Classic Models
A. Pair of Dice (Sample Space $n(S) = 6^2 = 36$):
Outcomes are ordered pairs $(a, b)$ where $a, b \in \{1, 2, 3, 4, 5, 6\}$.
• Sum of numbers = 7: $\{(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)\} \implies n(E) = 6 \implies P = \frac{6}{36} = \frac{1}{6}$.
• Doublets (both numbers same): $\{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)\} \implies n(E) = 6 \implies P = \frac{6}{36} = \frac{1}{6}$.
B. The Leap Year 53 Sundays Problem:
- A leap year contains $366$ days.
- $366 \div 7 = 52 \text{ weeks} + 2 \text{ extra days}$.
- The 52 weeks guarantee exactly 52 Sundays. The 53rd Sunday depends on the 2 extra days!
- Sample space of 2 consecutive days: $\{\text{(Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)}\} \implies n(S) = 7$.
- Favorable outcomes containing a Sunday: $\{\text{(Sat, Sun), (Sun, Mon)}\} \implies n(E) = 2$.
- $$\mathbf{P(\text{53 Sundays in a Leap Year}) = \frac{2}{7}}$$
- For a Non-Leap Year (365 days = 52 weeks + 1 day): $P(\text{53 Sundays}) = \mathbf{\frac{1}{7}}$.