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 calendar probability problems:
The probability of having 53 of any specific weekday (e.g., Sunday, Monday, etc.) is:

• $\frac{1}{7}$ in a non-leap year (because there is 1 extra day).

• $\frac{2}{7}$ in a leap year (because there are 2 extra days).

Memorizing these standard probabilities helps you evaluate Assertion-Reason questions instantly without writing down the sample space.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The topic is Probability, specifically dealing with calendar-based probability problems.
We need to analyze two independent statements:

• Assertion (A) regarding the probability of having 53 Mondays in a leap year.

• Reason (R) regarding the probability of having 53 Mondays in a non-leap year.

After assessing the truth value of each, we will choose the correct option.

Step 2: Key Formula or Approach:
We calculate the number of days in leap and non-leap years:

• A standard non-leap year has 365 days.

• A leap year has 366 days.

• We divide the total number of days by 7 to determine the number of complete weeks and the remaining extra days.

• We then find the probability of the extra days falling on a Monday.


Step 3: Detailed Explanation:

• Analysis of Assertion (A):
A leap year contains 366 days.
Let us divide 366 by 7 to find the number of weeks:
\[ 366 = 52 \times 7 + 2 \]
This means a leap year has 52 complete weeks and 2 extra days.
The 52 complete weeks will definitely contain 52 Mondays.
For the year to have 53 Mondays, one of the 2 remaining extra days must be a Monday.
The 2 extra days must be consecutive days of the week. The possible pairs of consecutive days are:
\[ S = \{\text{(Sunday, Monday)}, \text{(Monday, Tuesday)}, \text{(Tuesday, Wednesday)}, \text{(Wednesday, Thursday)}, \]
\[ \text{(Thursday, Friday)}, \text{(Friday, Saturday)}, \text{(Saturday, Sunday)}\} \]
The total number of possible outcomes is $n(S) = 7$.
The pairs that contain "Monday" are:
\[ E = \{\text{(Sunday, Monday)}, \text{(Monday, Tuesday)}\} \]
The number of favorable outcomes is $n(E) = 2$.
Therefore, the probability of a leap year having 53 Mondays is:
\[ P(\text{53 Mondays in leap year}) = \frac{2}{7} \]
Thus, Assertion (A) is

true.

• Analysis of Reason (R):
A non-leap year contains 365 days.
Let us divide 365 by 7:
\[ 365 = 52 \times 7 + 1 \]
This means a non-leap year has 52 complete weeks and 1 extra day.
The 52 weeks contain exactly 52 Mondays.
For the year to have 53 Mondays, the single remaining extra day must be a Monday.
The single extra day can be any of the 7 days of the week:
\[ S' = \{\text{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}\} \] The total number of outcomes is $n(S') = 7$.
The only favorable outcome is $\{\text{Monday}\}$, so the number of favorable outcomes is 1.
Therefore, the probability of a non-leap year having 53 Mondays is:
\[ P(\text{53 Mondays in non-leap year}) = \frac{1}{7} \]
The Reason statement asserts that this probability is $\frac{5}{7}$.
Therefore, Reason (R) is

false.


Step 4: Final Answer:
Since Assertion (A) is true but Reason (R) is false, the correct option is (C).