Using addition theorem in conditional probability Two balls are drawn from an urn containing 2 white,3 red and 4 black balls one by one without replacement. What is the probability that at least one ball is red?
Your approach is valid, but there is a quicker way -- especially if we decide to increase the number of balls we draw.

Recall that $P(\text{at least one red ball}) = 1-P(\text{no red balls})$.

The probability that we don't draw a red ball the first time is $6/9$, since there are $6$ non-red balls, and $3$ red balls.

Now there are $8$ balls remaining: $5$ non-red balls and $3$ red balls. The probability that I don't draw a red ball the second time is $5/8$.

So $$P(\text{at least one red ball})=1-\left(\frac{6}{9}\right)\left(\frac{5}{8}\right).$$
Use the Hypergeometric distribution. If a random variable $X$ has Hypergeometric distribution: $X \stackrel{d}{=}Hg(n,R,N)$. Where $N$ is the size of the population, $n$ is the sample you take, and $R$ is the number of individuals from the population with the characteristic you are looking for. Then we have:

$$P(X=x)=\frac{\binom{R}{x}\binom{N-R}{n-x}}{\binom{N}{n}}.$$

You need to find $P(X\geq1)=1-P(X=0)$, $X \stackrel{d}{=}Hg(n=2,R=3,N=9).$