Question:
The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, is:
The probability that a leap year selected at random contains either 53 Sundays or 53 Mondays, is:
Updated On: Aug 19, 2025
- \(\frac{17}{53}\)
- \(\frac{1}{53}\)
- \(\frac{3}{7}\)
- None of these
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The Correct Option is C
Solution and Explanation
To determine the probability that a randomly selected leap year contains either 53 Sundays or 53 Mondays, we first need to understand the structure of a leap year. A leap year contains 366 days which equals 52 weeks and 2 extra days.
These extra days can be any of the following combinations:
- Sunday and Monday
- Monday and Tuesday
- Tuesday and Wednesday
- Wednesday and Thursday
- Thursday and Friday
- Friday and Saturday
- Saturday and Sunday
Out of these 7 combinations, "Sunday and Monday" and "Saturday and Sunday" combinations will result in 53 Sundays, while "Sunday and Monday" and "Monday and Tuesday" combinations will result in 53 Mondays.
The required outcomes for either 53 Sundays or 53 Mondays are:
- Sunday and Monday
- Monday and Tuesday
- Saturday and Sunday
We have 3 favorable outcomes out of 7 possible outcomes for the 2 extra days.
Therefore, the probability \(P\) that a leap year has either 53 Sundays or 53 Mondays is given by:
\(P=\frac{3}{7}\)
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