1. **Problem Statement:** Two standard dice are rolled. A. What is the probability that you roll doubles 3 times in a row? B. What is the probability of getting a sum of 4, given at least one die shows a 3? C. What is the probability of rolling a sum of 4 or 5? D. What is the probability of rolling a sum of 8 or doubles? E. What is the probability that you roll a sum of 7 or exactly one 2, 4 times in a row? --- 2. **Formulas and Rules:** - Probability of an event $A$ is $P(A) = \frac{\text{number of favorable outcomes}}{\text{total number of outcomes}}$. - For independent events $A$ and $B$, $P(A \text{ and } B) = P(A) \times P(B)$. - Conditional probability: $P(A|B) = \frac{P(A \text{ and } B)}{P(B)}$. - For mutually exclusive events $A$ and $B$, $P(A \text{ or } B) = P(A) + P(B)$. - For non-mutually exclusive events, $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$. --- 3. **Calculations:** **A. Probability of rolling doubles 3 times in a row:** - Probability of doubles in one roll: There are 6 doubles out of 36 total outcomes, so $P(\text{doubles}) = \frac{6}{36} = \frac{1}{6}$. - For 3 independent rolls: $P(\text{doubles 3 times}) = \left(\frac{1}{6}\right)^3 = \frac{1}{216}$. **B. Probability of sum 4 given at least one die shows a 3:** - Total outcomes with at least one 3: Calculate total outcomes where die1=3 or die2=3 minus double counting. - Number of outcomes with at least one 3: $6 + 6 - 1 = 11$ (since (3,3) counted twice). - Outcomes with sum 4 and at least one 3: Possible pairs summing to 4 are (1,3), (3,1), (2,2). Only (1,3) and (3,1) have a 3. - Number of favorable outcomes: 2. - Conditional probability: $P(\text{sum}=4|\text{at least one 3}) = \frac{2}{11}$. **C. Probability of rolling a sum of 4 or 5:** - Sum 4 outcomes: (1,3), (2,2), (3,1) = 3 outcomes. - Sum 5 outcomes: (1,4), (2,3), (3,2), (4,1) = 4 outcomes. - These sets are mutually exclusive. - Total favorable outcomes: $3 + 4 = 7$. - Probability: $\frac{7}{36}$. **D. Probability of rolling a sum of 8 or doubles:** - Sum 8 outcomes: (2,6), (3,5), (4,4), (5,3), (6,2) = 5 outcomes. - Doubles outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes. - Intersection (sum 8 and doubles): (4,4) only = 1 outcome. - Use inclusion-exclusion: $$P(8 \text{ or doubles}) = P(8) + P(\text{doubles}) - P(8 \text{ and doubles}) = \frac{5}{36} + \frac{6}{36} - \frac{1}{36} = \frac{10}{36} = \frac{5}{18}$$ **E. Probability of rolling a sum of 7 or exactly one 2, 4 times in a row:** - Sum 7 outcomes: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) = 6 outcomes. - Exactly one 2 outcomes: Pairs where exactly one die is 2. - Number of outcomes with exactly one 2: For die1=2 and die2≠2: 5 outcomes; die2=2 and die1≠2: 5 outcomes; total 10. - Intersection of sum 7 and exactly one 2: (2,5) and (5,2) = 2 outcomes. - Probability of sum 7 or exactly one 2: $$P = \frac{6}{36} + \frac{10}{36} - \frac{2}{36} = \frac{14}{36} = \frac{7}{18}$$ - For 4 independent rolls: $$P(\text{4 times}) = \left(\frac{7}{18}\right)^4 = \frac{2401}{104976}$$ --- 4. **Final answers:** A. $\boxed{\frac{1}{216}}$ B. $\boxed{\frac{2}{11}}$ C. $\boxed{\frac{7}{36}}$ D. $\boxed{\frac{5}{18}}$ E. $\boxed{\frac{2401}{104976}}$