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.

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.
