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Discrete Probability. Rosen 5.1. Finite Probability. An experiment is a procedure that yields one of a given set of possible outcomes. The sample space of the experiment is the set of possible outcomes. An event is a subset of the sample space.

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## Discrete Probability

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**Discrete Probability**Rosen 5.1**Finite Probability**• An experiment is a procedure that yields one of a given set of possible outcomes. • The sample space of the experiment is the set of possible outcomes. • An event is a subset of the sample space. • The probability of an event E, which is a subset of a finite sample space S of equally likely outcomes, is p(E) = |E|/|S|**Dice**What is the probability that when two dice are rolled, the sum of the numbers on the two dice is 7? • By the product rule |S| = 6*6 = 36 • |E| = 6, namely • (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) • |E|/|S| = 6/36 = 1/6**Power Ball**Pick the value, in order, for 4 balls whose values each range between 1 and 20 (and values can be repeated). • P(E) = 1/(20*20*20*20) = 1/160,000**The Big Game**Pick the correct value of six balls whose possible values range between 1 and 50. Order not important, but there can be no balls with duplicate values. C(50,6) = 50!/6!44! = 50*49*48*47*46*45/6*5*4*3*2 =15,890,700 There is one correct answer, so p(E) = 1/15,890,700**Poker**What is the probability that, if you are dealt five cards, you have a full house, three of one kind and two of another kind? (e.g., 3 Aces and 2 Kings) • How many possible poker hands? • C(52,5) = 2,598,960 • How many ways to get a full house? • Number of ways to pick two kinds in order: P(13,2) • Number of ways to pick 3 out of 4 of the first kind: C(4,3) • Number of ways to pick 2 out of 4 of the second kind: C(4,2) P(13,2)C(4,3),C(4,2)/2,598,960 = 13*12*4*6/2,598,960 = ~0.0014**Probability of Combinations of Events**• Let E be an event in a sample space S. The probability of the event E’, the complementary event of E, is given by p(E’) = 1 - p(E). • Let E1 and E2 be events in the sample space S. Then p(E1E2) = p(E1) + p(E2) - p(E1E2).**Let’s Make a Deal!**In a game show you are asked to select one of three doors. Behind one of the doors is a large prize. Once you select a door, the game show host, who knows what is behind each door, does the following. First, whether or not you select the winning door, he opens one of the other two doors that he knows is a losing door. Then he asks you whether you would like to switch doors. Should you switch doors?**Let’s Make a Deal solution**• YES. You should always switch doors. • With three doors, the probability that you have selected the correct door is 1/3. Therefore, the probability that you have selected incorrectly is 2/3. • You will always win if your initial choice was incorrect and you switch doors. So by switching doors your probability of winning is 2/3.**Example Problems**• What is the probability that a 5 card poker hand does not contain the queen of hearts. • What is the probability that a 5 card poker hand contains at least one ace? • What is the probability that a 5 card poker hand contains two pairs? (two of each of two different kinds and a fifth card of a third kind)? • What is the probability that a five card poker hand contains a straight (5 in sequence)?**Example Problems**• What is the probability that a 5 card poker hand does not contain the queen of hearts? Number of poker hands that do not contain the queen of hearts: C(51,5) Total number of poker hands: C(52,5)**Example Problems**• What is the probability that a 5 card poker hand contains at least one ace? Let’s compute the probability of a hand containing no aces and subtract that from 1. What is the probability of a hand containing no aces? p(at least one ace) = 1 - 0.659 = ~0.33**Example Problems**• What is the probability that a 5 card poker hand contains exactly two pairs? (two of each of two different kinds and a fifth card of a third kind)? First choose what the cards in each pair will be: C(13,2) = 78 ways to do this. Which two cards of each of the kinds of pairs are included: C(4,2) = 6 for each pair. How many choices for the fifth card (must be different or we have a full house): C(44,1) = 44 Using product rule: 78*6*6*44 ways to have 2 pairs p(2 pairs) = 78*6*6*44/C(52,5) = ~0.0475**Example Problems**• What is the probability that a five card poker hand contains a straight (5 in sequence)? How many ways can we start a straight? C(10,1) = 10 (can start at A, 2, 3, 4, 5, 6, 7, 8, 9, 10--assuming that the Ace can be either high or low) How many choices for each of the five cards? C(4,1) =4 for each card, 45 10*45/C(52,5) = ~0.00355 What if we don’t count straight flushes? (A flush is 5 of the same kind.) There are 10*4 = 40 straight flushes, so (10*45 - 40)/C(52,5)

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