If a random variable $X$ follows the Binomial distribution $B(10, p)$ such that $5 P(X=0)=P(X=1)$, then the value of $\frac{P(X=5)}{P(X=6)}$ is equal to

Given $X \sim B(10, p)$ with the condition $5P(X=0) = P(X=1)$. The probability $P(X=0) = (1-p)^{10}$ and $P(X=1) = 10p(1-p)^9$.

Substituting into the given relation yields $5(1-p)^{10} = 10p(1-p)^9$. For $p \neq 1$, dividing both sides by $(1-p)^9$ gives $5(1-p) = 10p$, leading to $p = \frac{1}{3}$.

The ratio $\frac{P(X=5)}{P(X-6)}$ simplifies to $\frac{\binom{10}{5}}{\binom{10}{6}} \cdot \frac{1-p}{p}$. Using the identity $\frac{\binom{10}{5}}{\binom{10}{6}} = \frac{6}{5}$ and substituting $p = \frac{1}{3}$, $1-p = \frac{2}{3}$, we obtain:

$\frac{P(X=5)}{P(X=6)} = \frac{6}{5} \cdot \frac{2/3}{1/3} = \frac{6}{5} \cdot 2 = \frac{12}{5}$.

Final answer: $\frac{12}{5}$