A jar contains 7 balls numbered from 1 to 7. If a ball is selected at random, and the number is noted down, and a second ball is selected without replacing the first, what is the probability that the two-digit number is even?

My Solution: if the number has to be even then its last digit is either 2,4,6.

Odd Set = {1,3,5,7}

Even Set = {2,4,6}

There are two possibilities to have our desired outcome:

I) Either both balls selected are even numbers therefore (C(1,3)*C(1,2)) / 7*6 = 6/42

II) The first one is odd but the other is even therefore (C(1,4)(1,3)) / 7*6 = 12/42

The result is : 6/42 + 12/42 = 18/42

Is this correct? Is there any easy way? Thank you