1. **Problem statement:** You roll a fair six-sided die twice. Calculate the probabilities for the following events: a) Event A: Rolling the number 6 twice. b) Event B: Rolling at least one 2. c) Event C: Rolling a pair (Pasch), i.e., both dice show the same number. 2. **Formulas and rules:** - The probability of an event is the number of favorable outcomes divided by the total number of possible outcomes. - Since the die rolls are independent, the probability of multiple events occurring together is the product of their individual probabilities. - Total possible outcomes when rolling two dice: $6 \times 6 = 36$. 3. **Calculations:** **a) Probability of rolling two 6s:** - Probability of rolling a 6 on one roll: $\frac{1}{6}$. - For two rolls: $$P(A) = \frac{1}{6} \times \frac{1}{6} = \frac{1}{36}.$$ **b) Probability of rolling at least one 2:** - It's easier to calculate the complement: no 2 on either roll. - Probability of not rolling a 2 on one roll: $\frac{5}{6}$. - Probability of no 2 in two rolls: $$\left(\frac{5}{6}\right)^2 = \frac{25}{36}.$$ - Therefore, probability of at least one 2: $$P(B) = 1 - \frac{25}{36} = \frac{11}{36}.$$ **c) Probability of rolling a pair (Pasch):** - Possible pairs: (1,1), (2,2), ..., (6,6) = 6 outcomes. - Total outcomes: 36. - Probability: $$P(C) = \frac{6}{36} = \frac{1}{6}.$$