Question

The probability that a leap year, selected at random, contains either 53 Sundays or 53 Mondays, is:

• A. 1/7
• B. 2/7
• C. 3/7
• D. None of these

Hint:

List all 7 possible “extra day” pairs in a leap year and count how many include Sunday or Monday, without double-counting the Sun-Mon pair.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept.
A leap year has 366 days, which is 52 complete weeks (364 days) plus 2 extra days. Every day of the week is guaranteed to occur at least 52 times; whether a day occurs a 53rd time depends only on these 2 extra days.

Step 2: List the possible extra-day pairs.
The 2 extra days must be consecutive days of the week: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun) — 7 equally likely possibilities.

Step 3: Find the probability of 53 Sundays.
Sunday appears in 2 of these 7 pairs: (Sat, Sun) and (Sun, Mon). So \(P(53\text{ Sundays})=\dfrac{2}{7}\).

Step 4: Find the probability of 53 Mondays.
Monday appears in 2 pairs: (Sun, Mon) and (Mon, Tue). So \(P(53\text{ Mondays})=\dfrac{2}{7}\).

Step 5: Avoid double-counting.
The pair (Sun, Mon) gives both 53 Sundays and 53 Mondays at once, and appears in both counts above, so \(P(\text{both})=\dfrac{1}{7}\).

Step 6: Apply the addition rule.
\(P(53\text{ Sun or }53\text{ Mon})=\dfrac{2}{7}+\dfrac{2}{7}-\dfrac{1}{7}=\dfrac{3}{7}\).

Step 7: Final Answer.
The probability is \(\dfrac{3}{7}\), so option C is correct.