Question 1195531
Certainly, let's analyze the probability distribution of selecting I-phones.

**a. Probability Distribution of X in Tabular Form**

* **Possible Values of X:** 
    * X can take on values 0, 1, 2, or 3.

* **Total Phones:** 6 Samsung + 6 iPhones + 6 Techno = 18 phones

* **Calculate Probabilities:**

    * **P(X = 0):** Probability of selecting no iPhones:
        * (12/18) * (11/17) * (10/16) = 0.2710

    * **P(X = 1):** Probability of selecting exactly one iPhone:
        * (6/18) * (12/17) * (11/16) + (12/18) * (6/17) * (11/16) + (12/18) * (11/17) * (6/16) = 0.4662

    * **P(X = 2):** Probability of selecting exactly two iPhones:
        * (6/18) * (5/17) * (12/16) + (6/18) * (12/17) * (5/16) + (12/18) * (6/17) * (5/16) = 0.2268

    * **P(X = 3):** Probability of selecting all three iPhones:
        * (6/18) * (5/17) * (4/16) = 0.0357

* **Probability Distribution Table:**

| X (Number of iPhones) | P(X) |
|---|---|
| 0 | 0.2710 |
| 1 | 0.4662 |
| 2 | 0.2268 |
| 3 | 0.0357 |

**b. Probability of at least one iPhone**

* P(X ≥ 1) = P(X = 1) + P(X = 2) + P(X = 3) 
* P(X ≥ 1) = 0.4662 + 0.2268 + 0.0357 = 0.7287

**c. Probability of no iPhones**

* P(X = 0) = 0.2710 (already calculated in part a)

**d. Expected Number of iPhones (E[X])**

* E[X] = Σ [X * P(X)] 
* E[X] = (0 * 0.2710) + (1 * 0.4662) + (2 * 0.2268) + (3 * 0.0357) 
* E[X] = 0 + 0.4662 + 0.4536 + 0.1071 
* E[X] = 1

**e. Variance of X (Var[X])**

* Var[X] = E[X²] - (E[X])² 
* E[X²] = Σ [X² * P(X)] 
    * E[X²] = (0² * 0.2710) + (1² * 0.4662) + (2² * 0.2268) + (3² * 0.0357) 
    * E[X²] = 0 + 0.4662 + 0.9072 + 0.3213 
    * E[X²] = 1.6947

* Var[X] = 1.6947 - (1)² = 0.6947

**Expected Value of (4X - 2)**

* E[4X - 2] = 4 * E[X] - 2 
* E[4X - 2] = 4 * 1 - 2 = 2

**Summary:**

* Probability Distribution Table (see part a)
* P(X ≥ 1) = 0.7287
* P(X = 0) = 0.2710
* E[X] = 1
* Var[X] = 0.6947
* E[4X - 2] = 2

I hope this comprehensive explanation is helpful!