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

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

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

The given condition is $9P(X = 4) = P(X = 2)$.

Substituting the probabilities, we get:

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

Since ^{6}C_{4} = ^{6}C_{2} = 15$, we can cancel 15 k^2 (1-k)^2$ from both sides (as $k \neq 0$ and $k \neq 1$):

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

Taking the square root on both sides (since $k > 0$ and $1-k > 0$):

$3k = 1 - k$

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

Now, we need to find $P(X = 3)$:

$P(X = 3) = ^{6}C_{3} k^3 (1-k)^3$

$P(X = 3) = 20 \left(\dfrac{1}{4}\right)^3 \left(\dfrac{3}{4}\right)^3$

$P(X = 3) = 20 \times \dfrac{1}{64} \times \dfrac{27}{64}$

$P(X = 3) = \dfrac{540}{4096} = \dfrac{135}{1024}$

Answer: $\dfrac{135}{1024}$