11.4 Tree Diagrams 695 R G O B R G O B R G O B R G O B Sample Space Second Selection First Selection RR RG RO RB BR BG BO BB GR GG GO GB OR OG OO OB R G O B Figure 11.9 Solution a) The first selection can be any of the four balls. Since the first ball is replaced, all four balls remain for the second selection. Thus, there are ⋅ 4 4, or 16, sample points in the sample space. b) The first ball selected can be red, blue, green, or orange. Since this experiment is done with replacement, the same colored ball can be selected twice. The tree diagram and sample space are shown in Fig. 11.9. The sample space contains 16 points. That result checks with the answer obtained in part (a) using the fundamental counting principle. Now try Exercise 21 c) Of the 16 points in the sample space, one has the red ball chosen twice, RR. = P(two red balls) 1 16 d) Of the 16 points in the sample space, nine do not have an orange ball. They are RR, RB, RG, BR, BB, BG, GR, GB, and GG. = P(neither ball selected is orange) 9 16 e) Selecting the orange ball at least once means the orange ball is selected one or more times. There are 7 points in the sample space which list the orange ball at least once (RO, BO, GO, OR, OB, OG, OO). = P(at least one orange ball is selected) 7 16 7 Learning Catalytics Keyword: Angel-SOM-11.4 (See Preface for additional details.) In Example 4, if you add the probability of no orange ball being selected (part d) with the probability of the orange ball being selected at least once (part e), you get + = , 9 16 7 16 16 16 or 1. In any probability problem, if E is a specific event, then either E happens at least one time or it does not happen at all. Thus, P E( happening at least once) + P E( does not happen) 1, = which leads to the following rule. Probability of an Event Happening at Least Once The probability of an event happening at least once can be determined by the following formula. = − P P (event happening at least once) 1 (event does not happen) Applets Simulation Urn Sampling StatCrunch
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