What is the probability that an ordinary year has 53 Sundays?
A. $\dfrac{{53}}{{365}}$
B. $\dfrac{1}{7}$
C. $\dfrac{2}{7}$
D. $\dfrac{{48}}{{53}}$

Answer Verified Verified
Hint: Before attempting this question, one should have prior knowledge about the year and also remember that in an ordinary year there are 52 weeks, this information will help you to approach the solution of the problem.

Complete step by step answer:
According to the given it is an ordinary year.
As we know that in an ordinary year there are 365 days and 52 weeks.
So, we know that in a week there is only 1 Sunday therefore in 52 weeks there are 52 Sundays.
Also, we know that 52 weeks there are 364 days so out of 365 days we only have only one day left and there is a possibility that the remaining day can be any day of the week.
And we know that in a week there are 7 days also we know that in 1 week there is 1 Sunday.
Therefore, the possibility of getting 53 Sundays is = $\dfrac{1}{7}$

So, the correct answer is “Option B”.

Note: In the above solution we came across the term “year” which can be of 2 different types that are leap year and ordinary year. In the ordinary year, there are 365 days but in a leap year, there are 366 days. The special about the leap year is it occurs after every 4 years basically in ordinary there are only 28 days in the months of February whereas in a leap year there are 29 days which is the reason behind leap year having 366 days.
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