Step 1:

Let $A$ be the event of not getting a red ball in first draw and $B$ be the event of not getting a red ball in second draw,

Then the required probability=Probability that at least one ball should be red.

$\Rightarrow$ 1-Probability that none is red.

$\Rightarrow 1-P(A\; and\; B)$

$\Rightarrow 1-P(A\cap B)$

$\Rightarrow 1-P(A)P( B/A)$

Step 2:

Let $P(A)$ =Probability of not getting a red ball in first draw.

(i.e)$P(A)$=Probability of getting another colour(white or black) ball in first draw.

$\Rightarrow P(A)=\large\frac{6}{9}=\frac{2}{3}$

Now that are 5 other color balls (white or black) and 3 red balls.

The other color ball can be drawn in $5C_1$ ways

$\therefore P(B/A)=\large\frac{5}{8}$

Step 3:

Hence the required probability is $1-P(A)P(B/A)$

$\Rightarrow 1-\large\frac{2}{3}\times \frac{5}{8}$

$\Rightarrow \large\frac{7}{12}$