(i) Non leap year
No of days = 365
= 365/7 weeks = 52 weeks + 1 day
In 52 week we have 52 Sundays. So we have to find the probability of getting the remaining one day as Sunday. The remaining 1 day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday
(i.e.,) n(S) = 7
In the n (Sunday) = A {Saturday to Sunday or Sunday to Monday}
(i.e.,) n(A) = 1
So, P(A) = 1.
∴ Probability of getting 53 Sundays n(A)/n(S) = 1/7
(ii) Leap Year:
In 52 weeks we have 52 Sundays. We have to find the probability of getting one Sunday form the remaining 2 days the remaining 2 days can be a combination of the following
S = {Saturday to Sunday, Sunday to Monday, Monday, to Tuesday, Tuesday to wednes¬day, Wednesday to Thursday, Thursday to Friday, Friday and Saturday}.
(i.e) n(s) = 7
In this = A {Saturday to Sunday, Sunday to Monday}
(i.e) n(A) = 2
So, P(A) = 2/7
