Assertion (A) : The probability that a leap year has 53 Mondays is .
Reason (R) : The probability that a non-leap year has 53 Mondays is .
(a) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
Assertion (A) : The probability that a leap year has 53 Mondays is .
Reason (R) : The probability that a non-leap year has 53 Mondays is .
(a) Both, Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both, Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.
Options
(c) Assertion (A) is true, but Reason (R) is false — a non-leap year gives , not .
Marking Scheme
- 11 mark: correctly selecting option (c) — A true, R false; no partial credit in MCQ format.
Hint
Leap year = 52 weeks + 2 extra days (7 pairs, 2 favourable); ordinary year = 52 weeks + 1 extra day (7 outcomes, 1 favourable).
Quick Oral Answer
A leap year has 366 days, giving 2 extra days beyond 52 full weeks, so the probability of 53 Mondays is — the Assertion is true. But a non-leap year has only 1 extra day out of 365, so the probability of 53 Mondays there is , not — making the Reason false.
Analysis & Explanation
Tests probability of extra weekdays in leap vs. non-leap years.
Concept
- A leap year (366 days) = 52 weeks + 2 extra days; a non-leap year (365 days) = 52 weeks + 1 extra day.
- The extra day(s) decide which weekday occurs 53 times.
Key points
- Leap year: 2 extra days form 7 equally likely consecutive pairs — (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun), (Sun,Mon). Monday appears in 2 of these ⟹ . So A is TRUE.
- Non-leap year: the 1 extra day is equally likely to be any of the 7 weekdays, only 1 of which is Monday ⟹ , not . So R is FALSE.
- Hence option (c): A true, R false.
Common mistakes
- Confusing the '2 extra days' rule (leap year) with the '1 extra day' rule (ordinary year).
- Misremembering the ordinary-year probability as 5/7 instead of 1/7.
Common Mistakes
- 1Assuming both a leap year and an ordinary year have the same chance for an extra weekday, ignoring that leap years have 2 extra days while ordinary years have only 1.
- 2Writing the Reason's probability as instead of correctly computing it as , and then wrongly marking option (a) or (b).
- 3Not listing out the 7 possible day-pairs/outcomes and instead guessing the favourable count.
Interesting Facts
A leap year occurs every 4 years (except century years not divisible by 400) specifically to correct the extra ~0.2422 days the Earth takes to orbit the Sun beyond 365 days.
The same '2 extra days' logic used for Mondays in this question applies to finding the probability of 53 of any specific day (e.g., 53 Sundays or 53 Fridays) in a leap year — always .
This assertion-reason format on leap year probability has appeared repeatedly in CBSE board papers since 2015, making it one of the most predictable 1-mark probability questions.
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Frequently Asked Questions
Why does a leap year give 2/7 probability but a non-leap year gives 1/7?
A leap year has 366 days = 52 weeks + 2 extra days, so there are 7 possible pairs of extra days and 2 contain Monday, giving . A non-leap year has 365 days = 52 weeks + 1 extra day, so there are only 7 single-day outcomes and 1 is Monday, giving .
Would the probability change if we asked for 53 Sundays instead of Mondays?
No — by symmetry, the probability of 53 occurrences of any specific weekday in a leap year is always , and in a non-leap year is always .