There are three bags containing natural numbers less than 10. First b...
Divisibility rule of 11 says that the alternate sum of the numbers should be 0 or a multiple of 11
Let the number be abc, then a - b + c = 0 or multiple of 11, then abc is divisible by 11
First number is selected from odd numbers, second from all numbers and third from even numbers. Total possible numbers are 5 × 9 × 4 = 180
a is odd and c is even
odd - odd + even = even, so only possible way is when alternate sum is 0
odd - even + even = odd, only possible way is when alternate sum is 11
Case 1:
a = 1, b is odd, c is even -> c-b = -1 ⇒ {b,c} = {(3,2),(5,4),(7,6),(9,8)}
a = 1, b is even, c is even -> c-b = 10, no such combination of numbers
Hence, 4 cases
Case 2:
a = 3, b is odd, c is even -> c-b = -3 ⇒ {b,c} = {(5,2),(7,4),(9,6)}
a = 3, b is even, c is even -> c-b = 8, no such combination of numbers
Hence, 3 cases
Case 3:
a = 5, b is odd, c is even -> c-b = -5 ⇒ {b,c} = {(7,2),(9,4)}
a = 5, b is even, c is even -> c-b = 6 ⇒ {b,c} = {(8,2)}
Hence, 3 cases
Case 4:
a = 7, b is odd, c is even -> c-b = -7 ⇒ {b,c} = {(9,2)}
a = 7, b is even, c is even -> c-b = 4 ⇒ {b,c} = {(8,4),(6,2)}
Hence, 3 cases
Case 5:
a = 9, b is odd, c is even -> c-b = -9, no such combination of numbers
a = 9, b is even, c is even -> c-b = 2⇒ {b,c} = {(8,6),(6,4),(4,2)}
Hence, 3 cases
Total Possible cases are 16 out of 180.
Probability = 16/180 = 4/45
Hence, the correct option is (c).