Question

find the probability of getting 53 Sunday's in. (1)non leap year. (2)leap year

Answer 1

(1) 

There are 365 days in a non-leap year.

The number of weeks in a non-leap year = 7.

= 365/7

= 52 * 1/7

That means 52 weeks + 1- day.

This 1-day can be:

{Mon, Tues, Wed, Thurs, Fri, Sat, Sun}.

Favorable cases = 1.

Therefore, the probability that a non-leap year contains 53 Sundays

= 1/7.



(2) 

There are 366 days in a leap year.

The number of weeks in a leap year = 7.

= 365/7

= 52 2/7.

That means 52 weeks + 2 days.

These 2-days can be

{Mon,Tues},{Tues,Wed},{Wed,Thurs},{Thurs,Fri}, {Fri,Sat},{Sat,Sun},{Sun,Mon} = 7

In order to have 53 Sundays, we should have either {Sat,Sun},{Sun,Mon}.

Favorable cases  = 2.

Therefore the probability that a leap-year contains 53-Sundays

= 2/7.


Hope this helps!

Answer 2

1)A leap year has 366 days, therefore 52 weeks i.e. 52 Sunday and 2 days.

The remaining 2 days may be any of the following :

(i) Sunday and Monday

(ii) Monday and Tuesday

(iii) Tuesday and Wednesday

(iv) Wednesday and Thursday

(v) Thursday and Friday

(vi) Friday and Saturday

(vii) Saturday and Sunday

For having 53 Sundays in a year, one of the remaining 2 days must be a Sunday.

n(S) = 7

n(E) = 2

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

2)In a year there are 365 days

In which there are 52 weeks (364 days) and 1 more day.

So there are already 52 sundays.

The last day can be any one of the seven days, i.e , sunday or monday or tuesday ....

But favourable posibillitie is only one : Sunday

Therefore probability of 53 sundays is 1/7