Question

Find the probability of getting $53$ Sundays in a non-leap year.
A. $\dfrac{2}{7}$
B. $\dfrac{1}{7}$
C. $\dfrac{1}{5}$
D. $\dfrac{1}{9}$

Answer 1

The given question revolves around the concepts of probability. We know that there are a total of $365$ days in a non-leap year. So, a non-leap year consists of a total of $52$ complete weeks and one odd day. We know that a week consists of seven days. So, we know that there are definitely $52$ Mondays, Tuesdays, Wednesdays, Thursdays, Fridays, Saturdays and Sundays in a non-leap year as it consists of $52$ perfect weeks. Now, we have to find the probability that the remaining one odd day is Sunday.

Complete step by step answer:
In the given question, we have to find the probability of getting $53$ Sundays in a non-leap year. So, number of days in a non-leap year $= 365$
Number of complete weeks in a non-leap year $= 52$
Number of odd days in a leap year $= 1$
Now, we have to make sure that the one odd day is also a Sunday so that the non-leap year consists of a total of $53$ Sundays.

We know that there are seven days in a week: Monday, Tuesday, Wednesday, Thursday, Friday, Saturday and Sunday. So, there are seven options of days for the one odd day left. So, the odd day may be Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday.Hence, the total number of options is $7$. Also, the number of favorable options is $1$. Now, we know that the probability of an event to happen is $\dfrac{{Number\,of\,favourable\,outcomes}}{{Number\,of\,total\,outcomes}}$
Therefore, we get the probability of getting $53$ Sundays in a non-leap year as $\dfrac{{Number\,of\,favourable\,outcomes}}{{Number\,of\,total\,outcomes}} = \dfrac{1}{7}$.
So, the probability of getting seven Sundays in a non-leap year is $\dfrac{1}{7}$.

Hence, option B is the correct answer.

Note: We must know the basic concepts and formulae of probability to attempt and solve the given problem. We should be careful while carrying out the calculations so as to be sure of the answer. We must know the concepts of odd days and complete weeks in order to tackle such problems.