Question

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.

Hint:

In any year, there are at least 52 of every day of the week.
The probability of having a 53rd occurrence of any specific day is:
- \(\frac{1}{7}\) for a non-leap year (because of 1 extra day).
- \(\frac{2}{7}\) for a leap year (because of 2 extra days).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A year consists of 52 full weeks plus some extra days. A leap year has 366 days, and a non-leap year has 365 days.
Step 2: Detailed Explanation:
Evaluating Assertion (A):
Leap year = 366 days = 52 weeks + 2 extra days.
52 weeks contain 52 Mondays. We get a 53rd Monday if one of the 2 extra days is a Monday.
Possible pairs for the 2 extra days:
1. (Sun, Mon)
2. (Mon, Tue)
3. (Tue, Wed)
4. (Wed, Thu)
5. (Thu, Fri)
6. (Fri, Sat)
7. (Sat, Sun)
Total outcomes = 7.
Favorable outcomes (containing Monday) = 2 (Pairs 1 and 2).
\(P(\text{53 Mondays in leap year}) = \frac{2}{7}\).
Thus, Assertion (A) is true.
Evaluating Reason (R):
Non-leap year = 365 days = 52 weeks + 1 extra day.
52 weeks contain 52 Mondays. We get a 53rd Monday if the extra day is a Monday.
Possible extra days: {Sun, Mon, Tue, Wed, Thu, Fri, Sat}.
Total outcomes = 7. Favorable outcome = 1 (Monday).
\(P(\text{53 Mondays in non-leap year}) = \frac{1}{7}\).
The reason states the probability is \(\frac{5}{7}\), which is incorrect.
Thus, Reason (R) is false.
Step 3: Final Answer:
Assertion (A) is true but Reason (R) is false.