Question

What is the chance that (i) non leap year (ii) leap year should have 53 Sundays.

Answer 1

Hint : To answer this question first divide the total number of days in a year by 7 such that you will get the number of weeks. Then multiply the number of weeks and find the number of days such that we can answer the number of days after all the complete weeks have completed. Then find the probability in the remaining days.

Complete step-by-step answer :
In a non leap year:
A non-leap year has 365 days
A year has 52 weeks. Hence there will be 52 Sundays for sure.
52 weeks $= {\text{ }}52 \times 7 = 364$ days .
$365-364 = 1$ day extra.
In a non-leap year there will be 52 Sundays and 1day will be left.
This 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, friday, Saturday, Sunday.
Of these total 7 outcomes, the favourable outcomes are 1.
Hence the probability of getting 53 sundays $\dfrac{{1}}{{7}}$ .
In a leap year:
The two odd days can be {Sunday,Monday},{Monday,Tuesday},{Tuesday,Wednesday}, Wednesday,Thursday},{Thursday,Friday},{Friday,Saturday},{Saturday,Sunday}.
So there are 7 possibilities out of which 2 have a Sunday. So the probability of 53 Sundays in a leap year is $\dfrac{{2}}{{7}}$ .

Note : We can use the exact same concept for all other days of a week all will have the same chance. However the results depend on combinations of the days that are probable for the event.