53 Sundays in an ordinary year — Aptitude Question
53 Sundays in an ordinary year: What is the probability that a non-leap year has 53 Sundays? Answer and worked solution on CodeKairo.
- Topic: Probability
- Section: Quantitative Aptitude
- Difficulty: Medium
- Time target: 60 seconds
Question
What is the probability that a non-leap year has 53 Sundays?
Options
- 1/7
- 2/7
- 3/7
- 1/2
Show the answer and the worked solution
Correct answer: A. 1/7
Worked solution
- 365 = 52 × 7 + 1, so one weekday occurs 53 times.
- That day is equally likely to be any of the 7.
- Probability = 1/7.
Answer: 1/7.
The idea
One spare day in an ordinary year, two in a leap year. Everything else about the calendar is irrelevant.
Shortcut. Divide the days by 7 and put the remainder over 7: 365 = 7 × 52 + 1, so the probability is 1/7.
Common trap. "2/7" — borrowing the leap-year answer; 365 = 52 × 7 + 1 leaves only one spare day, not two.