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's denote a three-digit number as $abc$ where:

$= a$ is the hundreds digit (1, 2, or 3)

$= b$ is the tens digit (0-9)

$= c$ is the units digit (0-9)

We need $a + b + c = 10$

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When $a = 1$, we need $b + c = 9$

Possible numbers: $109, 118, 127, 136, 145, 154, 163, 172, 181, 190$

When $a = 2$, we need $b + c = 8$

Possible numbers: $208, 217, 226, 235, 244, 253, 262, 271, 280$

When $a = 3$, we need $b + c = 7$

Possible numbers: $307, 316, 325, 334, 343, 352, 361, 370$

In total, there are $10 + 9 + 8 = 27$ numbers between 100 and 400 with the digit sum $= 10$

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A number is divisible by 4 if its last two digits form a number divisible by 4.

Numbers divisible by 4: $136, 172, 208, 244, 280, 316, 352$

That gives us 7 numbers (divisible by 4) out of 27 total numbers.

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$\dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{7}{27}$