Question

find the probability that a leap year selected at random will contain 53 Sundays

Answer 1

Hey friend!!

Here's ur answer...


========================

Leap year = 366 days = (52×7 + 2) days = 52 weeks and 2days.

Thus a leap year always has 52 Sundays.

So, remaining 2 days can be :

(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

Out of these 7 cases, Sunday is in 2 cases

So, P (53 Sundays) = 2/7.



================================



Hope it may help you....


Thank you :) :)))

Answer 2

Hey there!

Number of days in a leap year = 365 days.

If we divide them into subsequent weeks ,

then, 366 days = 7(52) + 2 = 52 weeks, 2 days.

So, As leap year has 52 weeks, It will surely consist of 52 Sundays .

Now, We need one more Sunday in the remaining two days .

The remaining two days can be

1)Saturday ,Sunday
2) Sunday, Monday.
3) Monday, Tuesday
4) Tuesday, Wednesday
5) Wednesday, Thursday
6) Thursday, Friday
7) Friday , Saturday.

We see that Sample space = 7 days.

Probability of 53rd Sunday = 2/7 .

Therefore, There is a 2/7 probability that A leap year selected at random would have 53 Sundays .