The logic Behind solving this question is quite simple. the even number 2,4,6,8,10,12,14,16,18,20 are 10.
and the possible positions we can place “2” relative to any even number(any permutation of the rest of the 9 numbers). is 10. Now just think about it in a broad sense that if we put the probability of fiding 2 before any even number occurs is X. Then the probability of “2” occuring before the “8 even number and after first even number” will be the X too.
Similarly, 2-9even, 1even-2-8even, 2even-2-7even, 3even-2-6even, 4even-2-5even………... all will have probability X.
total number of such arrangement are 10, so 10X and the one that favours our condition is only one so X.
so the probability would be= X/10X
=1/10
Answer is B