Intro to Stats problem So i have this problem here:

A student is interested in whether or not their fellow students can
  correctly identify​ Coke, Dr.​ Pepper, and Pepsi in a taste test. Each
  subject in the test is given 3 unmarked​ cups: One with​ Coke, one
  with Dr.​ Pepper, and another with Pepsi. After tasting all​ three,
  the subject is asked to identify which cup belongs to which brand.
1) Assuming that they randomly​ guess, what is the probability that they
  correctly identify exactly one of the three​ cups? ?
2) What is the expected number of cups that will be correctly​
  identified? ?

So i believe i have to do whats called a Bri.. Trial. Forgot the name sorry. But here is my calculations to determine the answers.
1) (3 choose 1) * (1/3)^3 * (2/3)^2

(3 choose 1) = 3

3*(1/27)*(4/9) = (4/81)

2) (4/81)*3 = (4/27)

Can anyone check that and make sure im doing it right?
 A: Comment:  This is so easy to solve by writing out the sample space,
that I just couldn't resist. (And, to agree that @JamesPak is 'up-votably' right if his Comment becomes and Answer.)
Suppose the true order of presentation is Coke, Pepsi, Dr. Pepper.
The sample space of possible answers has six elements.
For simplicity, I will use capital letters for correct identifications.
CPD, Cpd, pcD, pdc, dcp, dPc

So there is 1 chance in six of 3 correct IDs, there are 3 chances
of exactly 1 correct ID, and no possibility of getting exactly 
2 correct IDs.
If the random variable $X$ is the number of correct IDs, then
$P(X = 0) = 1/3,\;$ $P(X = 1) = 1/2,\;$ $P(X = 3) = 1/6.$
Hence 
$$E(X) = 0(1/3) + 1(1/2) + 2(0) + 3(1/6) = 1.$$
Note: A serious disadvantage of this experiment is that it does not give someone who claims ability to make correct IDs the chance convincingly
to demonstrate his/her ability. The p-value of a perfect $X = 3$
is 1/6 (not below 5% = 1/20, or anywhere near).
What about a test in which participants are asked to do this
test of ability three times? An individual's score $S$ is the
total number of correct IDs in three trials; perfect score $S = 9.$
Is that experiment designed so that someone has a convincing
chance to show ability? (Under the null hypothesis that a
participant is purely guessing, what is the p-value for $S = 9$, or of the
next best score $S = 7?$)
Added later: I suppose the distribution table of $S$ is as below, where each
'prob' is divided by 216.
 Values:   0  1  2  3  4  5  6  7  9 
 'Probs':  8 36 54 39 36 27  6  9  1 

