A number is selected at random from $100$ to $199$. What is the probability that the number is divisible by $2$ and has exactly two even digits?

Total numbers from $100$ to $199$ = $199-100+1=100$.
Let the three-digit number be represented as $1ab$, where the first digit (hundreds place) is always $1$ (which is an odd digit).
For the number to have exactly two even digits, both $a$ (tens place) and $b$ (units place) MUST be even digits, because the hundreds digit is odd.

Even digits available: $\{0,2,4,6,8\}$ (total $5$ even digits).

Also, the number must be divisible by 2, which means the units digit $b$ must be even. Since $b$ is already required to be even to fulfill the digit condition, any selection where both $a$ and $b$ are even will automatically be divisible by $2$.

Let's calculate the number of favorable combinations for digits $a$ and $b$:

* Number of choices for $a$ (even) = $5$ (can be $0,2,4,6,8$)
* Number of choices for $b$ (even) = $5$ (can be $0,2,4,6,8$)

Total favorable outcomes = $5\times 5=25$.

The required probability is:

$$\text{Probability}=\frac{\text{Favorable Outcomes}}{\text{Total Outcomes}}=\frac{25}{100}=\frac{1}{4}$$