Answer is 1/10 .

(read above made easy solution image attached and read Amitabh tiwari comment )

Now My solution-

Made easy solution is correct , if you solve this ( a big calculation) you'll get 1/10 as answer.

So, here calculation is so big so we will solve a small example and try to find out pattern.

Example 1- if i have four numbers 1,2,3,4. then total permutation is 4! .

Now, same condition is here " What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?"

now , we will solve using method

Number of permutations with 2 in first position --> 2 _ _ _ = 3!= 6

Number of permutations with 2 in second position --> _ 2 _ _ = 2 *2!= 4

Number of permutations with 2 in third position --> _ _ 2 _ = 2*1*1= 2

so probability is = 6+4+2 / 4! = 12 /24 = 1/2

So Answer is 1/2 .

Second method as Arjun Sir explained -

There are 2 even numbers (2,4) possible as the one in the earliest position and all of these are equally likely. So, the

probability of 2 becoming the earliest is simply 1/2.

Example 2- if i have Six numbers 1,2,3,4,5,6. then total permutation is 6! .

Now, same condition is here " What is the probability that 2 appears at an earlier position than any other even number in the selected permutation?"

now , we will solve using method

Number of permutations with 2 in first position --> 2 _ _ _ _ _ = 5!

Number of permutations with 2 in second position --> _ 2 _ _ _ _ = 3 *4!

Number of permutations with 2 in third position --> _ _ 2 _ _ _= 3*2*3!

Number of permutations with 2 in Fourth position --> _ _ _ 2 _ _= 3*2*1*2!

so probability is = 5! + 3*4! + 3*2 *3! + 3*2*1*2! / 6! = 40 / 120 = 1/3

So Answer is 1/3 .

Second method as Arjun Sir explained -

There are 3 even numbers (2,4,6) possible as the one in the earliest position and all of these are equally likely. So, the

probability of 2 becoming the earliest is simply 1/3.

Please check ....