The probability that a leap year will have53 Sundays and 53 Mondays is

1.	The probability that a leap year will have53 Sundays and 53 Mondays is
Answer» A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess week days in a leap year. The two excess week days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). So, the sample space S has 7 pairs of excess week days. i.e. n(S) = 7. Now we want the desired event E to have 53 Sundays and 53 Mondays. E consists of only one pair in S which is (Sunday, Monday). So n(E) = 1 Hence, the probability that a leap year will contain 53 Sundays and 53 Mondays = n(E)/n(S) = 1/7 normally year consists of 365 days .but a leap year has 365+1days. 365 days divided into weekly days has 365/7 i.e 52days+2extra days.

Answer»

A leap year has 366 days.

Now 364 is divisible by 7 and therefore there will be two excess week days in a leap year.

The two excess week days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday).

So, the sample space S has 7 pairs of excess week days. i.e. n(S) = 7.

Now we want the desired event E to have 53 Sundays and 53 Mondays. E consists of only one pair in S which is (Sunday, Monday). So n(E) = 1

Hence, the probability that a leap year will contain 53 Sundays and 53 Mondays = n(E)/n(S) = 1/7

normally year consists of 365 days .but a leap year has 365+1days. 365 days divided into weekly days has 365/7 i.e 52days+2extra days.

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