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 ) :ac ( g ) x ( 5 )
300,,then the probability P{full house} 2 %'l = .0335 {i} We can understand that #Hﬁour alike” 2 ( E) :ac ( 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.) ...
View
Full Document
 Winter '08
 Buckingham
 Addition, Probability

Click to edit the document details