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?

Let $E_k$ be the event that the $k$th egg is good.
$P[E_2|E_1 ] = {P[E_1 \cap E_2] \over P[E_1]}$.
$P[E_1] = {7 \over 10}$, $P[E_1 \cap E_2] = { \binom{7}{2}\over \binom{10}{2}} = {7 \over 15}$.
Hence $P[E_2|E_1 ] = {10 \over 15} = {2 \over 3}$.
Comment:
The question is ambiguous, I interpreted "one egg is found to be good" as meaning the first checked egg
is good.
Another interpretation is to compute $P[N=2|N\ge 1]$ which is easily
computed to be ${ 1\over 2}$ (where $N$ is the number of good eggs in 
the selection).
However, I think this interpretation is less likely (unless elaborated otherwise) because if we assert $N\ge 1$, it means both eggs must have been
checked, in which case the probability is academic.
