Find the probability that a leap year selected at random will contain $53$ Sundays.
Step-by-Step Solution
Key Concept: A leap year has 366 days $= 52 \text{ weeks} + 2 \text{ extra days}$. The 2 extra days can be (Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun). Total $= 7$ pairs. Favourable pairs $= 2$.
Stepwise Solution:
A leap year has $366$ days $= 52$ complete weeks $+$ $2$ extra days. [1.0 Mark]
Possible pairs for 2 extra days: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Total $= 7$ outcomes. [1.0 Mark]
Favourable outcomes containing Sunday: (Sun, Mon) and (Sat, Sun) $= 2$ outcomes.
$P(53 \text{ Sundays in a leap year}) = \dfrac{2}{7}$. [1.0 Mark]
Marking Scheme:
• Stating 2 extra days in a leap year: 1.0 Mark
• Listing 7 possible day pairs: 1.0 Mark
• Identifying 2 favourable pairs and calculating $P = 2/7$: 1.0 Mark
Correct Answer: