What is the probability that an ordinary year has 53 Sundays?
Answers
here in ques it is not specifically said that the year is leap year or non leap year . so i am writing the answer for both conditions.
considering a non leap year
see we have 365 days in a non-leap year.
52 weeks and one extra day= 365 days
52 weeks means definitely there are 52 sundays (this is true for all other days also).
that means if the extra day comes out to be sunday then we will have 53 sundays.
so now the ques boils down to what is the prob of this extra day to be a sunday .
this extra one day can be {monday or tuesday or wednesday or thursday or friday or saturday or sunday }= samplespace (s)
i.e n(s) = 7
so prob of this extra one day to be a sunday is 1/7.
[note - this is also the answer for having 53 mondays or 53 tuesdays or 53 wednesdays or 53 thursdays or 53 fridays or 53 saturdays.]
considering a leap year
leap year contains 366 days
52 weeks plus two two extra days
52 weeks means definitely there are 52 sundays (this is true for all other days also).
if either of these two is sunday then we will have 53 sundays
these two days can be {mon, tue} or { tue , wed} or { wed , thurs} or {thurs , fri} or { fri, sat} or { sat , sun} or {sun , mon} i.e total =7
out of these only two outcomes i.e { sat , sun} and {sun , mon} is having sunday with them .
so our desired prob is 2/7.
Answer. An ordinary year has 365 days A year has 52 weeks. Hence there will be 52 Sundays for sure. 52 weeks = 364 days 365 – 364 = 1 day In an ordinary year, there will be 52 Sundays and 1 day will be left. This one day can be, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday. Of these total 7 outcomes, the favourable outcome is 1. Hence the probability of getting 53 Sundays = 1/7