Question 1192309: there are 12 CNG kits out of which 4 are defective if 4 kits are selected at random then find the probability that at least 3 are defective??
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! **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.**
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