Lat $A$ be one of the fair coins. Let $p$ be the probability that the number of "heads" among the rest (i.e., the coins $\ne A$) is even. Then the probability for a total number of even "heads" is $$P(A\text{ tails})P(\text{rest even})+P(A\text{ heads})P(\text{rest odd})=\frac12p+\frac12(1-p)=\frac12.$$
Let's denote the event in which the fair coin lands on heads $A$, and the event in which the coin that was picked landed on heads $B$.
We want to find $P$($A$|$B$).
By Bayes' rule, we have that
$$P(A|B)= \frac{P(B|A) \,P(A)}{P(B)}.$$
The event $B$|$A$ is equivalent to the event in which the coin that was picked out of the $6$ landed on heads, given the fact that the fair coin landed on heads.
Therefore,
$$P(B|A)=3/6=1/2.$$
Also, $P$($A$) (the probability that the fair coin landed on heads) = $1$/$2$.
Now, we will calculate $P$($B$) using the Law of Total Probability.
$$P(B)=P(A)\,P(B|A)+P(A^C)\,P(B|A^C)$$
$$=1/2*3/6 + 1/2*2/6$$
$$=5/12.$$
Now, we have all the unknowns necessary to solve for
$$P(A|B)=\frac{P(B|A)\,P(A)}{P(B)}=\frac{\frac 12 \frac 12}{\frac {5} {12}}=\frac 35$$
Best Answer
Pick a coin, suppose it comes up heads with probability $p$. Suppose the other $n-1$ coins give an even number of heads with probability $q$. Then we need $$p(1-q)+(1-p)q=\frac{1}{2}\implies p+q-2pq=\frac{1}{2}\implies(2p-1)(2q-1)=0.$$ So either $p=\frac{1}{2}$ or $q=\frac{1}{2}$. Inducting downwards implies that the probability of an even number of heads is $\frac{1}{2}$ iff at least one of the coins is fair.