The problem you're asking about involves a binomial distribution $X \sim B(n=15, p=\frac{3}{4})$. We need to solve two parts:

(a) $P(X = 8)$
(b) $P(2 \leq X < 6)$

Let's solve these step by step.

(a) $P(X = 8)$

For a binomial distribution, the probability of getting exactly $X = k$ successes in $n$ trials is given by the binomial probability formula:

$P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}$

where:

• $n = 15$ is the number of trials,
• $p = \frac{3}{4}$ is the probability of success on a single trial,
• $k = 8$ is the number of successes.

We can compute:

$P(X = 8) = \binom{15}{8} \left( \frac{3}{4} \right)^8 \left( \frac{1}{4} \right)^{15-8}$

Now let's compute that.

(b) $P(2 \leq X < 6)$

This is asking for the cumulative probability for $X$ between 2 and 5 inclusive. So we need to calculate:

$P(2 \leq X < 6) = P(2 \leq X \leq 5) = P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

We will calculate each of these probabilities using the binomial formula from part (a) and sum them up.

Let me now compute both parts (a) and (b).