Copyright © 2009 Pearson Education, Inc.

Copyright © 2009 Pearson Education, Inc.
6.2 Basics of Probability LEARNING GOAL Know how to find probabilities using theoretical and relative frequency methods and understand how to construct basic probability distributions. Page 238 Copyright © 2009 Pearson Education, Inc.

Let’s begin by considering a toss of two coins. Figure 6.1 shows that there are four different ways the coins could fall. We say that each of these four ways is a different outcome of the coin toss. Figure 6.1 But suppose we are interested only in the number of heads. Because the two middle outcomes in Figure 6.1 each have 1 head, we say that these two outcomes represent the same event. Pages Figure 6.1 shows that there are four possible outcomes for the two-coin toss, but only three possible events: 0 heads, 1 head, and 2 heads. Copyright © 2009 Pearson Education, Inc. Slide

Definitions Outcomes are the most basic possible results of observations or experiments. An event is a collection of one or more outcomes that share a property of interest. Page 239 Copyright © 2009 Pearson Education, Inc. Slide

Expressing Probability The probability of an event, expressed as P(event), is always between 0 and 1 inclusive. A probability of 0 means that the event is impossible, and a probability of 1 means that the event is certain. Page 239 Figure 6.2 The scale shows various degrees of certainty as expressed by probabilities. Copyright © 2009 Pearson Education, Inc. Slide

Theoretical Probabilities
Theoretical Method for Equally Likely Outcomes Step 1. Count the total number of possible outcomes. Step 2. Among all the possible outcomes, count the number of ways the event of interest, A, can occur. Step 3. Determine the probability, P(A), from P(A) = Pages number of ways A can occur total number of outcomes Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 1 Guessing Birthdays Suppose you select a person at random from a large group at a conference. What is the probability that the person selected has a birthday in July? Assume 365 days in a year. Solution: If we assume that all birthdays are equally likely, we can use the three-step theoretical method. Step 1. Each possible birthday represents an outcome, so there are 365 possible outcomes. Step 2. July has 31 days, so 31 of the 365 possible outcomes represent the event of a July birthday. Page 240 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 1 Guessing Birthdays Solution: (cont.) Step 3. The probability that a randomly selected person has a birthday in July is P(July birthday) = ≈ which is slightly more than 1 in 12. 31 365 Page 240 Copyright © 2009 Pearson Education, Inc. Slide

Theoretical Probabilities Counting Outcomes Suppose we toss two coins and want to count the total number of outcomes. The toss of the first coin has two possible outcomes: heads (H) or tails (T). The toss of the second coin also has two possible outcomes. The two outcomes for the first coin can occur with either of the two outcomes for the second coin. So the total number of outcomes for two tosses is 2 × 2 = 4; they are HH, HT, TH, and TT, as shown in the tree diagram of Figure 6.4a (next slide). Page 240 Copyright © 2009 Pearson Education, Inc. Slide

Figure 6.4 a Tree diagram showing the outcomes of tossing two coins. TIME OUT TO THINK Explain why the outcomes for tossing one coin twice in a row are the same as those for tossing two coins at the same time. Pages Copyright © 2009 Pearson Education, Inc. Slide

We can now extend this thinking. If we toss three coins, we have a total of 2 × 2 × 2 = 8 possible outcomes: HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT, as shown in Figure 6.4 b. Pages Figure 6.4 b Tree diagram showing the outcomes of tossing three coins. Copyright © 2009 Pearson Education, Inc. Slide

Counting Outcomes Suppose process A has a possible outcomes and process B has b possible outcomes. Assuming the outcomes of the processes do not affect each other, the number of different outcomes for the two processes combined is a × b. This idea extends to any number of processes. For example, if a third process C has c possible outcomes, the number of possible outcomes for the three processes combined is a × b × c. Page 241 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 2 Some Counting How many outcomes are there if you roll a fair die and toss a fair coin? Solution: The first process, rolling a fair die, has six outcomes (1, 2, 3, 4, 5, 6). The second process, tossing a fair coin, has two outcomes (H, T). Therefore, there are 6 × 2 = 12 outcomes for the two processes together (1H, 1T, 2H, 2T, , 6H, 6T). Page 241 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 2 Some Counting What is the probability of rolling two 1’s (snake eyes) when two fair dice are rolled? Solution: Rolling a single die has six equally likely outcomes. Therefore, when two fair dice are rolled, there are 6 × 6 = 36 different outcomes. Of these 36 outcomes, only one is the event of interest (two 1’s). So the probability of rolling two 1’s is Page 241 P(two 1’s) = = = number of ways two 1’s can occur total number of outcomes 1 36 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 3 Counting Children What is the probability that, in a randomly selected family with three children, the oldest child is a boy, the second child is a girl, and the youngest child is a girl? Assume boys and girls are equally likely. Solution: There are two possible outcomes for each birth: boy or girl. For a family with three children, the total number of possible outcomes (birth orders) is 2 × 2 × 2 = 8 (BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG). The question asks about one particular birth order (boy-girl-girl), so this is 1 of 8 possible outcomes. Therefore, this birth order has a probability of 1 in 8, or 1/8 = Page 241 Copyright © 2009 Pearson Education, Inc. Slide

By the Way ... Births of boys and girls are not equally likely. Naturally, there are approximately 105 male births for every 100 female births. However, male death rates are higher than female death rates, so female adults outnumber male adults. Page 241 Copyright © 2009 Pearson Education, Inc. Slide

TIME OUT TO THINK How many different four-child families are possible if birth order is taken into account? What is the probability of a couple having a four-child family with four girls? Page 242 Copyright © 2009 Pearson Education, Inc. Slide

Theoretical Probabilities Relative Frequency Probabilities The second way to determine probabilities is to approximate the probability of an event A by making many observations and counting the number of times event A occurs. This approach is called the relative frequency (or empirical) method. Page 242 Copyright © 2009 Pearson Education, Inc. Slide

Relative Frequency Method
Step 1. Repeat or observe a process many times and count the number of times the event of interest, A, occurs. Step 2. Estimate P(A) by P(A) = number of times A occurred total number of observations Page 242 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE Year Flood Geological records indicate that a river has crested above a particular high flood level four times in the past 2,000 years. What is the relative frequency probability that the river will crest above the high flood level next year? Solution: Based on the data, the probability of the river cresting above this flood level in any single year is Because a flood of this magnitude occurs on average once every 500 years, it is called a “500-year flood.” The probability of having a flood of this magnitude in any given year is 1/500, or number of years with flood total number of years 4 2000 1 500 = Page 242 Copyright © 2009 Pearson Education, Inc. Slide

Theoretical Probabilities Relative Frequency Probabilities Subjective Probabilities The third method for determining probabilities is to estimate a subjective probability using experience or intuition. Page 242 Copyright © 2009 Pearson Education, Inc. Slide

Three Approaches to Finding Probability A theoretical probability is based on assuming that all outcomes are equally likely. It is determined by dividing the number of ways an event can occur by the total number of possible outcomes. A relative frequency probability is based on observations or experiments. It is the relative frequency of the event of interest. A subjective probability is an estimate based on experience or intuition. Page 242 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 5 Which Method? Identify the method that resulted in the following statements. a. The chance that you’ll get married in the next year is zero. b. Based on government data, the chance of dying in an automobile accident is 1 in 7,000 (per year). c. The chance of rolling a 7 with a twelve-sided die is 1/12. Solution: a. This is a subjective probability because it is based on a feeling at the current moment. b. This is a relative frequency probability because it is based on observed data on past automobile accidents. c. This is a theoretical probability because it is based on assuming that a fair twelve-sided die is equally likely to land on any of its twelve sides. Page 243 Copyright © 2009 Pearson Education, Inc. Slide

Probability of an Event Not Occurring Probability of an Event Not Occurring If the probability of an event A is P(A), then the probability that event A does not occur is P(not A). Because the event must either occur or not occur, we can write P(A) + P(not A) = or P(not A) = 1 – P(A) Note: The event not A is called the complement of the event A; the “not” is often designated by a bar, so Ā means not A. Page 244 Copyright © 2009 Pearson Education, Inc. Slide

Probability Distributions Suppose you toss two coins simultaneously. The outcomes are the various combinations of a head and a tail on the two coins. Because each coin can land in two possible ways (heads or tails), the two coins can land in 2 × 2 = 4 different ways. Table 6.2 has a row for each of these four outcomes. Page 245 Copyright © 2009 Pearson Education, Inc. Slide

Notice that the four outcomes represent only three different events: 2 heads (HH), 2 tails (TT), and 1 head and 1 tail (HT or TH). The probability of two heads is P(HH) = 1/4 = 0.25. The probability of two tails is P(TT) = 1/4 = 0.25. The probability of one head and one tail is P(H and T) = 2/4 =0.50. Page 245 Copyright © 2009 Pearson Education, Inc. Slide

These probabilities result in a probability distribution that can be displayed as a table (Table 6.3) or a histogram (Figure 6.6). Page 245 Figure 6.6 Histogram showing the probability distribution for the results of tossing two coins. Copyright © 2009 Pearson Education, Inc. Slide

By the Way ... Another common way to express likelihood is to use odds. The odds against an event are the ratio of the probability that the event does not occur to the probability that it does occur. For example, the odds against rolling a 6 with a fair die are (5/6)/(1/6), or 5 to 1. The odds used in gambling are called payoff odds; they express your net gain on a winning bet. For example, suppose that the payoff odds on a particular horse at a horse race are 3 to 1. This means that for each $1 you bet on this horse, you will gain $3 if the horse wins (and get your original $1 back). Page 245 Copyright © 2009 Pearson Education, Inc. Slide

Making a Probability Distribution A probability distribution represents the probabilities of all possible events. Do the following to make a display of a probability distribution: Step 1. List all possible outcomes. Use a table or figure if it is helpful. Step 2. Identify outcomes that represent the same event. Find the probability of each event. Step 3. Make a table in which one column lists each event and another column lists each probability. The sum of all the probabilities must be 1. Page 245 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 8 Tossing Three Coins Make a probability distribution for the number of heads that occurs when three coins are tossed simultaneously. Solution: We apply the three-step process. Step 1. The number of different outcomes when three coins are tossed is 2 × 2 × 2 = 8. All eight possible outcome are HHH, HHT, HTH, HTT, THH, THT, TTH, and TTT. Step 2. There are four possible events: 0 heads, 1 head, 2 heads, and 3 heads. By looking at the eight possible outcomes, we find the following: • Only one of the eight outcomes represents the event of 0 heads, so its probability is 1/8. • Three of the eight outcomes represent the event of 1 head (and 2 tails): HTT, THT, and TTH. This event therefore has a probability of 3/8. Page 246 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 8 Tossing Three Coins Solution: (cont.) • Three of the eight outcomes represent the event of heads (and 1 tail): HHT, HTH, and THH. This event also has a probability of 3/8. • Only one of the eight outcomes represents the event of 3 heads, so its probability is 1/8. Step 3. We make a table with the four events listed in the left column and their probabilities in the right column. Table 6.4 shows the result. Page 246 Copyright © 2009 Pearson Education, Inc. Slide

TIME OUT TO THINK When you toss four coins, how many different outcomes are possible? If you record the number of heads, how many different events are possible? Page 246 Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 9 Two Dice Distribution Make a probability distribution for the sum of the dice when two dice are rolled. Express the distribution as a table and as a histogram. Solution: Because there are six ways for each die to land, there are 6 × 6 = 36 outcomes of rolling two dice. We enumerate all 36 outcomes in Table 6.5 by listing one die along the rows and the other along the columns. In each cell, we show the sum of the numbers on the two dice. Pages Copyright © 2009 Pearson Education, Inc. Slide

EXAMPLE 9 Two Dice Distribution Solution: (cont.) The possible events are the sums from 2 to 12. These are the events of interest in this problem. We find the probability of each event by counting all the outcomes for each sum and then dividing the number of outcomes by 36. For example, the five highlighted outcomes in Table 6.5 (previous slide) have a sum of 8, so the probability of a sum of 8 is 5/36. Table 6.6 shows the complete probability distribution, and Figure 6.7 (next slide) shows the distribution as a histogram. Pages Copyright © 2009 Pearson Education, Inc. Slide

0.02 0.04 0.06 0.08 0.10 0.12 0.14 0.16 0.18 Probability Sum of two dice 2 3 4 5 6 7 8 9 10 11 12 Page 247 Figure 6.7 Histogram showing the probability distribution for the sum of two dice. Copyright © 2009 Pearson Education, Inc. Slide

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