what is the probability that a leap year has 53 Sundays and 53 Mondays
Anonymous:
47 upon 7
Answers
Answered by
195
A leap year has 366 days. Now 364 is divisible by 7 and therefore there will be two excess week days in a leap year. The two excess week days can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). So, the sample space S has 7 pairs of excess week days. i.e. n(S) = 7.
Now we want the desired event E to have 53 Sundays and 53 Mondays . E consists of only one pair in S which is (Sunday, Monday). So n(E) = 1
Hence, the probability that a leap year will contain 53 Sundays and 53 Mondays = n(E)/n(S) = 1/7
Now we want the desired event E to have 53 Sundays and 53 Mondays . E consists of only one pair in S which is (Sunday, Monday). So n(E) = 1
Hence, the probability that a leap year will contain 53 Sundays and 53 Mondays = n(E)/n(S) = 1/7
Answered by
53
The answer is 47 upon 7
Divide 365 by 53, and reminder will be 47
So, we know that a week has 7 days, then the probi. of both will be 47 upon 7
Divide 365 by 53, and reminder will be 47
So, we know that a week has 7 days, then the probi. of both will be 47 upon 7
Similar questions