Find the probability that a Leap Year has
i) 52 sundays ii)53 sundays
Total number of days in a leap year are 366 days
366 days = 364 days + 2 days
= 52 × 7 days + 2 days
= 52 weeks + 2 days
(i) There are 52 weeks in a leap
∴ 52 weeks must have 52 Sundays.
Hence the probability of 52 Sundays = 1
(ii) A leap year has 366 days which is equal to 52 weeks and 2 days.
These 2 day can 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 Saturday
Number of favourable outcomes = 2 (Sunday and Monday, Saturday and Sunday)
Total number of outcomes = 7
∴ Probability that a leap year has 53 Sundays
Number of days in a leap year = 366
= 52 weeks and 2 days
52 weeks have 52 Sundays. 2 Days are remaining. These 2 days can be arranged as follows:
1: Sunday Monday
2: Monday Tuesday
3: Tuesday Wednesday
4: Wednesday Thursday
5: Thursday Friday
6: Friday Saturday
7: Saturday Sunday
So, there 7 possible ways to arrange these 2 days.
As there are 5 such cases, where neither of the days are Sunday, so number of favorable events(not Sunday) = 5.
(We already have 52 Sundays)
∴ Required probability
thumbs up for the first ans... plz
A leap year has 365 days= 52weeks=7days( 1 day is extra)
Favourable Outcone - Sunday=1
Possible Outcome - Mon., Tue., Wed., Thu., Fri., Sat., Sun.=7
So the probability that, that one extra day is Sunday is simply 1/7.
It will be 1/7 as by dividing 365 days by 7 we will get only 1 as remainder so this one can be any of the day of the week.therefore it will be 1/7. Therefore probability of 53 Sundays is 1/7.
Hope it helps!!
in a leap year there are 366 days
366days=52 weeks and 2 days
thus, a leap year has 52 weeks, then the no. of sundays in one of the week selected from the 52 weeks will be1
(sunday, monday, tuesday, wednesday, thursday, friday, saturday)
thus the probability of 52 sundays in a leap year is 1/7
2. 366days=52 weeks 2 days
the remaining 2 days can be:
sunday and monday, mon and tues, tues and wed, wed and thurs, thurs and fri, fri and sat, sat and sun
thus, favourable no. of elementary events=2
hence, required probability=2/7