What is the probability that a leap year has 53 Sundays?
What is the probability that a leap year has 53 Sundays?
Show Hint
- $\frac{1}{7}$
- $\frac{2}{7}$
- $\frac{3}{7}$
$\frac{4}{7}$
The Correct Option is B
Solution and Explanation
Step 1: Understanding Days in a Leap Year
A leap year contains 366 days instead of the usual 365. This extra day is added to the month of February (29 days instead of 28) to keep our calendar in alignment with Earth's revolutions around the Sun.
When we divide 366 days by 7 (the number of days in a week), we get:
366 = 52 weeks + 2 extra days
This means that in a leap year, there are exactly 52 full weeks and 2 leftover days.
Step 2: Guarantee of 52 Sundays
Since each week has one Sunday, the 52 complete weeks ensure there are exactly 52 Sundays in every leap year without exception. The possibility of having 53 Sundays depends entirely on the 2 extra days.
Step 3: Analyzing the Extra 2 Days
The two extra days can occur in the following combinations:
(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).
Step 4: Favorable Cases for 53 Sundays
For a leap year to have 53 Sundays, one of the two extra days must be a Sunday. This happens in exactly two favorable cases:
- (Saturday, Sunday)
- (Sunday, Monday)
Step 5: Calculating the Probability
The total possible arrangements of the 2 extra days is 7 (one for each possible starting day of the year). The number of favorable arrangements is 2.
Therefore, the probability is:
Probability = (Number of favorable cases) / (Total possible cases) = 2 / 7
Step 6: Conclusion
The probability that a leap year contains 53 Sundays is:
⅔ / 7 or 2/7. This matches Option (2) in multiple-choice formats
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