Fair die: Probability of rolling $2$ before rolling $3$ or $5$ Independent trials consisting of rolling a fair die are performed, what is the probability that $2$ appears before $3$ or $5?$
There are $36$ cases if we take two trials like $11 12 13 14 15 16 
..21 22 23 24 25 26..31 32 33 34 35 36$ like this . But two has occurred before , so total $6$ cases , favourable just two$(23 25)$ so ${2\over 6} ={1\over 3}$
what is wrong in this approach , answer given is $3\over 8$ .

Your answer $1/3$ is correct. The easiest way to see it is to consider the first of $2$, $3$, and $5$ to occur.  (With probability $1$, one of them will eventually occur, so there is a first occurrence.) All three of $2$, $3$, and $5$ have the same probability of occurring first (of those three), so that probability must be $1/3$.
