5.5 PROBABILITY AS A THEORETICAL CONCEPT

Transcription

1 5.5 PROAILIY AS A EOREICAL CONCEP So far, we have solved probability problems by estimating the required probability after conducting some sort of experiment and collecting data. ut probability may be approached from another point of view: in many cases we can calculate the theoretical probability in advance by determining the number of ways a successful outcome can occur and dividing by the total number of possible outcomes in the trial (no data collection needed for this). We have already used this viewpoint in simple cases for example, when we decided which model to use in a given experiment. We first need to discuss the idea of equally likely outcomes. When the outcomes are equally likely, the method for computing the theoretical probability just proposed in the preceding paragraph is justified. First, two definitions. he sample space is the set of all possible outcomes in a trial. An event is merely a collection of outcomes. For example, the event that a tossed die gives an even number consists of the subset of outcomes, 4, 6 from the sample space of all possible outcomes,,, 4, 5, 6. Equally Likely Outcomes Many real-world trials or experiments have a finite set of outcomes possible on each trial. In many situations involving probability, all the outcomes are equally likely. hat is, if the situation were to be repeated a large number of times, each of the outcomes would occur approximately the same proportion of times. A coin toss, for example, has two equally likely outcomes: heads and tails (see Figure 5. a). A roll of an ordinary die has six equally likely outcomes:,,, 4, 5, and 6 (see Figure 5. b). Suppose we seek the theoretical probability of an event, such as the event that a rolled die produces an even number of spots. hen of the possible outcomes (,,, 4, 5, 6) we see that there are three ways (, 4, 6) of

2 (a) wo outcomes of tossing a coin (b) Six outcomes of rolling a die Figure 5. Possible outcomes of tossing a coin and rolling a die. obtaining a successful outcome. y successful we simply mean an outcome that causes the event of interest namely, the die being even to occur. For such probability problems in which we are willing to assume equally likely outcomes, the following rule is extremely useful. heoretical probability of an event when outcomes are equally likely Suppose we have a set of N equally likely number of ways of obtaining outcomes, of which S of them are successes a successful outcome p(event) in the sense that each successful outcome total number of outcomes causes the event to occur. hen the theoretical probability of the event is S N For the toss of a coin, we can say p (heads) 0. 5 For the probability of getting aoraontherollofadie, we can say p( or ) 6 because of the possible outcomes (, ) cause the event to happen and there are 6 possible outcomes. We did not do an experiment to find this probability. Instead, we used some basic ideas about equally likely outcomes and successful outcomes. Observe that we use a lowercase p, as in p (heads) 0. 5

3 to indicate a theoretical probability. Earlier in this chapter we referred to the coin-tossing experiment of Kerrich and found, based on 0,000 simulated tosses of the coin, that P (heads) he capital P shows that we are talking about an experimental probability based on data obtained from a certain number of trials. So P(heads) is an estimate of p(heads). hat is, the experimental probability of getting heads is an estimate of the theoretical probability of getting heads on the toss of a fair coin. Remember this p, P convention; it is used throughout the book! For a roll of a die, the probability of gettingais p() 6 As a slightly more complicated example, suppose we want to find the probability of getting an even number when rolling a die, as was briefly mentioned above. According to Figure 5. b, three of the outcomes give even numbers. So p (even number when rolling a die) Now let s see how to find the theoretical probability of getting two heads in the toss of two coins. Figure 5., called a tree diagram, shows the possible outcomes. he tree diagram helps us keep track of the possible outcomes when there are multiple subtrials, as in the tossing of two coins. Along the top branch of the tree, we find the successful outcome. Along the bottom branch we have one of the other outcomes,, and so on. From this diagram we count a total of four equally likely outcomes; only one of these is two heads. So the theoretical probability is p (two heads) First coin Second coin Possible outcomes heads head 0 heads Figure 5. Possible outcomes of tossing two coins, showing outcome of two heads.

4 First coin Second coin Outcomes One or more heads Figure 5. more heads. Possible outcomes yielding one or First child Second child Possible families girls girl 0 girls One or more girls Figure 5.4 Possible outcomes yielding one or more girls. If we want the theoretical probability of getting one or more heads when tossing two coins, we can count four possible outcomes from the tree diagram (Figure 5.). We find that three of the outcomes give one or more heads. So p (one or more heads) If we want to find the theoretical probability of one or more girls in a two-child family, for example, we can draw a tree diagram that is just like the diagram for the two coins (see Figure 5.4). Again, counting the number of families with one or more girls, we find three. So p (one or more girls) his process can be extended. For example, for a three-child family, as we considered in Section 5., the number of possible family combinations is found as shown in Figure 5.5. You can see that a tree diagram is not required to carry out the counting process, but it is a very helpful tool. We can use a rule to find the total number of outcomes (families, in this case) possible in a trial. In our example, the trial an observation of a three-child family can be divided into three subtrials, one for each child in the family. o find the total number of trials, we find the product of the numbers of possible outcomes of all individual subtrials. Since in our

5 First child Second child hird child Possible families girls girls girl girls girl 0 girls Figure 5.5 Possible family combinations for three children. example each child can be either a boy or a girl and there are three children, we have 8 possible family combinations. Drawing the tree diagram also makes this fact clear. Suppose one of two subtrials can occur in four ways and the second subtrial can occur in three ways. hen drawing a tree diagram would tell us that the total number of possible ways the trial can occur is 4, as we see in Figure 5.6. his rule is sometimes called the multiplication rule for counting the total number of ways a trial made up of subtrials can occur. Subtrial Subtrial a b c d Figure 5.6 Possible outcomes for subtrials and.

6 Returning to the example of the three-child family, we find that the case of three girls happens in one of the eight possible family combinations. So p (three girls) Some problems involving equally likely outcomes require complex reasoning involving permutations, combinations, and other counting formulas. For example, what is the theoretical probability of being dealt five hearts in a five-card hand from an ordinary deck of cards (a nice hand if one is playing poker)? We would need to study permutations and combinations before being able to answer a problem like this. One important theoretical probability model involving equally likely outcomes the binomial probability model will be discussed in Chapter 8. heoretical Independence We have already learned the law of experimental independence, namely that PA ( and ) PA ( ) P ( ) It should not surprise us that this is an approximate empirical stand-in for an analogous law for theoretical probabilities for which exact equality holds. he law of theoretical independence If two events are independent, then the the- If we denote the two independent events oretical probability that they both occur is as A and, then according to this law, exactly equal to the product of the two individual probabilities. pa ( and ) pa ( ) p ( ) Suppose, for example, we toss a coin and roll a six-sided die. What is the theoretical probability of heads on the coin and a6onthedie? Since p(heads) and p(6 on die) 6 and since the tossing of a coin and the rolling of a die are independent, we have p (heads and 6)

7 Example 5.8 Suppose we choose a pair of random digits independently, each taken from the digits 0 to 9. What is the theoretical probability of getting the pair, (that is, first and then )? (Note that this is what sampling two digits from able. amounts to.) Solution We have and So p() 0 p() 0 p( and then ) p() p() his property of independence also applies to more than two events for both estimated and theoretical probabilities. Example 5.9 hree Independent Events Involving Estimated Probabilities Suppose we collect weather data on three cities San Juan, Chicago, and Calcutta and find these estimated probabilities of rain on a certain day: P (rain in San Juan) P (rain in Chicago) P (rain in Calcutta) 0. 4 We make the model assumption that rainfall in each city is independent of rainfall in the other cities. So, using the law of experimental independence, the experimental probability of rain simultaneously in all three cities must obey P(rain in all three cities) P(rain in San Juan) P(rain in Chicago) P(rain in Calcutta) We know that p(rain in all three cities) P (rain in all three cities) hus we have an experimental estimate of p(rain in all three cities), obtained by assuming experimental independence! Let s now consider the theoretical probability of three independent events all occurring together.

8 Example 5.0 hree Independent Events Involving heoretical Probabilities What is the theoretical probability of three heads occurring in three tosses of a fair coin? Solution For each toss, herefore, p(heads) p(three heads) p(heads on first toss) p(heads on second toss) p(heads on third toss) Compare this answer with the one obtained earlier in this section for p(three girls) using a tree diagram (see Figure 5.5) and treating the eight outcomes possible in the trial as equally likely. We can thus reach the same correct answer of 0.5 in two different ways, one using the idea of eight equally likely outcomes and the other using the idea of three independent subtrials. What is the probability of three heads occurring if the coin is loaded with p (heads) 0. 6 on each trial? ere the tree diagram approach fails, and you must use the multiplication rule of theoretical independence. Working with probability as a theoretical concept with rules and formulas for computing the probability of events is further developed in Chapter 8 and in the beginning sections of Chapter 4.