Probability that either the word BAD or CAD can be formed. From the six letters A, B, C, D, E and F, three letters are chosen at random with replacement.What is the probability that either the word BAD or the word CAD can be formed from the chosen letters?

I think this is a easy problem but as I am weak in probability I am not sure whether I have done it correctly or not.

What I have done is probability that letter B is chosen is ${1\over6}$ similarly probability for A and D will also be ${1\over6}$ (due to replacement) and multiply them which give me ${1\over216}$, now multiply it with 2 (for CAD) and my final answer will be ${2\over216}$.
Since there are $6^3$ ways to choose the 3 letters, 

and there are $3!$ ways to obtain B,A,D and similarly for C,A,D,

the probability is $\displaystyle\frac{2\cdot3!}{6^3}=\frac{1}{18}$.
You state that "three letters are chosen at random with replacement", as you also ask "What is the probability that ... can be formed from the chosen letters?" what we want to look at is unordered sampling with replacement.

Take the set $\Omega=\left\{A,B,C,D,E,F\right\}$, then we need to see how many multisets of size 3 can be formed from $\Omega$. It is a known fact that the number of multisets of length $k$ from a set of size $n$ is given by $\binom{n+k-1}{k}$.

From this, we know that there are 56 different possible multisets of length 3 with elements taken from $\Omega$. Now, As each of these multisets has no particular ordering, there is exactly one that corresponds to $BAD$ and exactly one which corresponds to $CAD$, thus, the probability of selecting the words $BAD$ or $CAD$ (i.e. the multisets $\left\{\left\{A,B,D\right\}\right\}$ or $\left\{\left\{A,C,D\right\}\right\}$) is $\frac{2}{56}$.