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.
 A: The number of ways of having two A's, two B's, and one C is
$$\pmatrix{5\\2}\pmatrix{3\\2}=10\cdot 3= 30.$$
There are five positions for the A's.  Once they are filled, there are three positions for the B's.  The C will go in the remaining spot.
So $$p= \frac{30}{3^5} = \frac{30}{243} =\frac{10}{81}=0\overline{.123456790}$$
A: Some hints.

*

*Assume you have an arrangement with the required letters. How many ways can you place the $A$s? From there, how many ways can you place the $B$s? What about the $C$?


*It may be better to count all of the ways that there are no $A$s, and subtract from the total $243$.
Can you make some headway now?
Spoiler:

1) There are $5\choose 2$ ways to place the $A$s. Then there are $3\choose 2$ ways to place the $B$s. The $C$ goes in the last place. So $10 \times 3 \times 1 = 30$ possibilities. 2) There are $2^5 = 32$ messages that contain only $B$s and $C$s, so all of the others contain at least one $A$.

A: 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 $$P(\text{one }A)=\frac{80}{243}\approx 0.329218$$
A: Sometimes I like to approximate answers to such problems by using simulation. Sometimes one can overlook or double-count possibilities using combinatorial methods.
Simulations with a million iterations usually give two or three decimal places of accuracy--good enough to catch some mistakes.
Some initial confusion here due to change in the question, but no errors remain: (+1's to all).
To simulate your problem, I will have three kinds of tokens instead of A, B, and C, Instead, Tokens have numbers 100, 10, and 1. So two As, two Bs and a C means a total of 221. And "At least one A means a total greater than 100.
set.seed(2021)
t = replicate(10^6, 
              sum(sample(c(100,10,1),5,rep=T)))

mean(t==221);  mean(t >= 100)
[1] 0.123124   # two A's, two B'2, one C. Aprx 10/81
[1] 0.868125   # at least one A

mean(t > 100 & t < 200);  mean(t < 100)
[1] 0.329126   # exactly one A.  Aprx 80/243
[1] 0.131875   # no A.  Aprx. (2/3)^5 =  0.1317

Notes on R code.  Vector t has a million totals between 5 and 500. Logical vector t==221 has
a million TRUEs and FALSEs; its mean is the
proportion of its TRUEs. Symbol & means intersection.
