Question 1192309
**1. Define the Event:**

* Let's define the event "A" as "selecting at least 3 defective CNG kits out of 4."

**2. Possible Scenarios for Event A:**

* **Scenario 1:** Selecting exactly 3 defective kits.
* **Scenario 2:** Selecting all 4 kits as defective.

**3. Calculate Probabilities for Each Scenario:**

* **Scenario 1 (3 defective, 1 non-defective):**
    * Number of ways to select 3 defective kits from 4 defective kits: 4C3 = 4 
    * Number of ways to select 1 non-defective kit from 8 non-defective kits: 8C1 = 8
    * Total number of ways to select 4 kits from 12 kits: 12C4 = 495
    * Probability of Scenario 1: (4C3 * 8C1) / 12C4 = (4 * 8) / 495 = 32/495

* **Scenario 2 (all 4 defective):**
    * Number of ways to select all 4 defective kits: 4C4 = 1
    * Total number of ways to select 4 kits from 12 kits: 12C4 = 495
    * Probability of Scenario 2: 1C4 / 12C4 = 1 / 495

**4. Calculate the Probability of Event A:**

* Probability of Event A (at least 3 defective) = Probability of Scenario 1 + Probability of Scenario 2
* Probability of Event A = 32/495 + 1/495 = 33/495 = 1/15

**Therefore, the probability of selecting at least 3 defective CNG kits out of 4 selected at random is 1/15.**