Consider the following for the two (02) items that follow: Let X be...

The probability mass function of a binomial distribution is given by $P(X = r) = ^{n}C_{r} p^r q^{n-r}$, where $q = 1 - p$.

Given $n = 6$, $p = k$, and $q = 1 - k$.

We have $9P(X = 4) = P(X = 2)$.

Substituting the probabilities:

$9 (^{6}C_{4}) k^4 (1 - k)^2 = (^{6}C_{2}) k^2 (1 - k)^4$

Since ^{6}C_{4} = ^{6}C_{2} = 15$, we can cancel it from both sides:<br><br>9 k^4 (1 - k)^2 = k^2 (1 - k)^4$

Assuming $k \neq 0$ and $k \neq 1$, dividing both sides by $k^2 (1 - k)^2$ gives:

$9 k^2 = (1 - k)^2$

Taking the square root of both sides:

$3k = 1 - k \Rightarrow 4k = 1 \Rightarrow k = \dfrac{1}{4}$

The other root $3k = -(1 - k) \Rightarrow 2k = -1 \Rightarrow k = -\dfrac{1}{2}$ is rejected since probability cannot be negative.

Answer: $\dfrac{1}{4}$