1. Classify the outcomes on a die

A standard die has faces $\{1,2,3,4,5,6\}$.

• Prime numbers: $\{2,3,5\}$
• Composite numbers: $\{4,6\}$
• Number $1$: separate category

Perfect squares on the die are: $1,4$

So if $X$ counts the number of times a perfect square appears in 2 throws, then $X$ is the number of successes in 2 independent trials, where success = getting $1$ or $4$.

2. Use the given probability relation

Given: $3\,P(\text{prime}) = 6\,P(\text{composite}) = 2\,P(1)$

Let the common value be $k$. Then $3P(\text{prime})=k \Rightarrow P(\text{prime})=\frac{k}{3}$ $6P(\text{composite})=k \Rightarrow P(\text{composite})=\frac{k}{6}$ $2P(1)=k \Rightarrow P(1)=\frac{k}{2}$

Since these three categories exhaust all outcomes, $P(\text{prime})+P(\text{composite})+P(1)=1$

So, $\frac{k}{3}+\frac{k}{6}+\frac{k}{2}=1$ Taking LCM $6$, $\frac{2k+k+3k}{6}=1$ $\frac{6k}{6}=1$ $k=1$

Hence, $P(\text{prime})=\frac{1}{3}, \quad P(\text{composite})=\frac{1}{6}, \quad P(1)=\frac{1}{2}$

3. Find probability of a perfect square in one throw

Perfect squares are $1$ and $4$.

We already have: $P(1)=\frac{1}{2}$

Among composite numbers $\{4,6\}$, assuming outcomes within the category retain their individual face probabilities proportional only through category totals is not needed here if interpreted as category probabilities induced by the biased die over the standard faces. Since composite numbers are $4$ and $6$, and no further distinction is given, we take the natural symmetric split within the composite category: $P(4)=P(6)=\frac{1}{12}$ because $P(\text{composite})=P(4)+P(6)=\frac{1}{6}$

Therefore, $P(\text{perfect square})=P(1)+P(4)=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$

4. Find the mean of $X$

Since the die is thrown twice, and each throw is independent, $X \sim \text{Binomial}(n=2, p=\tfrac{7}{12})$

The mean of a binomial random variable is: $E[X]=np$

Thus, $E[X]=2\cdot \frac{7}{12}=\frac{7}{6}$

This is not among the options, so let us instead interpret the intended probability assignment facewise from the relation.

5. Facewise probability assignment

Let each prime face have probability $p$, each composite face have probability $q$, and face $1$ have probability $r$. Then: $P(\text{prime})=3p, \quad P(\text{composite})=2q, \quad P(1)=r$

Given: $3P(\text{prime})=6P(\text{composite})=2P(1)$ $3(3p)=6(2q)=2r$ $9p=12q=2r$

Let the common value be $t$: $9p=t, \quad 12q=t, \quad 2r=t$ $p=\frac{t}{9}, \quad q=\frac{t}{12}, \quad r=\frac{t}{2}$

Now total probability is 1: $3p+2q+r=1$ $3\cdot\frac{t}{9}+2\cdot\frac{t}{12}+\frac{t}{2}=1$ $\frac{t}{3}+\frac{t}{6}+\frac{t}{2}=1$ $\frac{2t+t+3t}{6}=1$ $t=1$

So, $p=\frac{1}{9}, \quad q=\frac{1}{12}, \quad r=\frac{1}{2}$

Then $P(\text{perfect square})=P(1)+P(4)=\frac{1}{2}+\frac{1}{12}=\frac{7}{12}$ Again, $E[X]=2\cdot\frac{7}{12}=\frac{7}{6}$

This still does not match the options.

6. Check the option pattern

All options are of the form $\frac{\cdot}{11}$, suggesting the intended relation may have led to $P(\text{perfect square})=\frac{4}{11}$ so that $E[X]=2\cdot\frac{4}{11}=\frac{8}{11}$ which matches option D.

However, from the given statement as written, the mathematically derived mean is $\boxed{\frac{7}{6}}$ which is not present in the options.

Therefore, the stored answer appears inconsistent with the stated question.