Let A denote the event that a 6-digit integer formed by 0,1,2,3,4,5,6 without repetitions, be divisible by 3 . Then, probability of event A is equal to
Given, digits ={0,1,2,3,4,5,6} Number of ways in which 6 digit number can be formed using these 7 digits =6⋅6! Number of ways ⇒
↓‌↓‌↓‌↓‌↓‌↓
6‌6‌5‌4‌3‌2
( 0 can't be filled here) If the required number is divisible by 3 , then the sum of the digits must be divisible by 3 . Sum of all 7 digits =0+1+2+3+4+5+6 =‌
6â‹…7
2
=21 Now, this further implies we have to remove anyone digit from the given digits. We can remove only the multiple of 3 as the sum of all 7 digits is already 21 . So, removing any other digit will lead to a number of 6 digit which will not be a multiple of 3 . Possible digits ⇒{1,2,3,4,5,6} {0,1,2,4,5,6} {0,1,2,3,4,5} Case I {1,2,3,4,5,6} Number of 6 digit numbers =6! Case II {0,1,2,4,5,6} Case III {0,1,2,3,4,5} Number of 6 digit numbers =5⋅5! ∴ Probability =(‌