The probability that a leap year has 53 Sundays
A) $ \dfrac{7}{{266}} $
B) $ \dfrac{{26}}{{183}} $
C) $ \dfrac{1}{7} $
D) $ \dfrac{2}{7} $

Hint: A leap year has 366 days, we can find the number of weeks and the remaining days for the same. We can calculate the probability for 53 Sundays using this information and the formula for probability given as:
 $ P = \dfrac{f}{T} $ where, f and T are favorable outcomes and total outcomes respectively.

Complete step by step solution:
A year in general has 365 days while a leap year has a day extra as 29 Feb and the number of days are 366.
A week consists of 7 days and every week has a Sunday. We can calculate the number of weeks in 366 days as:
 $ Weeks = \dfrac{{366}}{7} $
It is exactly divided at 52 leaving the remainder 2, so 366 days have 52 weeks and 2 days.
52 weeks means that 52 Sundays will be there, for $ {53^{rd}} $ Sunday, among the remaining 2 days, one should be Sunday.
The possible outcomes for the two days are:
[ (Sunday, Monday) (Monday, Tuesday) (Tuesday, Wednesday) (Wednesday, Thursday) (Thursday, Friday) (Friday, Saturday) (Saturday, Sunday) ]
The favorable outcomes will be the one that has Sunday.
Number of favorable outcomes (f) = 2
Total outcomes (T) = 7
Using the formula of probability to find the required probability:
 $ \Rightarrow P = \dfrac{f}{T} $
Substituting the values, we get:
 $ \Rightarrow P = \dfrac{2}{7} $
Therefore, the probability that a leap year has 53 Sundays is $ \dfrac{2}{7} $ and the correct option is D).
So, the correct answer is “Option D”.

Note: General information to remember is that a normal year has 365 days, 52 weeks and a day extra while a leap year has 366 days, 53 weeks and two days extra.
We can write the probability in decimal or fraction but as the options were given in fractions so we stick to that part. There are no dimensions of the probability and thus is unit less as well.