lamin 發表於 2004-12-19 21:22:09

probability

box a contains three white balls and one red ball while box b contains four white balls.One ball is randomly drawn from each bow and the two balls are then randomly put back into the boxes so that each box still contains four balls.This process is performed n times.Let P(K) be the probability that the red ball is in box A after the process is performed k times,where
1<=k<=n(<=係大過等於)


a)find P(1)
ans: 1/8+3/4=7/8

b)Show that P(K)=(3/4) P(K-1)+1/8    ,where 2<=k<=n
hence find P(K) in terms of K

ans: (1/8+3/4)P(K-1)+1/8(1-P(K-1))
=3/4(P(k-1)+1/8

跟住點find P(k) in terms of k
唔該教下我:icon003:

CYBASTER 發表於 2004-12-19 21:48:57

好易ja....
你將P(k-1)變做P(k-2)....變到P(1)

呢d係applied?

jeff 發表於 2004-12-20 01:41:11

好想幫你
如果係半年前ge我應該唔駛諗就答到
不過宜家ge我真係完完全全忘記曬:icon069:

lamin 發表於 2004-12-20 16:35:19

CYBASTER在 19-12-2004 09:48 PM 發表:

好易ja....
你將P(k-1)變做P(k-2)....變到P(1)

呢d係applied?


yes

lamin 發表於 2004-12-20 17:52:32

CYBASTER在 19-12-2004 09:48 PM 發表:

好易ja....
你將P(k-1)變做P(k-2)....變到P(1)

呢d係applied?



甘係咪變成(3/4)^k-1*(1/8)^k+1/8

但個答案係(3/8)(3/4)^k-1   +1/2

Newcastle Fans 發表於 2004-12-20 17:57:26

嚇死... 我以為MS有D咁既野 :icon101:

lamin 發表於 2004-12-20 18:11:51

Newcastle Fans在 2004-12-20 05:57 PM 發表:

嚇死... 我以為MS有D咁既野 :icon101:


的確係ms

我係本new way本ms既revision exercise 搵到

係applied 既past paper,佢印得係到,應該我地都岩做掛

Newcastle Fans 發表於 2004-12-20 18:26:51

lamin在 2004-12-20 18:11 發表:

的確係ms

我係本new way本ms既revision exercise 搵到

係applied 既past paper,佢印得係到,應該我地都岩做掛

我阿sir話係出書o個D人見好似有關係就求其整埋入去......
大把文科人讀MS... 應該唔會出到咁深掛 :icon101:

CYBASTER 發表於 2004-12-20 20:25:23

咁即係你讀緊applied定ms?

CYBASTER 發表於 2004-12-20 20:31:48

lamin在 2004-12-20 05:52 PM 發表:

甘係咪變成(3/4)^k-1*(1/8)^k+1/8

但個答案係(3/8)(3/4)^k-1   +1/2

P(K)=(3/4) P(K-1)+1/8
=1/8+(3/4)((3/4)P(K-2)+1/8)
=1/8+(3/4)(1/8)+[(3/4)^2]P(K-2)
=1/8+(3/4)(1/8)+[(3/4)^2][(3/4)P(K-3)+1/8)]
=1/8+(3/4)(1/8)+[(3/4)^2](1/8)+[(3/4)^3]P(K-3)
=1/8+(3/4)(1/8)+[(3/4)^2](1/8)+...+[(3/4)^(k-2)](1/8)+[(3/4)^(k-1)]P(1)

CYBASTER 發表於 2004-12-20 20:38:08

P(K)
=1/8+(3/4)(1/8)+[(3/4)^2](1/8)+...+[(3/4)^(k-2)](1/8)+[(3/4)^(k-1)]P(1)
=1/8+(3/4)(1/8)+[(3/4)^2](1/8)+...+[(3/4)^(k-2)](1/8)+[(3/4)^(k-1)](7/8)
=1/8(1-3/4^(k-1))/(1-3/4)+[(3/4)^(k-1)](7/8)
=1/2(1-3/4^(k-1))+(7/8
=1/2+(-1/4+7/8)[(3/4)^(k-1)]
=1/2+(3/8)[(3/4)^(k-1)]

:D

Newcastle Fans 發表於 2004-12-20 20:40:03

CYBASTER在 2004-12-20 20:25 發表:

咁即係你讀緊applied定ms?

我學校連a.maths都摺埋 :icon101:

CYBASTER 發表於 2004-12-20 20:43:02

Newcastle Fans在 2004-12-20 08:40 PM 發表:

我學校連a.maths都摺埋 :icon101:

下?:confused:

lamin 發表於 2004-12-20 20:52:07

CYBASTER在 20-12-2004 08:25 PM 發表:

咁即係你讀緊applied定ms?


ms
其實呢題唔讀applied maths 識唔識

Newcastle Fans 發表於 2004-12-20 20:56:54

CYBASTER在 2004-12-20 20:43 發表:

下?:confused:

連a.maths都摺埋... 咁緊係讀緊ms啦
PS: a.maths = applied maths

CYBASTER 發表於 2004-12-20 20:57:26

lamin在 2004-12-20 08:52 PM 發表:

ms
其實呢題唔讀applied maths 識唔識

應該都識ga

CYBASTER 發表於 2004-12-20 20:58:39

Newcastle Fans在 2004-12-20 08:56 PM 發表:

連a.maths都摺埋... 咁緊係讀緊ms啦
PS: a.maths = applied maths

即係冇得修applied?

Newcastle Fans 發表於 2004-12-20 21:04:11

CYBASTER在 2004-12-20 20:58 發表:

即係冇得修applied?

係呀
我記得兩三年前都仲有既... 得三個人讀 :icon101:
我成日諗有冇人係3AL+1AS...
讀AL Pure Maths + AL Physics + AL Applied Maths + AS MS :icon101:

CYBASTER 發表於 2004-12-20 21:05:01

Newcastle Fans在 2004-12-20 09:04 PM 發表:

係呀
我記得兩三年前都仲有既... 得三個人讀 :icon101:
我成日諗有冇人係3AL+1AS...
讀AL Pure Maths + AL Physics + AL Applied Maths + AS MS :icon101:

咁應該唔會太辛苦ga ja ~~:icon064:

lamin 發表於 2004-12-20 21:06:23

CYBASTER在 2004-12-20 08:38 PM 發表:

P(K)
=1/8+(3/4)(1/8)+[(3/4)^2](1/8)+...+[(3/4)^(k-2)](1/8)+[(3/4)^(k-1)]P(1)
=1/8+(3/4)(1/8)+[(3/4)^2](1/8)+...+[(3/4)^(k-2)](1/8)+[(3/4)^(k-1)](7/8)
=1/8(1-3/4^(k-1))/(1-3/4)+[(3/4)^(k-1)](7/8) ...

唔該
終於明:icon012:
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