What is the probability that a randomly chosen leap year has 53 Sundays?
2/7. A leap year has 366 days = 52 weeks + 2 extra days. Those 2 consecutive days can be any of 7 pairs (Mon-Tue, ..., Sun-Mon), and 2 of them include a Sunday.
Explain it simply
Every year has 52 full weeks, so every day of the week appears at least 52 times. The extra days beyond those weeks decide which days get a 53rd appearance. In an ordinary year there is 1 extra day, so the chance that it is a Sunday is 1/7. In a leap year there are 2 consecutive extra days.
Example
Question: P(a leap year has 53 Sundays)?
- 366 = 52 x 7 + 2. Two extra consecutive days.
- Possible pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). That is 7 equally likely pairs.
- Pairs containing Sunday: (Sun, Mon) and (Sat, Sun) = 2.
- P = 2/7.
For a non-leap year: 365 = 52 x 7 + 1, so P(53 Sundays) = 1/7.
Variation: P(a leap year has 53 Sundays or 53 Mondays)? Pairs with Sunday or Monday: (Sat, Sun), (Sun, Mon), (Mon, Tue) = 3/7.
const days = ["Sun", "Mon", "Tue", "Wed", "Thu", "Fri", "Sat"];
const pairs = days.map((d, i) => [d, days[(i + 1) % 7]]);
console.log(pairs.filter(p => p.includes("Sun")).length); // Output: 2What to say in the interview
"A leap year is 52 weeks plus 2 consecutive days. There are 7 possible pairs and 2 contain a Sunday, so 2 over 7. For an ordinary year it's 1 over 7."
Common mistakes
- Counting 14 cases by treating the two extra days as independent.
- Answering 1/7 by forgetting the year is a leap year.