SOLUTION: A balanced die is tossed twice. If A=Event (an even number comes up on the first toss), B = Event( an even number comes up on the 2nd toss, and C = Event ( both tosses result in th

Question 1192443: A balanced die is tossed twice. If A=Event (an even number comes up on the first toss), B = Event( an even number comes up on the 2nd toss, and C = Event ( both tosses result in the same number), are the events A B and C
(a) pairwise Independent?
(b) independent?
Answer by CPhill(1987) (Show Source): You can put this solution on YOUR website!
**1. Define Events**
* **A:** Event of an even number on the first toss (A = {2, 4, 6})
* **B:** Event of an even number on the second toss (B = {2, 4, 6})
* **C:** Event of both tosses resulting in the same number (C = {(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)})
**2. Calculate Probabilities**
* **P(A):** Probability of an even number on the first toss = 3/6 = 1/2
* **P(B):** Probability of an even number on the second toss = 3/6 = 1/2
* **P(C):** Probability of both tosses being the same = 6/36 = 1/6
* **P(A ∩ B):** Probability of even numbers on both tosses = 3/6 * 3/6 = 1/4
* **P(A ∩ C):** Probability of even number on first toss and both tosses being the same = 3/36 = 1/12
* **P(B ∩ C):** Probability of even number on second toss and both tosses being the same = 3/36 = 1/12
**3. Check for Pairwise Independence**
* **A and B:**
* P(A ∩ B) = 1/4
* P(A) * P(B) = (1/2) * (1/2) = 1/4
* Since P(A ∩ B) = P(A) * P(B), events A and B are independent.
* **A and C:**
* P(A ∩ C) = 1/12
* P(A) * P(C) = (1/2) * (1/6) = 1/12
* Since P(A ∩ C) = P(A) * P(C), events A and C are independent.
* **B and C:**
* P(B ∩ C) = 1/12
* P(B) * P(C) = (1/2) * (1/6) = 1/12
* Since P(B ∩ C) = P(B) * P(C), events B and C are independent.
**4. Check for Mutual Independence**
* For events to be mutually independent, the following condition must hold:
* P(A ∩ B ∩ C) = P(A) * P(B) * P(C)
* P(A ∩ B ∩ C) = Probability of even numbers on both tosses and both tosses being the same = 3/36 = 1/12
* P(A) * P(B) * P(C) = (1/2) * (1/2) * (1/6) = 1/24
* Since P(A ∩ B ∩ C) ≠ P(A) * P(B) * P(C), the events A, B, and C are **not mutually independent**.
**Conclusion**
* **(a) Pairwise Independent:** Yes, events A, B, and C are pairwise independent.
* **(b) Independent:** No, events A, B, and C are not mutually independent.

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