A natural number n lies between 100 and 400, and the sum of its digits is 10. The probability that n is divisible by 4, is

Let us assume the number as $$xyz$$

Total outcomes = All the numbers between 100 and 400 for which the sum of digits = 10

Favourable outcomes = All the numbers between 100 and 400 for which the sum of digits = 10, and the number is divisible by 4.

$$P=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}$$

Now, for the total outcomes, we have 3 cases as per the hundredth place digit.

Case-1: When the hundredth place digit = 1 (x = 1)

Since the sum of digits = 10, (y + z) = 9. Thus, there are 10 possible pairs for (y,z) = (0,9), (1,8), (2,7),......,(9,0)

=> 10 possibilities

Case-2: When the hundredth place digit = 2 (x = 2)

Since the sum of digits = 10, (y + z) = 8. Thus, there are 9 possible pairs for (y,z) = (0,8), (1,7), (2,6),......,(8,0)

=> 9 possibilities

Case-3: When the hundredth place digit = 3 (x = 3)

Since the sum of digits = 10, (y + z) = 7. Thus, there are 8 possible pairs for (y,z) = (0,7), (1,6), (2,5),......,(7,0)

=> 8 possibilities

For 400, the sum of the digits is 4; thus, we don't need to worry about 400.

Thus, the total outcomes = 10 + 9 + 8 = 27

For the favourable outcomes, we need to check whether the number's last two digits are divisible by 4, that is, yz is divisible by 4 or not.

For Case-1 => the only possible values for yz = 36 and 72. Thus, 2 possibilities.

For Case-2 => the only possible values for yz = 08, 44, and 80. Thus, 3 possibilities.

For Case-3 => the only possible values for yz = 16 and 52. Thus, 2 possibilities.

So, number of favourable outcomes = 2 + 3 + 2 = 7 

=> $$P=\dfrac{Favourable\ Outcomes}{Total\ Outcomes}=\dfrac{7}{27}$$