Q:

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?

A:

Am I right with my understanding?

You're overlooking a subtle point here, one that is often overlooked. You do get the right result, but due to a bit of luck.
The problem is you're overlooking a choice — in addition to having selected a pair of eggs from the basket, you are also choosing to assign the labels "one" and "other" to the two eggs.
Correctly modeling how that choice is made extremely important.
Fortunately, the problem surely means for the choice to be made in the easier-to-understand fashion: either one of the following two equivalent models was used

The labels were assigned randomly, with each of the two choices having equal probability
The eggs were selected not as a pair, but one at a time, and the first egg selected being called "one" and the second egg selected being called "other"

The reason that this is surely right is that the problem appears to indicate either of the following typical procedures, which directly correspond to the two bullet points above:

Take a pair of eggs from the basket
Pick one of the two eggs
Check whether the chosen egg is good or rotten
Guess whether the other is good or rotten

or

Take one egg from the basket
Check whether it is good or rotten
Take another egg from the basket
Guess whether it is good or rotten

An example of a different procedure requiring a different model, incidentally, could be described as follows:

Take a pair of eggs from the basket
Your colleague checks whether they are good or rotten
If both are rotten, things stop here. Otherwise, your colleague tells you one is good
You ask your colleague to point to a good egg. 
You now guess whether the other egg is good or rotten.

(the actual problem being modeled doesn't have to include a colleague; that character can just be a fictional entity that enables the analysis to actually pick out a good egg)

A:

After picking one good egg.
Rotten eggs = 3
Good ones = 6
Total = 9
Probability (second is also good) = $\frac{6}{9}$ = $\frac{2}{3}$  

Edit -

This question has some assumptions also.
How is it found? 
Both are picked together not in succession. 
When you are picking two eggs together there are two ways to select eggs. Do you check a specific one (then how?) and you discover it to be good, or you are told that one of the two eggs is good?  These different interpretations yield different results. So I think question should be more specific to get answer accurately. 
See this link also.

A:

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.

