If in $6$ trials, X is a binomial random variable which follows the relation $9P(x = 4) = P(x = 2)$, then the probability of failure is...

Let $p$ be the probability of success and $q$ be the probability of failure in a single trial, where $p + q = 1$.

For a binomial distribution with $n = 6$ trials, the probability of $x$ successes is given by $P(X = x) = ^{6}C_{x} p^{x} q^{6-x}$.

Given the relation:

$9P(X = 4) = P(X = 2)$

Substituting the binomial probabilities:

$9 \times ^{6}C_{4} p^{4} q^{2} = ^{6}C_{2} p^{2} q^{4}$

Since $^{6}C_{4} = ^{6}C_{2} = 15$, we can cancel it from both sides:

$9 p^{4} q^{2} = p^{2} q^{4}$

Dividing both sides by $p^{2} q^{2}$ (assuming $p, q \neq 0$):

$9 p^{2} = q^{2}$

Taking the square root on both sides (since probabilities are non-negative):

$3p = q$

We know that $p + q = 1$. Substituting $p = \dfrac{q}{3}$:

$\dfrac{q}{3} + q = 1$

$\dfrac{4q}{3} = 1$

$q = \dfrac{3}{4} = 0.75$

Thus, the probability of failure is $0.75$.