Question

In a non leap year ,find probability that there are 53 tuesdays ?

In a leap year , find probability that there are 53 tuesdays in the year?

Answer 1

some logic must be require for solving this question,
  as you know,
    1 year = 365 days
    1year has 52 week , it means there are 52 tuesdays  must come.
so, number of days in 52 week = 52 * 7 = 364 days
hence, days left  = number of days in 1 year - number of days in 52 week
                  =365 days - 364 days = 1 days
hence, we don't know left 1 days what will be possible,
but possibilities are , that day may be sunday, monday,tuesday,wednesday,thursday,friday,saturday
e.g., possible outcomes n(s)= 7
favourable outcomesn(E) = 1 [dark word shows]
so, propability = n(E)/n(S) = 1/7


now in case of leap year
number of days in 1 year = 366 days
number of days in  52 week = 364 days
so, number of day left = 366days - 364 days = 2days
so, possible outcome may be
(SUN, MON),(mon,tue),(tue,wed),(WED,THU),(THU,FRI),(FRI,SAT),
(SAT,SUN) e.g., possible outcomes n(S) = 7
favourable outcomes n(E) = 2 [dark word  shows favourable ]
so, probability = 2/7

Answer 2

(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 Tuesdays

= 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 Tuesdays, we should have either {Mon, Tues},{Tues, Wed}.

Favorable cases  = 2.

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

= 2/7.


Hope this helps!