Probability - Making a word out of letters 
If I have the letters A, B, C, D, E, F. I am making a world out of 5 letters. X is the number of times that the letter $A$ is in the word. What is the probability that X is Odd?

So the probability of $A$ to be chosen is $1/6$. Then, to be in each position in the word the probability is $(1/6)^5$. The probability to be odd is $P(X=1) + P(X=3) + P(X=5)$. So is it the binomial formula?    
 A: You can think of it as a binomial situation as follows : you are doing $5$ trials , each trial is independent and consists of picking one of $A,B,C,D,E,F$ uniformly. So if we call the appearance of $A$ as a success, then the number of successes in five independent trials, with the probability of success in each  trial being $\frac 16$, is the binomial random variable.(Of course, at the end of five trials, you make a word by putting all the letters together in the order they were picked.)
So the answer would be $P(X=1) + P(X = 3) + P(X = 5)$ where $X \sim \mbox{Bin}(5,\frac 16)$.
A: Note that the binomial expansions of $(5+1)^5$ and $(5-1)^5$ contain the same terms, the only difference being the signs of the terms that contain an odd number of factors of $\pm1$. Thus, $\frac12\left((5+1)^5-(5-1)^5\right)$ yields the sum of the terms in $(5+1)^5$ in which the $1$ occurs an odd number of times. This is the number of ways to form a word out of $5+1$ letters with the $1$ letter appearing an odd number of times, so the answer is $\frac12\left(6^5-4^5\right)\cdot6^{-5}=\frac12-\frac12\left(\frac46\right)^5=\frac{211}{486}$.
A: You are on the right track calculate $P(X=1) + P(X=3) + P(X=5)$. $P(X=k)= 5C_k (1/6)^k(5/6)^{5-k}$.
