53 Sundays in a leap year — Probability Aptitude Question
53 Sundays in a leap year: What is the probability that a 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 leap year has 53 Sundays?
Options
- 3/7
- 1/2
- 1/7
- 2/7
Show the answer and the worked solution
Correct answer: D. 2/7
Worked solution
- 366 = 52 × 7 + 2, so every weekday occurs 52 times plus two consecutive extra days.
- The extra pair is one of 7 equally likely pairs: (Sun,Mon), (Mon,Tue), …, (Sat,Sun). Two of them contain a Sunday.
- Probability = 2/7.
Answer: 2/7.
The idea
A leap year has 2 spare days, a normal year 1. So P(53 of a given day) = 2/7 in a leap year and 1/7 otherwise.
Shortcut. The 2 spare days hand an extra appearance to 2 of the 7 weekdays, each weekday equally likely to be one of them — so 2/7 for Sunday.
Common trap. 1/7 — the answer for an ordinary year, which has one spare day; a leap year has two.