Question 1193590
**1. Determine the Probability Mass Function (PMF) of X**

* **Possible Values of X:** 
    * X = 0 (no heads): TTT 
    * X = 1 (one head): HTT, THT, TTH
    * X = 2 (two heads): HHT, HTH, THH
    * X = 3 (three heads): HHH

* **Probabilities:**
    * P(X = 0) = 1/8 
    * P(X = 1) = 3/8 
    * P(X = 2) = 3/8 
    * P(X = 3) = 1/8

**2. Define the Cumulative Distribution Function (CDF)**

* The CDF, denoted as F(x), gives the probability that the random variable X is less than or equal to a specific value (x).

**3. Calculate the CDF for Each Possible Value of X**

* **F(x) = P(X ≤ x)**

    * F(x) = 0 for x < 0 
    * F(x) = P(X = 0) = 1/8 for 0 ≤ x < 1
    * F(x) = P(X = 0) + P(X = 1) = 1/8 + 3/8 = 1/2 for 1 ≤ x < 2
    * F(x) = P(X = 0) + P(X = 1) + P(X = 2) = 1/8 + 3/8 + 3/8 = 7/8 for 2 ≤ x < 3
    * F(x) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1/8 + 3/8 + 3/8 + 1/8 = 1 for x ≥ 3

**Therefore, the cumulative distribution function (CDF) of X is:**

* F(x) = 0 for x < 0
* F(x) = 1/8 for 0 ≤ x < 1
* F(x) = 1/2 for 1 ≤ x < 2
* F(x) = 7/8 for 2 ≤ x < 3
* F(x) = 1 for x ≥ 3