# Chapter 2 addition - PgEEI#15 Solution Let S be the sample...

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Unformatted text preview: PgEEI #15. Solution Let S be the sample space1then #{S} = 55 since we get 5 possible outcomes by rolling one die. {a} #{{no two alike” : ( g ) X 51 because we have ( g ) ways to choose the points of dioe to be diﬂ'erent from each other and 51 diii'erent orderings 1with respect to each selection.Hence the probability P{no two alike} = £5!- = .0025. {b}Wecanunderstandthat#Honepair”: ( E ) >c: ( g ) x31): ( g ) = 35001then the probability P}one pair} = ﬁn}; = .4530 x g ) x {c} We can understand that #[{two pairs” = ( g ) ( ( i ) = 1300,then the probability P}two pairs} = 1—399 = 2315 {d}Wecan understand that #chree aunt} = ( E x( g ) x ( : )xﬂl = 12001then the probability P}three alike} 2 13”” = .1543 [e] We can understand that #[{full house” 2 ( E ) :a-c ( g ) x ( 5 ) 300,,then the probability P{full house} 2 %'l = .0335 {i} We can understand that #Hﬁour alike” 2 ( E) :a-c ( i) x ( f ) 150,,then the probability P{four alike} 2 Lg; = .0103 {g} We knowthat #“ﬁve alike”: ( E ) = 5,thenthe probabilityP{ﬁve alike} 2 EFF = .0005 Pgﬁ? #30. Solution For each person, there are 5 choices to get accon'lodated in the town. so there are totally 5 x 5 x 5 1i'i'ays for the 3 persons to check into the hotels. Now let‘s think about the pomible ways for the 3 people to check into different hotels, there are ( 5 ways to choose the diﬂ'erent hotels for accornodation and ﬁor 3 each 3 chosen hotels1we have 3! arrangements to get the 3 persons checked into. So there are totally ('15 x 31 = 50 ways for them to check into diii'erent 3 hotels. So the probab ity that 3 persons check into 3 diﬂ'erent hotels is given no _ 12 E! — 25 ‘ {mumptiom each person is equally likely to check into any of the hotels and they arrived at the town separately.) ...
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