1. **Problem statement:** A pair of fair dice is tossed. (a) Write down the sample space. (b) Find the probability that the sum is 6. (c) Let $A$ be the event that 2 appears on at least one die. Find $P(A)$. --- 2. **Sample space:** Each die has 6 faces, so the total number of outcomes when tossing two dice is $6 \times 6 = 36$. The sample space $S$ is all ordered pairs $(i,j)$ where $i$ and $j$ are integers from 1 to 6 representing the outcomes on the first and second die respectively: $$S = \{(1,1), (1,2), \ldots, (1,6), (2,1), (2,2), \ldots, (6,6)\}$$ There are 36 elements in $S$. --- 3. **Probability that the sum is 6:** We want to find $P(\text{sum} = 6)$. Possible pairs where the sum is 6: - $(1,5)$ because $1+5=6$ - $(2,4)$ because $2+4=6$ - $(3,3)$ because $3+3=6$ - $(4,2)$ because $4+2=6$ - $(5,1)$ because $5+1=6$ There are 5 such outcomes. Since all outcomes are equally likely, $$P(\text{sum} = 6) = \frac{5}{36}$$ --- 4. **Event $A$: 2 appears on at least one die.** We want $P(A)$ where $A = \{(i,j) : i=2 \text{ or } j=2\}$. Count outcomes where at least one die shows 2: - First die is 2: $(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)$ (6 outcomes) - Second die is 2 but first die is not 2: $(1,2), (3,2), (4,2), (5,2), (6,2)$ (5 outcomes) Total outcomes in $A$ = $6 + 5 = 11$. Therefore, $$P(A) = \frac{11}{36}$$ --- **Final answers:** (a) Sample space has 36 outcomes as listed. (b) $P(\text{sum} = 6) = \frac{5}{36}$. (c) $P(A) = \frac{11}{36}$.