If a random variable $X$ follows the Binomial distribution $B(33, p)$ such that $3 P(X=0)=P(X=1)$, then the variance of $X$ is

A binomial random variable $X$ with parameters $n=33$ and unknown $p$ has its probability mass function defined as $P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}$.

The given condition $3P(X=0) = P(X=1)$ leads to the equation $3(1-p)^{33} = 33p(1-p)^{32}$. Since $p < 1$, we can divide both sides by $(1-p)^{32}$, yielding $3(1-p) = 33p$. Simplifying gives $3 - 3p = 33p$, then $3 = 36p$, so $p = \frac{1}{12}$.

The variance of a binomial distribution is $\text{Var}(X) = np(1-p)$. Substituting $n=33$ and $p=\frac{1}{12}$ gives $\text{Var}(X) = 33 \cdot \frac{1}{12} \cdot \frac{11}{12} = \frac{363}{144} = \frac{121}{48}$.

Final answer: $\boxed{\frac{121}{48}}$