If $X \sim B(35, p)$ such that $7 P(X=0)=P(X=1)$ then the value of…

Given that $X \sim B(35, p)$ with the condition $7P(X=0) = P(X=1)$, we determine $p$ from the binomial probability mass function:

$7(1-p)^{35} = 35p(1-p)^{34}$

Dividing both sides by $(1-p)^{34}$ (since $0 < p < 1$):
$7(1-p) = 35p$
$7 - 7p = 35p$
$7 = 42p$
$p = \frac{1}{6}$

For the ratio $\frac{P(X=15)}{P(X=20)}$, substitute $p = \frac{1}{6}$:

$\frac{P(X=15)}{P(X=20)} = \frac{\binom{35}{15}\left(\frac{1}{6}\right)^{15}\left(\frac{5}{6}\right)^{20}}{\binom{35}{20}\left(\frac{1}{6}\right)^{20}\left(\frac{5}{6}\right)^{15}}$

Using the identity $\binom{35}{15} = \binom{35}{20}$, the binomial coefficients cancel:
$\frac{P(X=15)}{P(X=20)} = \frac{\left(\frac{1}{6}\right)^{15}\left(\frac{5}{6}\right)^{20}}{\left(\frac{1}{6}\right)^{20}\left(\frac{5}{6}\right)^{15}} = \left(\frac{1}{6}\right)^{-5}\left(\frac{5}{6}\right)^{5} = 6^5 \cdot \left(\frac{5}{6}\right)^5 = 5^5$

$5^5 = 3125$, which corresponds to option B.