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**Hint:**

In this question given that urn has 5 Red, 4 black and 3 White marbles. And we draw 3 balls in a single draw so that at most 3 are red. At most 3 red balls either contain zero red ball, one red ball, 2 red ball or 3 red ball. We can either calculate this by calculating a single case every time. Or we can proceed it by other ways like calculation total selected – no of ways to select 4 balls.

**Complete step by step solution:**

A urn contains 5 red , 4 black and 3 white balls . And in a single draw we get at most 3 red balls.

We have to find the no of ways to select at most 3 red ball

= ( 0 Red ball + other 3 balls ) + ( 1 Red ball + other 3 balls) + ( 2 Red ball + other 2 balls )

+ ( 3 Red ball + 1 other )

= no of ways to select 4 balls – no of select 4 red balls

\[ = {}^{12}{C_4} - {}^5{C_4}\]

\[\begin{array}{l}

= \dfrac{{12 \times 11 \times 10 \times 9}}{{4 \times 3 \times 2 \times 1}} - 5\\

= 495 - 5\\

= 490

\end{array}\]

Hence the required number of ways is 490.

**Note:**

In this type of question students generally confuse at most and at least , more than and less than so take care of it during attempting these types of questions . Almost means max you get 3 red ball in a draw or minimum you can get zero

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