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

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      As and are disjoint events, then from condition 3, we obtain

      (3.3.3)

      But since and , we have the following:

      Theorem 3.3.1 (Rule of complementation) If E is an event in a sample space S, then

(3.3.4)

      The law of complementation provides a simple method of finding the probability of an event , if E is an event whose probability is easy to find. We sometimes say that the odds in favor of E are

      (3.3.4a)

which from 3.3.4 takes the form . The reader may note that .

Example 3.3.1 (Tossing coins) Suppose that 10 coins are tossed and we ask for the probability of getting at least 1 head. In this example, the sample space S has sample points. If the coins are unbiased, the sample points are equally likely (sample points are called equally likely if each sample point has the same probability of occurring), so that to each of the sample points the probability is assigned. If we denote by E the event of getting no heads, then E contains only one sample point, and , of course, has 1023 sample points. Thus

      The odds on E and are clearly and .

      Referring to the statement in Theorem 3.3.1 that and are disjoint events whose union is S, we have the following rule.

      Theorem 3.3.2 (General rule of complementation) If are events in a sample space S, then we have

      (3.3.5)

      Another useful result follows readily from (3.3.2c) by mathematical induction

      Theorem 3.3.3 (Rule of addition of probabilities for mutually exclusive events) If are disjoint events in a sample space S, then

(3.3.6)

      Example 3.3.2 (Determination of probabilities of some events) Suppose that a nickel and a dime are tossed, with H and T denoting head and tail for the nickel and h and t denoting head and tail for the dime. The sample space S consists of the four elements Hh, Ht, Th, and Tt. If these four elements are all assigned equal probabilities and if E is the event of getting exactly one head, then , and we have that

      Now suppose that and are arbitrary events in S. Then from Figure 3.2.2, with and , it can be easily seen that , are three disjoint events whose union is . That is,

(3.3.7)

Also, and are disjoint sets whose union is . Hence,

(3.3.8)

      Similarly

(3.3.9)

      Solving (3.3.8) for and (3.3.9) for and substituting in (3.3.7), we obtain the following.

      Theorem 3.3.4 (Rule for addition of probabilities for two arbitrary events) If and are any two events in a sample space S, then

(3.3.10)

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