11.2. Counting Techniques and Probability
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1 11.2. Counting Techniques and Probability 1
2 Objectives A. Draw and use a tree diagram to find the probability of an event. B. Use permutations and combinations to find the probability of an event. C. Solve applications involving probability. 2
3 3
4 Poker Improbabilities Introduction Do you know the probability of getting four aces in a 5-card poker hand? Since there are four aces in a deck, there is only 1 way of getting four aces if the fifth card can be any of the 48 remaining; hence, there are 1 C(48, 1) = 48 ways of getting four aces of the C(52, 5) = 2,598,960 possible poker hands. Thus, the probability of four aces in poker is only 4
5 Can you find the probability of getting three clubs and two diamonds? It is Introduction 5
6 Using Tree Diagrams 6
7 Example Have you heard of the witches of Wall Street? These are people who use astrology, tarot cards, or other supernatural means to predict whether a given stock will go up, go down, or stay unchanged. Not being witches, we assume that a stock is equally likely to go up (U), go down (D), or stay unchanged (S). A broker selects two stocks at random from the New York Stock Exchange list. (a) What is the probability that both stocks go up? (b) What is the probability that both stocks go down? (c) What is the probability that one stock goes up and one goes down? 7
8 Solution In order to find the total number of equally likely possibilities for selecting the two stocks, we draw the tree diagram shown in Figure Figure
9 Solution (a) There is only 1 outcome (UU) out of 9 in which both stocks go up. Thus, the probability that both stocks go up is (b) There is only 1 outcome (DD) in which both stocks go down, so the probability that both go down is (c) There are 2 outcomes (UD, DU) in which one stock goes up and one down. Hence, the probability of this event is (Notice that the tree diagram shows that there are 4 outcomes in which one stock stays unchanged and the other goes either up or down. The probability of this event is thus ) 9
10 Using Permutations and Combinations The next example deals with a problem involving ordinary playing cards. Note that in solving part (a), you can use combinations, permutations, or the Sequential Counting Principle. The important thing is to be consistent in the computation. 10
11 Example Two cards are drawn in succession and without replacement from an ordinary deck of 52 cards. Find the probability that (a) the cards are both aces. (b) an ace and a king, in that order, are drawn. 11
12 Solution (a) Here, the order is not important because we are simply interested in getting 2 aces. We can find this probability by using combinations. The number of ways to draw 2 aces is C(4, 2) because there are 4 aces and we want a combination of any 2 of them. The number of combinations of 2 cards picked from the deck of 52 cards is C(52, 2). Thus, the probability of both cards being aces is 12
13 Solution (b) In this part of the problem, we want to consider the order in which the 2 cards are drawn, so we use permutations. The number of ways of selecting an ace is P(4, 1) and the number of ways of selecting a king is P(4, 1). By the Sequential Counting Principle, the number of ways of doing these two things in succession is P(4, 1)P(4, 1). The total number of ways of drawing 2 cards is P(52, 2), so the probability of drawing an ace and a king, in that order, is 13
14 Example An oil company is considering drilling an exploratory oil well. If the rocks under the drilling site are characterized by what geologists call a dome structure, the chances of finding oil are 60%. The well can be dry, a low producer, or a high producer of oil. The probabilities for these outcomes are given in the table. (a) Draw a tree diagram for the data given in the table. (b) What is the probability that the well is dry? 14
15 Solution (a) Since the probability of finding oil when there is a dome structure is 60%, the probability of finding oil when there is no dome structure is 100% 60% = 40%. We draw the tree diagram in Figure 11.8 and label the first 2 branches Dome (0.6) and No dome (0.4). The total probability for a dry well is = Figure
16 Solution We then label 3 branches starting from the end of the Dome outcome and 3 branches starting from the end of the No dome outcome with the probabilities for a dry, low-producing, and high-producing well. (b) To find the probability that the well is dry, we start at the branch in Figure 11.8 labeled Dome (0.6) and continue through the branch labeled Dry (0.6). The probability of that path is =
17 Solution The other possibility for a dry well is to start at the branch labelled No dome (0.4) and continue through the branch labelled Dry (0.850). The probability for that branch is = The total probability for a dry well is the sum of the two probabilities, =
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