12. Toss a quarter and then toss a die. What is the probability of: a. Tossing a head and an even number? b. Tossing a head or an even? c. Tossing a head and a three? d. Tossing a head or a three? e. Tossing a tail and a number greater than two? f. Tossing a tail or a number greater than two?

To solve this problem, we need to analyze the outcomes of tossing a quarter and a die. 1. **Known Conditions:** - The quarter has 2 outcomes: Heads (H) or Tails (T). - The die has 6 outcomes: 1, 2, 3, 4, 5, or 6. - The total number of outcomes when tossing both the quarter and the die is \(2 \times 6 = 12\). Now, let's calculate the probabilities for each part: ### a. Tossing a head and an even number - Outcomes for heads: H - Even numbers on the die: 2, 4, 6 - Favorable outcomes: (H, 2), (H, 4), (H, 6) → 3 outcomes \[ P(\text{H and even}) = \frac{3}{12} = \frac{1}{4} \] ### b. Tossing a head or an even number - Outcomes for heads: H - Even numbers on the die: 2, 4, 6 - Total outcomes for heads: 6 (H with each die outcome) - Total outcomes for even numbers: 3 (T with 2, 4, 6) - Overlap (H and even): (H, 2), (H, 4), (H, 6) → 3 outcomes Using the formula for the probability of the union of two events: \[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \] Where: - \(P(A) = \frac{6}{12} = \frac{1}{2}\) (for heads) - \(P(B) = \frac{3}{12} = \frac{1}{4}\) (for even numbers) - \(P(A \cap B) = \frac{3}{12} = \frac{1}{4}\) (for heads and even) Calculating: \[ P(\text{H or even}) = \frac{1}{2} + \frac{1}{4} - \frac{1}{4} = \frac{1}{2} \] ### c. Tossing a head and a three - Favorable outcome: (H, 3) → 1 outcome \[ P(\text{H and 3}) = \frac{1}{12} \] ### d. Tossing a head or a three - Outcomes for heads: 6 (H with each die outcome) - Outcomes for three: 2 (T, 3) and (H, 3) → 1 outcome Using the union formula: \[ P(\text{H or 3}) = P(\text{H}) + P(\text{3}) - P(\text{H and 3}) \] Where: - \(P(\text{H}) = \frac{6}{12} = \frac{1}{2}\) - \(P(\text{3}) = \frac{2}{12} = \frac{1}{6}\) - \(P(\text{H and 3}) = \frac{1}{12}\) Calculating: \[ P(\text{H or 3}) = \frac{1}{2} + \frac{1}{6} - \frac{1}{12} = \frac{6}{12} + \frac{2}{12} - \frac{1}{12} = \frac{7}{12} \] ### e. Tossing a tail and a number greater than two - Outcomes for tails: T - Numbers greater than two on the die: 3, 4, 5, 6 → 4 outcomes - Favorable outcomes: (T, 3), (T, 4), (T, 5), (T, 6) → 4 outcomes \[ P(\text{T and >2}) = \frac{4}{12} = \frac{1}{3} \] ### f. Tossing a tail or a number greater than two - Outcomes for tails: 6 (T with each die outcome) - Outcomes for numbers greater than two: 4 (T with 3, 4, 5, 6) - Overlap (T and >2): (T, 3), (T, 4), (T, 5), (T, 6) → 4 outcomes Using the union formula: \[ P(\text{T or >2}) = P(\text{T}) + P(>2) - P(\text{T and >2}) \] Where: - \(P(\text{T}) = \frac{6}{12} = \frac{1}{2}\) - \(P(>2) = \frac{4}{12} = \frac{1}{3}\) - \(P(\text{T and >2}) = \frac{4}{12} = \frac{1}{3}\) Calculating: \[ P(\text{T or >2}) = \frac{1}{2} + \frac{1}{3} - \frac{1}{3} = \frac{1}{2} \] ### Summary of Probabilities: - a. \(P(\text{H and even}) = \frac{1}{4}\) - b. \(P(\text{H or even}) = \frac{1}{2}\) - c. \(P(\text{H and 3}) = \frac{1}{12}\) - d. \(P(\text{H or 3}) = \frac{7}{12}\) - e. \(P(\text{T and >2}) = \frac{1}{3}\) - f. \(P(\text{T or >2}) = \frac{1}{2}\)