Question: 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?
Attempt: probability that exactly one ball is red (red-any ball + any ball-red ball) + probability that both balls are red
$$\left[\frac 39\times\frac 68 + \frac 69\times\frac 38\right]+ \left[\frac 39\times\frac 28\right] $$
This gives the answer as $\frac 7{12}$, which is right but the book has used conditional probability formulas while solving the example. Is my thinking correct here?
Your method is indeed correct - you've done \begin{align} P(\text{at least one red ball}) &= P(1^{\text{st}}\text{red},2^{\text{nd}}\text{not red}) + P(1^{\text{st}}\text{not red},2^{\text{nd}}\text{red}) + P(1^{\text{st}}\text{red},2^{\text{nd}}\text{red}) \\&= \frac{3}{9}\frac{6}{8} + \frac{6}{9}\frac{3}{8} + \frac{3}{9}\frac{2}{8} \\&= \frac{7}{12} \end{align} which accounts for all possibilities involving at least 1 red ball.
I imagine the conditional probabilities you are referring to are: \begin{align} P(\text{at least one red ball}) =& P(1^{\text{st}}\text{red})\times P(2^{\text{nd}}\text{not red}|1^{\text{st}}\text{red})\\& + P(1^{\text{st}}\text{not red})\times P(2^{\text{nd}}\text{red}|1^{\text{st}}\text{not red})\\& + P(1^{\text{st}}\text{red})\times P(2^{\text{nd}}\text{red}|1^{\text{st}}\text{red}) \end{align} which is just the same as above just written differently.