Explanation:

Two cases:
Case 1: 0 at last place So 4 choices for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. 
So numbers = 4*3*2*1 = 24 
Case 2: 6 at last place For 5-digit number 0 cannot be placed at 1st place or cannot be 1st digit 
So 3 choices (1, 3, 5) for 1st digit, 3 for 2nd, 2 for 3rd and 1 for 4th. So numbers = 3*3*2*1 = 18 
So total choices = 24+18 = 42 Number 
total 5-digit numbers that can be formed from 0, 1, 3, 5, and 6 0 not allowed at 1st place 
so 4 choices for 1st place, 4 for 2nd, 3 for 3rd, 2 for 4th and 1 for 5th. 
Sp total = 4*4*3*2*1 = 96 
So required probability = 42/96 = 7/16