hw2vqn - Vinh Nguyen Stat 120A (Fall 2007): HW 2 Solutions...

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Unformatted text preview: Vinh Nguyen Stat 120A (Fall 2007): HW 2 Solutions October 10, 2007 Text: Probabilty and Statistics , by Degroot and Schervish (3 rd ed) Problems: 1.7.5, 1.7.8, 1.8.6, 1.8.18, 1.9.8, 2.1.9, 2.1.10, 2.2.8, 2.2.16, 2.3.2, 2.3.8 1.7.5 Four dice are rolled and would like to compute the probability that each of the four numbers that appear will be different. We would like the following to occur: Dice Result Comment Possibilities 1 D1 Any of 6 side appears 6 2 D2 Any side that is not D1 5 3 D3 Any side that is not D1 or D2 4 4 D4 Any side that is not D1, D2, or D3 3 Thus, the probability is 6 6 5 6 4 6 3 6 = 5 18 1.7.8 Each of the five elevator passenger will choose one of seven floors to get off. The probability for getting off at a floor is 1 7 for each consumer. We want the probability that the five passengers will choose a different floor to get off of (call this event E). This is exactly like Example 1.7.4 on p.25. The number of possible outcomes for event E is P 7 , 5 and the total number of possible outcomes is 7 5 (each person could get off at any of the 7 floors, and they are all independent). Thus we have p = P 7 , 5 7 5 = 7! (7- 5)!7 5 . 1.8.6 We have n people to be seated at random in a theater row of n seats. We want two particular people A and B to be seated next to each other. Since they are to be seated next to each other, we could view the two people as one when making the seating arrangements. That is, treat the arrangement as arranging n- 1 people. The number of possible arrangements is therefore ( n- 1)!. However, A could be sitting to1)!...
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This note was uploaded on 01/21/2010 for the course STATS 37850 taught by Professor Phelan during the Spring '09 term at UC Irvine.

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hw2vqn - Vinh Nguyen Stat 120A (Fall 2007): HW 2 Solutions...

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