1102 CHAPTER 11 Further Topics in Algebra For example, if the probability of rain can be established as 1 3 , the odds that it will rain are P1rain2 to P1no rain2 = 1 3 to 2 3 = 1 3 2 3 = 1 3 , 2 3 = 1 3 # 3 2 = 1 2 , or 1to2. On the other hand, the odds against rain are 2 to 1 Aor 2 3 to 1 3B . If the odds in favor of an event are, say, 3 to 5, then the probability of the event is 3 8 , and the probability of the complement of the event is 5 8 . Rules for Odds If m represents the number of outcomes in event E and n represents the number of outcomes in event E′, then the following hold true. P1E2 = m m+n and P1E′2 = n m+n The odds in favor of event E are P1E2 P1E′2 = m n , or m to n. The odds against event E are P1E′2 P1E2 = n m , or n to m. EXAMPLE 3 Finding Odds in Favor of an Event A shirt is selected at random from a dark closet containing 6 blue shirts and 4 shirts that are not blue. Find the odds in favor of a blue shirt being selected. SOLUTION Let E represent “a blue shirt is selected.” Then P1E2 = 6 6 + 4 = 6 10 = 3 5 and P1E′2 = 4 6 + 4 = 4 10 = 2 5 . Therefore, the odds in favor of a blue shirt being selected are found as follows. P1E2 P1E′2 = 3 5 2 5 = 3 5 , 2 5 = 3 5 # 5 2 = 3 2 , or 3to2 The odds in favor of a blue shirt being selected are 3 to 2, so we can quickly determine that the odds against selecting a blue shirt are 2 to 3. S Now Try Exercise 23(e). Compound Events A compound event involves an alternative, as in “H or K,” where H and K are events. For example, suppose a fair die is rolled. Let H be the event “the result is a 3,” and K the event “the result is an even number.” From earlier in this section, we have the following. H= 536 K = 52, 4, 66 H´K = 52, 3, 4, 66 P1H2 = 1 6 P1K2 = 3 6 = 1 2 P1H´K2 = 4 6 = 2 3 Notice that in this case, P1H2 + P1K2 = P1H´K2.
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