probability question

P(a sum of 2)
= P(1 , 1)
= 1/6 * 1/6
= 1/36
P(a sum of 4)
= P(1 , 3) + P(2 , 2) + P(3 , 1)
= 1/36 + 1/36 + 1/36
= 1/12
P(a sum of 2 is rolled before a sum of 4 is rolled)= (1/36) / (1/12 + 1/36)= 1 / (3 + 1)= 1/4 = 0.25

Alternatively :Consider the first two sum of 2 or 4 rolls can be :
(2 , 2) , (2 , 4) , (4 , 2) or (4 , 4) ,P(E)= P ( (2 , 2) or (2 , 4) ) / P ( (4 , 2) or (4 , 4) )= ( (1/36)(1/36) + (1/36)(1/12) ) / ( (1/12)(1/36) + (1/12)(1/12) )= (1 + 3) / (3 + 9)= 0.25


2011-10-10 14:09:47 補充：
Sorry , the calculation of alternatively is wrong ,

Correction :

Alternatively :

Consider the first two sum of 2 or 4 rolls can be :
(2 , 2) , (2 , 4) , (4 , 2) or (4 , 4) ,

2011-10-10 14:09:52 補充：
P(E)

= P ( (2 , 2) or (2 , 4) ) / [P ( (2 , 2) or (2 , 4) ) + P ( (4 , 2) or (4 , 4) )]

= ( (1/36)(1/36) + (1/36)(1/12) ) / ( (1/36)(1/36) + (1/36)(1/12) + (1/12)(1/36) + (1/12)(1/12) )

= (1 + 3) / (1 + 3 + 3 + 9)

= 4 / 16

= 1/4

2011-10-10 14:10:16 補充：
= 0.25