Assertion (A) : The probability that a leap year has 53 Mondays is $\frac{2}{7}$ . Reason (R) : The probability that a non-leap year has 53 Mondays is $\frac{5}{7}$ . (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.

Probability

Assertion-Reason: 53 Mondays in a year

Options

(A)Both A and R are true and R is the correct explanation of A

(B)Both A and R are true, but R is not the correct explanation of A

(C)A is true, but R is false

(D)A is false, but R is true

Official Answer

(c) Assertion (A) is true, but Reason (R) is false — a leap year's 2 extra days give $P(53 \text{ Mondays}) = \frac{2}{7}$ , while a non-leap year's 1 extra day gives $P(53 \text{ Mondays}) = \frac{1}{7}$ , not $\frac{5}{7}$ .

leap year 366 days52 weeks plus 2 extra days2/7 probabilitynon-leap year 1 extra day1/7 not 5/7assertion true reason falseoption C

Marking Scheme

11 mark: Correct option (c) selected.
2No partial marks in MCQ; full mark only for (c).
3Reasoning expected (if asked to justify): leap year 2 extra days → $\frac{2}{7}$ (A true); ordinary year 1 extra day → $\frac{1}{7} \ne \frac{5}{7}$ (R false).

Hint

Count the EXTRA days beyond 52 full weeks: leap year leaves 2, ordinary year leaves 1. A given weekday repeats 53 times only if it is among those extra days.

Quick Oral Answer

A leap year has 366 days, that is 52 weeks and 2 extra days, so any chosen weekday like Monday has probability $\frac{2}{7}$ of coming 53 times; a non-leap year has only 1 extra day, so that probability is $\frac{1}{7}$ and not $\frac{5}{7}$ — hence Assertion is true but Reason is false, option (c).

Analysis & Explanation

This Assertion–Reason item probes the classical probability of 53 Mondays in leap vs. non-leap years.

Concept

A year = whole weeks + leftover days; only the leftover days decide if a weekday occurs 53 times.
Leap year: 366 days = 52 weeks + 2 extra days → 7 equally likely consecutive pairs (Sun-Mon, Mon-Tue, ..., Sat-Sun).
Non-leap year: 365 days = 52 weeks + 1 extra day → 7 equally likely single days.

Key points

Monday occurs in 2 of the 7 pairs (Sun-Mon, Mon-Tue) → $P(53 \text{ Mondays in leap year}) = \frac{2}{7}$ , so A is true.
In a non-leap year the single extra day is equally likely to be any of 7 days → $P(53 \text{ Mondays}) = \frac{1}{7}$ , NOT $\frac{5}{7}$ → R is false.
Hence the correct choice is (c): A true, R false.

Common mistakes

Confusing $\frac{5}{7}$ (probability of NOT getting a given weekday 53 times in a leap year) with the non-leap year probability.
Forgetting that leap years have 2 extra days while ordinary years have only 1.

Common Mistakes

1Assuming a non-leap year also has 2 extra days and computing $\frac{2}{7}$ , which wrongly makes R look consistent with A.
2Reading $\frac{5}{7}$ as the probability of getting 53 Mondays instead of the probability of NOT getting 53 of a weekday — leading students to mark (A) or (B).
3Forgetting that the 2 extra leap-year days are CONSECUTIVE, so only pairs containing Monday count (2 out of 7), not any random pair.

Interesting Facts

In a leap year, EXACTLY two different weekdays occur 53 times (the two extra consecutive days), while the other five weekdays occur 52 times — a fact you can verify on any physical calendar.

Because 400 years contain exactly $146097 \text{ days} = 20871 \text{ weeks}$ with zero remainder, the Gregorian calendar's weekday pattern repeats perfectly every 400 years.

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Frequently Asked Questions

Why is the probability of 53 Mondays in a leap year $\frac{2}{7}$ and not $\frac{1}{7}$ ?

A leap year has 366 days = 52 weeks + 2 extra CONSECUTIVE days. These 2 days form 7 equally likely pairs, and Monday appears in two of them — (Sunday, Monday) and (Monday, Tuesday). So 2 favourable out of 7, giving $\frac{2}{7}$ .

What is the correct probability of 53 Mondays in a non-leap year?

A non-leap year has 365 days = 52 weeks + 1 extra day. That single day is equally likely to be any weekday, so the probability of it being Monday (giving 53 Mondays) is $\frac{1}{7}$ — the Reason's value of $\frac{5}{7}$ is wrong.

How do I answer Assertion-Reason questions quickly?

Evaluate the truth of A and R independently first. Here A is true ( $\frac{2}{7}$ is correct) and R is false ( $\frac{5}{7}$ is wrong), so the answer is (c). Only when both are true do you then check whether R actually explains A.