What is the probability that the message consists of $2 A$’s , $2 B$’s, and $1 C\;$? A message of length $5$ letters is to be sent. You can pick three letters $A$, $B$ and $C$ to make a message of length $5$. So the total sample space is $243 = 3^5$
Two Questions:

*

*What is the probability of that the message consists of $2 A$’s , $2 B$’s and $1 C$ ?


*And what is the probability that it contains at least $1 A$ in the message ?
First Question:
At first I thought it was ${5\choose2} + {5\choose2}+ 5 = 10 + 10 + 5 = 25$. So the probability  $25/243 = 0.1029 $ or ${5\choose2} + {5\choose2} = 20$, so the probability is $20/243 = 0.082$ but  that wasn’t correct.
Help wanted for both questions.
Hints, proof or answer.

NOTE:  The actual question has been changed.  Please find below an answer to the original question.  You can take this as a 'hint' to solving the new question.  Also, I agree, the new question is best solved by enumerating number of possibilities with "no A's".
The second question asks (asked!) "What if there is one A?".  If so, these are the possibilities:
$$\{A,C,C,C,C\}$$
$$\{ A, B, C,C,C \}$$
$$\{ A, B, B,C,C \}$$
$$\{ A, B,B,B,C\}$$
$$\{A, B,B,B,B\}$$
along with all the permutations of these.
That would be:
$$\begin{aligned}\pmatrix{5\\1}+ \pmatrix{5\\1}\pmatrix{4\\1}+\pmatrix{5\\1}\pmatrix{4\\2}+ \pmatrix{5\\1}\pmatrix{4\\3} + \pmatrix{5\\1} &= 5\cdot\left(1+4+6+4+1\right)\\&=80 \end{aligned}$$
so $$P(\text{one }A)=\frac{80}{243}\approx 0.329218$$
