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13 THE NATURE OF PROBABILITY Copyright © Cengage Learning. All rights reserved. 13.4 Calculated Probabilities Copyright © Cengage Learning. All rights reserved. Keno Games 3 Keno Games A very popular lotto game in several states, as well as in most casinos, is a game called Keno. The game consists of a player’s trying to guess in advance which numbers will be selected from a pot containing 80 numbers. 4 Keno Games The person’s choices are marked on a Keno card, and those choices are registered before the game starts. Twenty numbers are then selected at random. 5 Keno Games The player may choose from 1 to 15 spots and gets paid according to Table 13.2 on below. Keno Payoffs Table 13.2 6 Keno Games cont’d Keno Payoffs Table 13.2 7 Example 1 – Decide if a game is fair Suppose a person picks one number and is paid $3.00 if the number picked is among the 20 chosen for the game. The cost for playing this game (the cost is collected in advance) is $1.00. Is this a fair game? Solution: In this problem we are picking 1 out of 80 and we will win if the one number we pick is among the 20 numbers chosen from the pot. 8 Example 1 – Solution cont’d Since the order in which the 20 numbers are picked from the pot does not matter, we see this is a combination. 9 Example 1 – Solution cont’d We now calculate the mathematical expectation: EXPECTATION = $3.00 – $1.00 = –$0.25 Subtract because you pay first. No, it is not a fair game, since the expectation is negative. 10 Independent Events 11 Independent Events Consider the following problem dealing with four cards. Suppose four cards are taken from an ordinary deck of cards, and we form two stacks—one with an ace (one) and a deuce (two) and the other with an ace and a jack. If a card is drawn at random from each pile, what is the probability that a blackjack will occur (an ace and a jack)? 12 Independent Events The tree diagram illustrates the possibilities. 13 Independent Events Notice that the probability of obtaining an ace from the first stack is 1/2 and the probability of obtaining a jack from the second stack is also 1/2. 14 Independent Events The probability of a blackjack is . Notice for this example that probability of ace from first stack = probability of jack from second stack = probability of blackjack = 15 Independent Events If one event (draw from the first stack) has no effect on the outcome of the second event (draw from the second stack), then we say that the events are independent. Thus, if events E and F are independent, P(E|F) = P(E), since the occurrence of event F has no effect on the occurrence of E. 16 Independent Events To understand how independence is determined, consider the following alternative problem. Suppose that all four cards are put together into one pile. This is an entirely different situation. 17 Independent Events There are four possibilities for the first draw, and three possibilities for the second draw, as illustrated here. 18 Independent Events The probability of a blackjack is now . We find this by looking at the sample space of equally likely possibilities. We cannot find this probability by multiplying the component parts for the first and second draws, as we did when considering independent stacks. For this second situation we see that the first and second draws are dependent because the probabilities of selecting a particular second card are influenced by the draw on the first card. 19 Probability of an Intersection 20 Probability of an Intersection If two events are independent, then we can find the probability of an intersection by multiplication. This property is called the multiplication property of probability or the probability of an intersection. If there are n mutually exclusive and equally likely possibilities, then by definition By the conditional probability formula 21 Probability of an Intersection = P(E|F) P(F) By definition of probability = P(E) P(F) P(E|F) = P(E) since E and F are independent. This leads us to the following property. 22 Example 3 – Find a roulette probability What is the probability of black occurring on two successive plays on a U.S. roulette wheel? Solution: A U.S. roulette wheel has 38 compartments, 18 of which are black. Since the results of the spin of the wheel for one game are independent of the results of another spin, the plays are independent. Let B1 = {black on first spin} and B2 = {black on second spin}. Then 0.2243767313 23 Probability of an Intersection Independence and the multiplication property are often used with the idea of complementary events. We now consider the mathematics associated with an experiment called the birthday problem. 24 Example 4 – Birthday problem What is the probability that at least 2 of 4 unrelated persons share the same birthday? Solution: We use Pólya’s problem-solving guidelines for this example. Understand the Problem. We assume that the birthdays of the individuals are independent, and we ignore leap years. We also are not considering the birthday year. We want to know, for example, whether any 2 (or more) of the 4 persons have the same birthday—say, August 3. 25 Example 4 – Solution cont’d Devise a Plan. The first person’s birthday can be any day (365 possibilities out of 365 days). We will find the probability that the second person’s birthday is different from the first person’s birthday. This is The probability that the third person’s birthday is different from the first two birthdays is , and the probability that the fourth person’s birthday is different is . Since these are independent events, we use the multiplication property of probability. 26 Example 4 – Solution cont’d Carry Out the Plan. P(match) = 1 – P(no match) 0.0163559125 = 27 Example 4 – Solution cont’d Look Back. There are about 2 chances out of 100 that there will be a birthday match with 4 persons in the group. It is worth noting that if we consider this for 23 persons, we have This means that if you look at a group of more than 23 persons, it is more likely than not that there will be a birthday match. 28 Probability of a Union 29 Probability of a Union The multiplication property of probability is used to find the probability of an intersection (E and F); now let’s turn to the probability of a union (E or F). P(E or F) = P(E F) Translate the word “or” as a union. = Definition of probability = Cardinality of a union = P(E) + P(F) – P(E F) Definition of probability 30 Probability of a Union We have derived a property called the addition property of probability. 31 Example 6 – Find a calculated probability If A, B, and C are independent events so that P(A) = 0.2, P(B) = 0.6, and P(C) = 0.3, find P[(A B) C]. Solution: We need to do some algebra before we evaluate. P[(A B) C] = 1 – P[(A B) C] = 1 – P[(A B) C] Complementary probabilities Multiplication property of probability 32 Example 6 – Solution cont’d = 1 – [P(A) + P(B) – P(A B)] P(C) Addition property = 1 – [P(A) + P(B) – P(A) P(B)] P(C) Multiplication property = 1 – P(A) P(C) – P(B) P(C) + P(A) P(B) P(C) Distributive property (watch signs) We now evaluate: P[(A B) C] = 1 – (0.2)(0.3) – (0.6)(0.3) + (0.2)(0.6)(0.3) = 0.796 33 Drawing With and Without Replacement 34 Drawing With and Without Replacement To highlight the difference between a combination model and a permutation model, we focus on the idea of replacement. Drawing with replacement means choosing the first item, noting the result, and then replacing the item in the sample space before selecting the second item. Drawing without replacement means selecting the first item, noting the result, and then selecting a second item without replacing the first item. 35 Example 7 – Find probabilities with and without replacement Find the probability of each event when drawing two cards from an ordinary deck of cards. a. Drawing a spade on the first draw and a heart on the second draw with replacement b. Drawing a spade on the first draw or a heart on the second draw with replacement c. Drawing two hearts with replacement d. Drawing a spade on the first draw and a heart on the second draw without replacement 36 Example 7 – Find probabilities with and without replacement e. Drawing a spade on the first draw or a heart on the second draw without replacement f. Drawing two hearts without replacement Rank the events from the most probable to the least probable. Solution: We state the decimal approximations for each probability for ease of comparison. 37 Example 7 – Solution cont’d Let S1 = {draw a spade on the first draw} and H2 = {draw a heart on the second draw} With replacement a. With replacement, the events are independent. Thus P(S1 H2) = P(S1) P(H2) = = = 0.0625 b. P(S1 H2) = P(S1) + P(H2) – P(S1 H2) = = 0.4375 From part a 38 Example 7 – Solution cont’d c. With replacement, the events are independent. Then P(two hearts) = = = 0.0625 Without replacement d. We use the permutation model since the order is important. The number of possibilities is 52P2 = 52 51; the number of successes (fundamental counting principle) is 13 13. Thus, P(S1 H2) = = = 0.0637254902 39 Example 7 – Solution cont’d e. P(S1 H2) = P(S1) + P(H2) – P(S1 H2) From part a = 0.4362745098 f. Without replacement, draw both cards at once. We use combinations since the order is not important. P(two hearts) = = = 0.0588235294 40 Example 7 – Solution cont’d The ranking is: (1) part b, (2) part e (probabilities of parts b and e are about the same), (3) part d, (4) parts a and c (tie), (5) part f. The probabilities of parts a, c, d, and f are about the same. 41 Tree Diagrams 42 Tree Diagrams A powerful tool in handling probability problems is a device we have frequently used—a tree diagram. If the events are independent, we find the probabilities using the multiplication property of probability. 43 Example 8 – Deciding to play. Or not? Consider a game consisting of at most three cuts with a deck of cards. You win and the game is over if a face card turns up on any of the cuts, but you lose if a face card does not turn up. (A face card is a jack, queen, or king.) If you stand to win or lose $1 on this game, should you play? 44 Example 8 – Solution 45 Example 8 – Solution cont’d P(win) = 1 – P(lose) 1 – 0.455 = 0.545 EXPECTATION = Amount to win P(win) + Amount to lose P(lose) $1 (0.545) + (–$1) (0.455) = $0.09 Since the mathematical expectation is positive, the recommendation is to play the game. If you played this game 1,000 times, you could expect to be ahead about 1,000($0.09) = $90.00 46