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Homework 2, Econ 400

Homework 2, Econ 400 - Economics 70 Book Exercise 2...

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Hit .01 E 1 = HH Hit ? Miss E 2 = HM .09 E 3 = MH Hit Miss .81 Miss E 4 = MM Economics 70 Book Exercise 2 Exercise Title: Book Exercise 2 Error: Reference source not found Date: 11:52:36 Name (Last, First): ANSWER KEY PID: Section No. of Course: Sections 1 & 2 Sign the Honor Pledge: 3.1 This experiment involves tossing a single die and observing the outcome. The sample space for this experiment consists of the following simple events: E 1 : Observe a 1 E 4 : Observe a 4 E 2 : Observe a 2 E 5 : Observe a 5 E 3 : Observe a 3 E 6 : Observe a 6 A through F are compound events and are composed in the following manner: A: (E 2 ) D: Contains no simple events B: (E 1 , E 3 , E 5 ) E: (E 1 , E 2 , E 3 , E 5 ) C: (E 1 , E 2 , E 3 ) F: (E 2 ) To find the probability of an event, we sum the probabilities assigned to the simple events in that event. Since the simple events are equally likely, P(E i ) = 1/6 for i = 1,..6, so that P(A) = 1/6 P(B) = 3/6 = ½ P(C) = 3/6 = ½ P(D) = 0 P(E) = 4/6 = 2/3 P(F) = P(E 2 ) = 1/6 2. The assignment of probabilities is valid in this experiment, since each probability lies between 0 and 1 and since the sum of the four probabilities equals 1. 3.3. In order to have a valid assignment of probabilities, we must have Since P(E 1 ) + P(E 2 ) + P(E 3 ) + P(E 4 ) = 4(.15) = .6, we must have P(E 5 ) = 1 - .6 = .4. 3.4 It is given that P(E 1 ) = P(E 2 ) = .15 and P(E 3 ) = .40. Since , we know that P(E 4 ) + P(E 5 ) = 1 - .15 - .15 - .40 = .30 ( i ) Also, it is given that P(E 4 ) = 2P(E 5 ). ( ii ) 1
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