Statistics and Probability with Applications for Engineers and Scientists Using MINITAB, R and JMP. Bhisham C. Gupta

least two heads occur.At most one head occurs.Exactly two heads occur.No head occurs.Find the probability for the occurrence of each event.

      14 Two dice are rolled and the sum of the points that appear on the uppermost faces of the two dice is noted. Write all possible outcomes such that:The sum is seven.The sum is five or less.The sum is even or nine.Find the probability for the occurrence of each event you described in parts (a) through (c).

3.4 Techniques of Counting Sample Points

      The problem of computing probabilities of events in finite sample spaces where equal probabilities are assigned to the elements reduces to the operation of counting the elements that make up the events in the given sample space. Counting such elements is often greatly simplified by the use of a tree diagram and the rules for permutations and combinations.

      3.4.1 Tree Diagram

      A tree diagram is a tool that is useful not only in describing the sample points but also in listing them in a systematic way. The following example illustrates this technique.

Figure 3.4.1 Tree diagram for the experiment in Example 3.3.1

      Example 3.4.1 (Constructing a tree diagram) Consider a random experiment consisting of three trials. The first trial is testing a chip taken from the production line, the second is randomly selecting a part from the box containing parts produced by six different manufacturers, and the third is, again, testing a chip off the production line. The interest in this experiment is in describing and listing the sample points in the sample space of the experiment.

Solution: A tree‐diagram technique describes and lists the sample points in the sample space of the experiment consisting of three trials. The first trial in this experiment has two possible outcomes: the chip could be defective (D) or nondefective (N); the second trial has six possible outcomes because the part could come from manufacturer 1, 2, 3, 4, 5, or 6; and the third, again, has two possible outcomes (D, N). The problem of constructing a tree diagram for a multitrial experiment is sequential in nature: that is, corresponding to each trial, there is a step of drawing branches of the tree. The tree diagram associated with this experiment is shown in Figure 3.4.1.

The number of sample points in a sample space is equal to the number of branches corresponding to the last trial. For instance, in the present example, the number of sample points in the sample space is equal to the number of branches corresponding to the third trial, which is 24 (). To list all the sample points, start counting from o along the paths of all possible connecting branches until the end of the final set of branches, listing the sample points in the same order as the various branches are covered. The sample space S in this example is

      The tree diagram technique for describing the number of sample points is extendable to an experiment with a large number of trials, where each trial has several possible outcomes. For example, if an experiment has n trials and the ith trial has possible outcomes (), then there will be branches at the starting point o, branches at the end of each of the branches, branches at the end of the each of branches, and so on. The total number of branches at the end would be , which represents all the sample points in the sample space S of the experiment. This rule of describing the total number of sample points is known as the Multiplication Rule.

      3.4.2 Permutations

      Suppose that we have n distinct objects . We can determine how many different sequences of x objects can be formed by choosing x objects in succession from the n objects where . For convenience, we may think of a sequence of x places that are to be filled with x objects. We have n choices of objects to fill the first place. After the first place is filled, then with objects left, we have choices to fill the second place. Each of the n choices for filling the first place can be combined with each of the choices for filling the second place, thus yielding ways of filling the first two places. By continuing this argument, we will see that there are ways of filling the x places by choosing x objects from the set of n objects. Each of these sequences or arrangements of x objects is called a permutation of x objects from n. The total number of permutations of x objects from n, denoted by , is given by

      (3.4.1)

      Note that the number of ways of permuting all the objects is given by

      (3.4.2)

      where is read as n factorial.

      Expressed in terms of factorials, we easily find that

      (3.4.3)

      3.4.3 Combinations

It is easy to see that if we select any set of x objects from n, there are ways this particular set of x objects can be permuted. In other words, there are permutations that contain any set of x objects taken from the n objects. Any set of x objects from n distinct objects is called a combination of x objects from n objects. The number of such combinations is usually denoted by .