![]() If there are 16 flavors, there would be (16) × (15) × (14) ×. By the fundamental counting principle, there are The first position can be filled in ten ways, the second position in 9 ways, and the third position in 8 ways, and so on. Using a tree diagram, we can develop the sample space for an experiment consisting of tossing a fair coin and then rolling a die as follows:Ĭonsider how many ways the owner of an ice cream parlor can display ten ice cream flavors in a row along the front of the display case. Consequently, there are 3 × 2 × 1, or 6, different arrangement. Then, for each outcome in the second stage, there is only one possibility at the third stage. For each outcome at the first stage, there are two possibilities at the second stage. The tree starts with three branches in the first stage, representing the three possibilities for first stage. The following tree diagram list the different ways three different flavors of ice cream, chocolate vanilla and strawberry can be arranged on a cone, with no flavor used more than once. If an operation consists of a sequence of k separate steps of which the first can be performed in ways, followed by the second in ways, and so on until the can be performed in ways, then the operation consisting of k steps can be performed inĪ tree diagram is a device that can be used to list all possible outcomes of a sequence of experiments where each experiment can occur only in a finite number of ways. We can now generalize these results and state them formally as the fundamental principle or multiplication rule of counting. If we let be the event “going from New York to Chicago,” be the event “going from Chicago to Denver,” and be the event “going from Denver to San Francisco,” then because there are four ways to accomplish 3 ways to accomplish and 6 ways to accomplish the number of routes the man can follow is ![]() Suppose a man has four ways to travel from New York to Chicago, three ways to travel from Chicago to Denver, and six ways to travel from Denver to San Francisco, how many ways can he go from New York to San Francisco via Chicago and Denver? To deal with these situations in which the sample space contains a large number of points, we need to have an understanding of basic counting or combinatorial procedures. The total enumeration of the sample space becomes more complicated even if we increase the class size to 6 students. It would be most unusual for a class to consist of only three students. The number of outcomes in the event Then,įrequently, it may be possible to enumerate fully all the sample space points in S and then count how many of these correspond to the event For example, if a class consists of just three students, and the instructor always calls on each student once and only once during each class, then if we label the students 1, 2, and 3, we can easily enumerate the points in S asĪssume that the instructor chooses a student at random, it would seem reasonable to adopt a uniform probability model and assign probability to each point in If A is the event that John is selected last, then To determine the probability of an event we need,Ģ. , then using a uniform probability model, we assign probability for each point in S that is. ![]() For example, in the special instance of a uniform probability function, the probability of an event is known when the number of outcomes that comprises the event is known that is, as soon as the number of outcomes in the subset that defines the event is known. There are many instances in the application of probability theory where it is desirable and necessary to count the outcomes in the sample space and the outcomes in an event.
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