In a simultaneous throw of two dice, what is the probability of getting a total of 10 or 11 ?

### Concept & Logic When throwing two dice, the sample space consists of $6 \times 6 = 36$ equally likely outcomes. To find the probability of a compound event connected by "or" (where events are mutually exclusive), we add the individual favorable outcomes together. $$P(E) = \frac{n(E)}{n(S)}$$ ### Step-by-Step Solution 1. **Identify Sample Space:** The total number of possible outcomes when two dice are thrown is $n(S) = 36$. 2. **Find Favorable Outcomes for Sum 10:** The pairs that sum to 10 are $(4, 6), (5, 5),$ and $(6, 4)$. There are 3 outcomes. 3. **Find Favorable Outcomes for Sum 11:** The pairs that sum to 11 are $(5, 6)$ and $(6, 5)$. There are 2 outcomes. 4. **Calculate Total Favorable Outcomes:** Total ways to get 10 or 11 is $3 + 2 = 5$. So, $n(E) = 5$. 5. **Calculate Probability:** $P(E) = \frac{n(E)}{n(S)} = \frac{5}{36}$ ### Exam Strategy & Shortcut Memorize the frequency of sums for two dice. The number of ways to get a sum $S$ (for $S \ge 7$) is $13 - S$. Ways to get 10: $13 - 10 = 3$. Ways to get 11: $13 - 11 = 2$. Total ways = $5$. Probability = $\frac{5}{36}$. ### Common Pitfall Students sometimes miscount doubles (like $(5,5)$) twice or forget that $(4,6)$ and $(6,4)$ are distinct outcomes, leading to an incorrect count of favorable events. ### Final Answer Therefore, the correct answer is **$\frac{5}{36}$**.