1. Total marbles in the urn

There are:

• $5$ red marbles
• $4$ black marbles
• $3$ white marbles

So total marbles $= 5+4+3 = 12$.

We need the number of ways to draw $4$ marbles such that at most three are red.

2. Interpret “at most three red”

“At most three red” means the number of red marbles drawn can be:

$0,1,2,3$

An easier method is:

• Count total ways to draw any $4$ marbles from $12$
• Subtract the invalid case where all $4$ drawn marbles are red

3. Total number of ways to draw 4 marbles from 12

$\binom{12}{4} = \frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1} = 495$

4. Count the invalid case: all 4 marbles are red

There are $5$ red marbles, and we choose $4$ of them:

$\binom{5}{4} = 5$

5. Required number of ways

$\binom{12}{4} - \binom{5}{4} = 495 - 5 = 490$

6. Final Answer

$\boxed{490}$