Question 1202002
**1. Determine the Probability Mass Function (PMF) of X**

* **Good Mood (Probability q = 0.58):** Sends "11" 
    * P(X=2) = 0.58 
* **Bad Mood (Probability 1-q = 0.42):** Sends "01" 
    * P(X=1) = 0.42

**2. Determine the Possible Received Messages (Y) and their Probabilities**

* **"11" Sent:**
    * Received "11": Probability = 0.58 * (1-p) * (1-p) = 0.58 * 0.74 * 0.74 = 0.3175 
    * Received "10": Probability = 0.58 * (1-p) * p = 0.58 * 0.74 * 0.26 = 0.1129 
    * Received "01": Probability = 0.58 * p * (1-p) = 0.58 * 0.26 * 0.74 = 0.1129
    * Received "00": Probability = 0.58 * p * p = 0.58 * 0.26 * 0.26 = 0.0392

* **"01" Sent:**
    * Received "11": Probability = 0.42 * p * (1-p) = 0.42 * 0.26 * 0.74 = 0.0808
    * Received "10": Probability = 0.42 * p * p = 0.42 * 0.26 * 0.26 = 0.0283 
    * Received "01": Probability = 0.42 * (1-p) * p = 0.42 * 0.74 * 0.26 = 0.0808
    * Received "00": Probability = 0.42 * (1-p) * (1-p) = 0.42 * 0.74 * 0.74 = 0.2289

**3. Calculate E[X], E[Y], and E[XY]**

* **E[X] (Expected value of X):** 
    * E[X] = (2 * 0.58) + (1 * 0.42) = 1.58

* **E[Y]:** 
    * E[Y] = (2 * 0.3175) + (1 * 0.1129 + 0.1129 + 0.0808 + 0.0808) + (0 * 0.0392 + 0.0283 + 0.2289) 
    * E[Y] = 0.635 + 0.3874 + 0.2572 = 1.2796

* **E[XY]:** 
    * E[XY] = (2 * 2 * 0.3175) + (2 * 1 * 0.1129) + (1 * 2 * 0.0808) + (1 * 1 * 0.0808) + (0 * 2 * 0.0392) + (0 * 1 * 0.0283) + (0 * 1 * 0.2289) 
    * E[XY] = 1.27 + 0.2258 + 0.1616 + 0.0808 
    * E[XY] = 1.7382

**4. Calculate Covariance**

* **Cov(X, Y) = E[XY] - E[X] * E[Y]**
* Cov(X, Y) = 1.7382 - (1.58 * 1.2796) 
* Cov(X, Y) = 1.7382 - 2.0215 
* Cov(X, Y) = -0.2833

**Therefore, the covariance between X (number of ones sent) and Y (number of ones received) is -0.2833.**

This negative covariance indicates that there is a tendency for the number of ones in the sent message to be inversely related to the number of ones in the received message.