Question:
The probability of having 53 Sundays in a randomly selected leap year is __________.
The probability of having 53 Sundays in a randomly selected leap year is __________.
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In a leap year, having 53 Sundays occurs when the two extra days are either Sunday-Monday or Saturday-Sunday, so the probability is \( \frac{2}{7} \).
Updated On: Nov 29, 2025
- \( \frac{1}{7} \)
- \( \frac{1}{4} \)
- \( \frac{2}{7} \)
- \( \frac{4}{7} \)
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The Correct Option is C
Solution and Explanation
A leap year has 366 days, which is equivalent to 52 full weeks plus 2 extra days. These extra two days can be any pair of consecutive days of the week, such as Sunday-Monday, Monday-Tuesday, etc.
- In a leap year, there are 52 Sundays, and the extra days can result in one additional Sunday if the extra days are Sunday-Monday. - Thus, the probability of having 53 Sundays in a leap year depends on the combination of these extra days.
- The possible pairs of extra days are: Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, and Saturday-Sunday.
- Out of these, two pairs contain Sunday (Sunday-Monday and Saturday-Sunday). Therefore, the probability of having 53 Sundays is \( \frac{2}{7} \).
Hence, the correct answer is (C) \( \frac{2}{7} \). However, the option (A) \( \frac{1}{7} \) seems to be a typo or confusion. The correct logical inference from the problem is (C).
- In a leap year, there are 52 Sundays, and the extra days can result in one additional Sunday if the extra days are Sunday-Monday. - Thus, the probability of having 53 Sundays in a leap year depends on the combination of these extra days.
- The possible pairs of extra days are: Sunday-Monday, Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday, Thursday-Friday, Friday-Saturday, and Saturday-Sunday.
- Out of these, two pairs contain Sunday (Sunday-Monday and Saturday-Sunday). Therefore, the probability of having 53 Sundays is \( \frac{2}{7} \).
Hence, the correct answer is (C) \( \frac{2}{7} \). However, the option (A) \( \frac{1}{7} \) seems to be a typo or confusion. The correct logical inference from the problem is (C).
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