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 ${2\over216}$.
Best Answer
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 $\displaystyle\frac{2\cdot3!}{6^3}=\frac{1}{18}$.