If you want to read Chnese,let me know.Here I can't put inChines

When we play a die ,Sample space={1,2,3,4,5,6}, P(i)=1/6, i=1,2,3,4,5 or 6.

Event E={1,2,3,}, F={2,6} G={3,4,5,6}.We say E and F are independendt as P(E*F)=P(2)=1/6

P(E)=1/2,P(F)=1/3. P(E)=1/2,P(E*F)=P(E)*P(F)=1/6.E and G are not...Why...?

Back to the star...,A,B,C<D,E,F ..as show up,a,b,c,d,e,f as not.

Sample space= {Adcdef,aBcdef,abCdef,abcDef,abcdEf,abcdeF,abcdef}

P(notA*notB)=P(notA)*P(notB)=.54....so,they are independent......

your Question not clear,..

所有跟帖： 

• why P(notA*notB)=P(notA)*P(notB)? -wushuihe- ♂ (0 bytes) () 02/06/2011 postreply 08:46:37

• 回複：why P(notA*notB)=P(notA)*P(notB)? -jinjing- ♀ (311 bytes) () 02/06/2011 postreply 12:06:40

• why P(abCdef,abcDef,abcdEf,abcdef}=.535? -wushuihe- ♂ (0 bytes) () 02/06/2011 postreply 12:31:45

• P(abCdef,abcDef,abcdEf,abcdeF,abcdef}=.681^2*.319+.681^3*.319+.. -jinjing- ♀ (219 bytes) () 02/06/2011 postreply 17:43:32

• 回複：P(abCdef,..,f}=.681^2*.319+.681^3*.319+..=.464=.681*.681 -jinjing- ♀ (0 bytes) () 02/06/2011 postreply 17:47:52