result in an outcome of 1. So Pr (2) = Pr (0) =.25, and Pr (1) =.5. 4

5 p580 #38b. Consider the sample space S = {o 1, o 2, o 3, o 4 }. Suppose you are given Pr (o 1 ) + Pr (o 2 ) = Pr (o 3 ) + Pr (o 4 ). If Pr (o 1 ) =.15 and Pr (o 3 ) =.22, find the probability assignment. (So we need to find Pr (o 2 ) and Pr (o 4 )). Answer: Pr (o 1 ) + Pr (o 2 ) = Pr (o 3 ) + Pr (o 4 ) =.5 (since they are equal and add to 1), so Pr (o 2 ) = =.35 and Pr (o 4 ) = =.28. There are 8 players (call them P 1,, P 8 ) entered in a chess tournament. Our random experiment is to hold the tournament and note the winner. The sample space is S = { P 1,, P 8 }. We are told that P 1 has a 25% chance of winning, P 2 has a 15% chance, P 3 has a 5% chance, and all others have an equal chance. Find the probability assignment. Answer: We are given that Pr (P 1 ) =.25, Pr (P 2 ) =.15 and Pr (P 3 ) =.05. So the probability of one of these three players winning is Pr (P 1 ) + Pr (P 2 ) + Pr (P 3 ) = =.45. Thus the probability of one of the other players winning is Pr (P 4 ) + Pr (P 5 ) + Pr (P 6 ) + Pr (P 7 ) + Pr (P 8 ) = =.55. Also, Pr (P 4 ) = = Pr (P 8 ), so each is 1/5 of.55, that is Pr (P 4 ) = = Pr (P 8 ) =.55/5 =.11. 5

6 Events. An event is any subset of the sample space. The probability of an event is the sum of the probabilities of the outcomes in that event. If we toss a fair coin two times, and note on each toss whether it is heads or tails, then the sample space S = {HH, HT, TH, TT}. Here is a listing of some of the events: Description of event Set of outcomes Probability Toss two heads {HH}.25 Toss one head {HT, TH}.5 Toss at least one head {HH, HT, TH}.75 Toss two heads or two tails {HH, TT}.5 Toss at most two heads {HH, HT, TH, TT} 1 Toss three heads { } 0 The last two events listed are called the certain event and the impossible event. p.580 # 44. Consider the random experiment where a student takes a four-question true-false quiz. Write out the event described by each of the following statements: a) Exactly two of the answers given are T s (T for true). b) At least two of the answers given are T s. c) At most two of the answers given are T s. d) The first two answers given are T s. Answers: a) {TTFF, TFTF, TFFT, FTTF, FTFT, FFTT} b) {TTFF, TFTF, TFFT, FTTF, FTFT, FFTT, TTTF, TTFT, TFTT, FTTT, TTTT} c) {FFFF, TFFF, FTFF, FFTF, FFFT, TTFF, TFTF, TFFT, FTTF, FTFT, FFTT} d) {TTFF, TTTF, TTFT, TTTT} 6

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