what is the probability that a leap year has 53 sundays and 53 mondays?

- leap year has 366 days which is 52 weeks + 2 days there can be 53 sundays only when the last 2 days are
- (saturday , sunday or sunday monday) so P(53 sundays)= 2/7 as total there can be 7 cases similarly P(53 mondays) = 2/7 P(53 sundays or 53 mondays)= P(53 sundays union 53 mondays)
- = P(53 sundays)+ P(53 mondays) - P(53 sundays and 53 mondays) which is seen in only 1 case so P(53 sundays or 53 mondays)
- = 2/7+2/7-1/7 = 3/7

- 10

for a NON leap year: No of days = 365

No of week = 52Week +1 day

so first 52weeks will surely have a monday.

Event that a Monday will come on the remaining 1 day: n(E) = 1 [E= {Monday}]

Sample Space (no of Possibilities) n(S) = 7 [ S= {M,T,W,T,F,S,S} ]

Therefore P= n(E)/n(S) = 1/7

for a Non leap year: No of days = 366

No of week = 52Week + 2 days

so first 52weeks will surely have a monday.

Event that a Monday will come on remaining 2 days: n(E) = 2 [E = { (Sun,Mon) / (Mon,Tue) } ]

Sample Space (no of Possibilities) n(S) = 7

[S= { (Sun,Mon) (Mon,Tue) (Tue,Wed) (Wed,Thu) (Thu,Fri) (Fri,Sat) (Sat,Sun) }

Therefore P= n(E)/n(S) = 2/7..

No of week = 52Week +1 day

so first 52weeks will surely have a monday.

Event that a Monday will come on the remaining 1 day: n(E) = 1 [E= {Monday}]

Sample Space (no of Possibilities) n(S) = 7 [ S= {M,T,W,T,F,S,S} ]

Therefore P= n(E)/n(S) = 1/7

for a Non leap year: No of days = 366

No of week = 52Week + 2 days

so first 52weeks will surely have a monday.

Event that a Monday will come on remaining 2 days: n(E) = 2 [E = { (Sun,Mon) / (Mon,Tue) } ]

Sample Space (no of Possibilities) n(S) = 7

[S= { (Sun,Mon) (Mon,Tue) (Tue,Wed) (Wed,Thu) (Thu,Fri) (Fri,Sat) (Sat,Sun) }

Therefore P= n(E)/n(S) = 2/7..

- 37

A leap year has 366 days

So no: of weeks = 52

These 52 weeks surely have 52 sundays and 52 mondays

No: of remaining days = 2

These two days could be :

{(Sun,Mon), (Mon,Tue), (Tue, Wed), (Wed,Thur), (Thur,Fri), (Fri,Sat), (Sat,Sun)}

No: of favourable outcomes (i.e., the remaining 2 days are Sunday and Monday to make it 53 sundays and 53 mondays) = 1

Total possible ways = 7

Therefore, P(E) = 1/7

So no: of weeks = 52

These 52 weeks surely have 52 sundays and 52 mondays

No: of remaining days = 2

These two days could be :

{(Sun,Mon), (Mon,Tue), (Tue, Wed), (Wed,Thur), (Thur,Fri), (Fri,Sat), (Sat,Sun)}

No: of favourable outcomes (i.e., the remaining 2 days are Sunday and Monday to make it 53 sundays and 53 mondays) = 1

Total possible ways = 7

Therefore, P(E) = 1/7

- -1

In a leap year the no of days is 366, right. And we also know that in a year(365) there are 52weeks.

But there is one day extra in a leap year.

So that day can be any day of the week( sunday, monday, tuesday, wednesday, thursday, friday or saturday)

In a normal year we have 52mondays right.

The probability of having 53mondays in a leap year is 1/7

Because that one extra day may be any day of the week

But there is one day extra in a leap year.

So that day can be any day of the week( sunday, monday, tuesday, wednesday, thursday, friday or saturday)

In a normal year we have 52mondays right.

The probability of having 53mondays in a leap year is 1/7

Because that one extra day may be any day of the week

- 0