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

To determine the truthfulness of the Assertion (A) and Reason (R), let's analyze them separately:

1. Assertion (A): The probability that a leap year has 53 Mondays is \(\frac{2}{7}\).

A leap year has 366 days, which is 52 weeks and 2 extra days. These extra days could be any of the following combinations: (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), or (Saturday, Sunday).

Out of these 7 possibilities, "Monday" appears in 2 of them: (Sunday, Monday) and (Monday, Tuesday).

Thus, the probability that a leap year has 53 Mondays is \(\frac{2}{7}\), making Assertion (A) true.

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

A non-leap year has 365 days, equaling 52 weeks and 1 extra day. This extra day can be any of the 7 days in a week.

For a non-leap year to have 53 Mondays, the extra day must be a Monday. Hence, the probability is \(\frac{1}{7}\).

Therefore, Reason (R) is not correct as it states \(\frac{5}{7}\) instead of \(\frac{1}{7}\).

Since Assertion (A) is true and Reason (R) is false, the correct option is: Assertion (A) is true, but Reason (R) is false.