Probability

i)Let P(A will certainly win) = x

P(A will certainly win) = P(A win at the first throw) + P(A lose at the 1st throw and B lose at the 2nd throw) * P(A will certainly win) ---(starting at the 3rd throw)

x = 1/6 + (5/6)(5/6)(x)

x - (25x/36) = 1/6

11x/36 = 1/6

x =6/11

ii) P( B will certainly win)

= 1 - P(A win)

= 1 - 6/11

= 5/11

2009-12-04 19:14:12 補充：
method 2 :

P(A win at 1) + P(A win at 3) + P(A win at 5) + ....

i)1/6 + [(5/6)^2](1/6) + [(5/6)^4](1/6) + [(5/6)^6](1/6) + ...

= (1/6)(1 + (5/6)^2 + (5/6)^4 + (5/6)^6 + ......)

= (1/6) ( 1 / [1 - (5/6)^2] )

= (1/6) (36/11)

= 6/11

2009-12-04 19:14:29 補充：
method 2 :

P(A win at 1) + P(A win at 3) + P(A win at 5) + ....

i)1/6 + [(5/6)^2](1/6) + [(5/6)^4](1/6) + [(5/6)^6](1/6) + ...

= (1/6)(1 + (5/6)^2 + (5/6)^4 + (5/6)^6 + ......)

= (1/6) ( 1 / [1 - (5/6)^2] )

= (1/6) (36/11)

= 6/11