Note that there are a total of $6 \times 6 \times 6 = 216$ options. The possible sums are from $3$ to $18$.
Also note that the distribution has to be symmetric since if a roll gives $x,y,z$ adding to $n$, then $7-x,7-y,7-z$ adds to $21-n$.
Hence, the number of ways of getting $n$ is same as the number of ways of getting $21-n$. We would hence expect the maximum to occur for $10$ and $11$.
Number of ways to get $3$ is $1$ i.e. $\dbinom{2}{2}$, which is the same as the number of ways to get $21-3=18$.
Number of ways to get $4$ is $3$ i.e. $\dbinom{3}{2}$, which is the same as the number of ways to get $21-4=17$.
Number of ways to get $5$ is $6$ i.e. $\dbinom{4}{2}$, which is the same as the number of ways to get $21-5=16$.
Number of ways to get $6$ is $10$ i.e. $\dbinom{5}{2}$, which is the same as the number of ways to get $21-6=15$.
Number of ways to get $7$ is $15$ i.e. $\dbinom{6}{2}$, which is the same as the number of ways to get $21-7=14$.
Number of ways to get $8$ is $21$ i.e. $\dbinom{7}{2}$, which is the same as the number of ways to get $21-8=13$.
Number of ways to get $9$ is $25$ i.e. $\dbinom{8}{2}-3$, which is the same as the number of ways to get $21-9=12$.
Number of ways to get $10$ is $27$ i.e. $\dbinom{9}{2} - 3 - 6$, which is the same as the number of ways to get $21-10=11$.
Now by symmetry you can get the number of ways to get the sum between $11$ and $18$.
Hence, the distribution peaks at $10$ and $11$ as expected.
As a sanity check, we have $2 \left( 1 + 3 + 6 +10 +15 + 21 + 25 + 27\right) = 216$.
Below is the distribution of the number of times a number occurs as a sum of three dices, where the $X$-axis the sum of the three dice and $Y$-axis is the number of times the sum occurs.
This is correct, other than getting your computation wrong. Check wythatoras' comment for the correct value.
As there is only a single instance of 4 and 5 in the permuted set, 9! does not double count on the possible positions of 4 or 5.
The probability of never rolling 6 is similarly accounted for- if you roll three 1s, two 2s, two 3s, a 4 and a 5, no dice remain unrolled. If all the other conditions are satisfied, not rolling a 6 comes automatically.
Best Answer
$i)$ Since even numbers are twice as likely to appear as any odd numbers.
Note that in a die {$1,2,3,4,5,6$}, Prime$=${$2,3,5$},Non-Prime$=${$1,4,6$}
$ii)$Probability that a prime number appears$$=\frac29+\frac19+\frac19=\frac49$$
$iii)$Probability that odd number appears$$=\frac39 =\frac13$$
$iv)$Probability that an odd prime number appears $(3,5)$
$$=\frac19+\frac19=\frac29$$