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# Asst1-solved - Statistics 427 Homework#1 Solutions to...

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Statistics 427: Homework #1 Solutions to Assigned Problems 2.3 [8 pts (2,2,2,2) ] (a) The sample space for this experiment is S = { ( SSS ) , ( SSF ) , ( SFS ) , ( SFF ) , ( FSS ) , ( FSF ) , ( FFS ) , ( FFF ) } where S denotes success and F denotes failure. Event A contains all outcomes with exactly two S ’s: A = { ( SSF ) , ( SFS ) , ( FSS ) } . (b) The event where at least two of the components function contains the outcomes B = { ( SSS ) , ( SSF ) , ( SFS ) , ( FSS ) } . (c) The system will function if component one functions and at least one of components two and three functions. So C = { ( SSS ) , ( SSF ) , ( SFS ) } . (d) Event Outcomes C 0 { ( SFF ) , ( FSS ) , ( FSF ) , ( FFS ) , ( FFF ) } A C { ( SSS ) , ( SSF ) , ( SFS ) , ( FSS ) } A C { ( SSF ) , ( SFS ) } B C { ( SSS ) , ( SSF ) , ( SFS ) , ( FSS ) } B C { ( SSS ) , ( SSF ) , ( SFS ) } 1

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2.11 [6 pts (2,2,2)] (a) You can read this right oﬀ the table: 0 . 07. (b) There are three bond funds, so we compute 0 . 15 + 0 . 10 + 0 . 05 = 0 . 30. (c) Let event A be the event where the selected customer does not own stocks. Then we can just sum up all the possibilities 0 . 20+0 . 15+0 . 10+0 . 05+0 . 07 = 0 . 57. We could have done less arithmetic by ﬁnding the probability A 0 by summing 0 . 18 + 0 . 25 = 0 . 43 and then subtracting 1 - 0 . 42 = 0 . 57. 2.22 [9 pts (3,3,3)] (a) Let B 1 be the event that the person stops at the ﬁrst light and B 2 be the event that the person stops at the second light. To ﬁnd the probability that the person must stop at both signals, we can use P ( B 1 B 2 ) = P ( B 1 ) + P ( B 2 ) - P ( B 1 B 2 ) = 0 . 4 + 0 . 5 - 0 . 6 = 0 . 3 . (b)
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Asst1-solved - Statistics 427 Homework#1 Solutions to...

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