SOLUTION: 4) Suppose we roll a pair of dice and add the upmost facing numbers. Let E be the event that the sum is even and F be the even that the sum is a multiple of three. Fill in the char

Question 1191575: 4) Suppose we roll a pair of dice and add the upmost facing numbers. Let E be the event that the sum is even and F be the even that the sum is a multiple of three. Fill in the chart then find the required probabilities.
𝑃(𝐸) 𝑃(𝐹) 𝑃(𝐸 𝑎𝑛𝑑 𝐹)
𝑃(𝐸 ∪ 𝐹) 𝑃(𝐹^𝑐)
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Here's how to calculate the probabilities for rolling a pair of dice:
**1. Sample Space:** When rolling two dice, there are 6 x 6 = 36 possible outcomes.
**2. Event E (Sum is Even):** The sums that are even are: 2, 4, 6, 8, 10, 12. There are 18 ways to get an even sum (1-1, 1-3, 1-5, 2-2, 2-4, 2-6, 3-1, 3-3, 3-5, 4-2, 4-4, 4-6, 5-1, 5-3, 5-5, 6-2, 6-4, 6-6). Therefore, P(E) = 18/36 = 1/2 = 0.5
**3. Event F (Sum is a Multiple of 3):** The sums that are multiples of 3 are: 3, 6, 9, 12. There are 12 ways to get a sum that is a multiple of 3 (1-2, 2-1, 1-5, 5-1, 2-4, 4-2, 3-3, 3-6, 6-3, 4-5, 5-4, 6-6). Therefore, P(F) = 12/36 = 1/3 ≈ 0.3333
**4. Event E and F (Sum is Even AND a Multiple of 3):** The sums that meet both criteria are: 6, 12. There are 6 ways to get a sum that is even and a multiple of 3 (1-5, 5-1, 2-4, 4-2, 3-3, 6-6). Therefore, P(E and F) = 6/36 = 1/6 ≈ 0.1667
**5. Event E or F (Sum is Even OR a Multiple of 3):** We use the formula: P(E or F) = P(E) + P(F) - P(E and F) = 1/2 + 1/3 - 1/6 = 2/3 ≈ 0.6667
**6. Event Fᶜ (Sum is NOT a Multiple of 3):** This is the complement of F, so P(Fᶜ) = 1 - P(F) = 1 - 1/3 = 2/3 ≈ 0.6667
**Summary of Probabilities:**
* P(E) = 0.5
* P(F) ≈ 0.3333
* P(E and F) ≈ 0.1667
* P(E or F) ≈ 0.6667
* P(Fᶜ) ≈ 0.6667
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