The probability that in a year of the 22nd century chosen at random, there will be 53 Sundays is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{5}{28}\)
Note: The given answer is as per the official NTA Answer Key. However, the exact value for the 22nd century is \(\frac{31}{175}\), which is not among the options. Option (3) is the value found with the usual chance \(\frac{1}{4}\) of a leap year, as explained below.
A non-leap year has 365 days = 52 weeks + 1 day. It has 53 Sundays only if the extra day is a Sunday:
\( \displaystyle P(53 \text{ Sundays} \mid \text{non-leap}) = \frac{1}{7} \)
A leap year has 366 days = 52 weeks + 2 days. The 2 extra days are one of 7 equally likely pairs (Sun–Mon, Mon–Tue, …, Sat–Sun), and 2 of them contain a Sunday:
\( \displaystyle P(53 \text{ Sundays} \mid \text{leap}) = \frac{2}{7} \)
Taking a year to be a leap year with probability \(\frac{1}{4}\), the theorem of total probability gives
\( \displaystyle P = \frac{3}{4} \times \frac{1}{7} + \frac{1}{4} \times \frac{2}{7} = \frac{3}{28} + \frac{2}{28} = \frac{5}{28} \)
Exact value: the 22nd century (2101–2200) has 24 leap years, 2104, 2108, …, 2196. The year 2200 is not a leap year, because it is divisible by 100 but not by 400.
\( \displaystyle P = \frac{76}{100} \times \frac{1}{7} + \frac{24}{100} \times \frac{2}{7} = \frac{124}{700} = \frac{31}{175} \approx 0.177 \)
This is very close to \(\frac{5}{28} \approx 0.179\), and \(\frac{5}{28}\) is the nearest option.