If one egg is found to be good, then what is the probability that other is also good? A basket contains 10 eggs out of which 3 are rotten. Two eggs are taken out together at random. If one egg is found to be good, then what is the Probability that other is also good?

I applied conditional probability. It says that one of them is good, so the probability of the other one being good can be found in the 9 eggs left out of which 6 are good, so Probability = $6/9$
Am I right with my understanding?

I would interpret the problem statement somewhat differently. The space of all possible results when drawing two eggs of ten is
$$\Omega = \{ \{ x, y \} \subseteq \{ e_1, \ldots, e_{10} \} : x \neq y \}$$
with $|\Omega| = \displaystyle\binom{10}{2} = 45$. Then, let 
$$A = \{ \{ x, y \} \in \Omega : \text{both } x \text{ and } y \text{ are good} \}$$
be the set of results with both eggs good and let 
$$B = \{ \{ x, y \} \in \Omega : \text{at least one of } x, y \text{ is good} \}$$
be the condition that we know holds. The probability in question is therefore
$$P(A|B) = \frac{|A \cap B|}{|B|} = \frac{\binom{7}{2}}{45-\binom{3}{2}} = \frac{21}{42} = \frac{1}{2}.$$
This interpretation assumes that we don't know which one of the eggs is the good one, we just know at least one of them is.
