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We are given a random variable \( X \) with the cumulative distribution function \( F(x) \) defined as:

\[
F(x) = 
\begin{cases} 
0, & \text{if } x < 0 \\
1 - \frac{2}{3} e^{-\frac{x}{2}} - \frac{1}{3} e^{-\left\lfloor \frac{x}{2} \right\rfloor}, & \text{if } x \geq 0
\end{cases}
\]

where \( \left\lfloor x \right\rfloor \) denotes the greatest integer less than or equal to \( x \). The problem asks for the value of the probability \( P(4 \leq X < 6) \).

### Step 1: Understanding the distribution function
Since \( F(x) \) is the cumulative distribution function, we can compute probabilities using:
\[
P(a \leq X < b) = F(b) - F(a)
\]
Thus, in this case:
\[
P(4 \leq X < 6) = F(6) - F(4)
\]

### Step 2: Computing \( F(6) \)
For \( x = 6 \), the cumulative distribution function \( F(6) \) will be:
\[
F(6) = 1 - \frac{2}{3} e^{-\frac{6}{2}} - \frac{1}{3} e^{-\left\lfloor \frac{6}{2} \right\rfloor}
\]
\[
F(6) = 1 - \frac{2}{3} e^{-3} - \frac{1}{3} e^{-3}
\]
\[
F(6) = 1 - e^{-3}
\]

### Step 3: Computing \( F(4) \)
For \( x = 4 \), the cumulative distribution function \( F(4) \) will be:
\[
F(4) = 1 - \frac{2}{3} e^{-\frac{4}{2}} - \frac{1}{3} e^{-\left\lfloor \frac{4}{2} \right\rfloor}
\]
\[
F(4) = 1 - \frac{2}{3} e^{-2} - \frac{1}{3} e^{-2}
\]
\[
F(4) = 1 - e^{-2}
\]

### Step 4: Final calculation
Now, we can find \( P(4 \leq X < 6) \) by subtracting:
\[
P(4 \leq X < 6) = F(6) - F(4)
\]
\[
P(4 \leq X < 6) = (1 - e^{-3}) - (1 - e^{-2})
\]
\[
P(4 \leq X < 6) = e^{-2} - e^{-3}
\]

Thus, the correct answer is:

\[
\boxed{e^{-2} - e^{-3}}
\]

This corresponds to option **(a)**.

---

Would you like a deeper explanation of any step? Here are five related questions you might explore:

1. How do cumulative distribution functions (CDFs) relate to probability density functions (PDFs)?
2. What is the meaning of the floor function \( \left\lfloor x \right\rfloor \) in the context of probability distributions?
3. How would you compute the mean or variance of this random variable \( X \)?
4. What is the significance of exponential terms in probability distributions?
5. How can we derive the probability mass function from a given CDF?

**Tip:** Always verify the domain of the random variable when applying a CDF to compute probabilities.

Let X be a random variable with distribution function F(x) where F(x) = 0 if x < 0, and F(x) = 1 - 2/3 * e^(-x/2) - 1/3 * e^(-floor(x/2)) for x ≥ 0. Find P(4 ≤ X < 6).

This problem explores finding the probability P(4 ≤ X < 6) for a random variable X with a cumulative distribution function that includes exponential terms. Detailed steps include using the CDF to compute the probability and understanding the exponential nature of the distribution.

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