If a number is chosen at random from out of the four-digit numbers formed by using the digits 0 ,   1 ,…

We have to form four-digit number from the digits $0, 1, 2, 3, 4, 6$ without repetition such that the number is divisible by $4$.

Now, total number of four-digits number formed from $0, 1, 2, 3, 4, 6$ is

$=5\times 5\times 4\times 3=300$

Now, we will form the four-digit number such that it is divisible by $4$. Then, last two digits must be divisible by $4$.

Now, possible choices for last two digits are

$04, 12, 16, 20, 24, 32, 36, 40, 60, 64$.

So, there are total $10$ choices for the last two digits.

Now, if $0 \& 4$ are fixed at last two places, then other places can be filled in $4\times 3=12$ ways.

Now, if $1 \& 2$  are fixed at last two places, then other places can be filled in $3\times 3=9$.

Similarly, for other numbers we get

$$\begin{gathered} 16\to 9 \text{ways} \\ 20\to 12 \text{ways} \\ 24\to 9 \text{ways} \\ 32\to 9 \text{ways} \\ 36\to 9 \text{ways} \\ 40\to 12 \text{ways} \\ 60\to 12 \text{ways} \\ 64\to 9 \text{ways} \end{gathered}$$

So, favourable outcomes$=102$.

Required probability$=\frac{102}{300}=\frac{17}{50}$.