What is the probability,that a leap year selected at random will contain 53 Sundays ?
Answers
Answered by
3
Probability of getting 53 Sundays is 2/7
As in a simple year, we have 1 Sunday, 1 Monday etc.
but in a leap year, we have 1 day extra so a day will have its number = 53 in that year.
If we make pair of days - (Sunday-Monday, Monday-Tuesday)
we will be having 2 pairs having Sunday - (Sunday-Monday, Saturday-Sunday)
we have total days in a week = 7
Hence probability becomes 2/7
As in a simple year, we have 1 Sunday, 1 Monday etc.
but in a leap year, we have 1 day extra so a day will have its number = 53 in that year.
If we make pair of days - (Sunday-Monday, Monday-Tuesday)
we will be having 2 pairs having Sunday - (Sunday-Monday, Saturday-Sunday)
we have total days in a week = 7
Hence probability becomes 2/7
Answered by
1
Answer:
In a leap year there are 366 days. In 366 days, we have 522 weeks and 2 days, Thus we can say that leap year ah always 52 Sundays.
The remaining two days can be
(i) Sunday and Mondays
(ii) Mondays and Tuesdays
(iii) Tuesday and Wednesday
(iv) Wednesday and Thursday
(v) Thursday and Friday
(VI) Friday and Saturday
(vii) Saturady and Sunday.
From above it is clear that there are 7 elementary events associated with this random experiment.
Clearly the event A will happen if the last two days of the leap year are either Sunday and Monday or Saturday and Sunday.
∴ P (E) = n (E)/n (S) = 2/7
Similar questions