What is the probability that a leap year selected at random will contain 53 Thursdays or 53 Fridays?
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Answered by
4
in a leap year = 366 days
in which
366 days = 52wewks + 2days
probability of 53 Friday=number of favorable outcomes/ total number of outcomes
= 2/7
I hope it will help u ok
in which
366 days = 52wewks + 2days
probability of 53 Friday=number of favorable outcomes/ total number of outcomes
= 2/7
I hope it will help u ok
Answered by
3
There are 53 Tuesdays in a leap year if it starts on a Monday or Tuesday.
This can't happen in 2/7 of all leap years, because the distribution of leap years and first-days-of-the year repeats every 400 years. There are 97 leap years in each such period, which is not a multiple of 7.
More precisely, in every 400-year period,
13 leap years start on a Monday
14 leap years start on a Tuesday
14 leap years start on a Wednesday
13 leap years start on a Thursday
15 leap years start on a Friday
13 leap years start on a Saturday
15 leap years start on a Sunday
So the probablility that a leap year chosen uniformly among the leap years in a cycle starts on a Monday or Tuesday (and so contains 53 Tuesdays) is
13+14/97=27/97
This can't happen in 2/7 of all leap years, because the distribution of leap years and first-days-of-the year repeats every 400 years. There are 97 leap years in each such period, which is not a multiple of 7.
More precisely, in every 400-year period,
13 leap years start on a Monday
14 leap years start on a Tuesday
14 leap years start on a Wednesday
13 leap years start on a Thursday
15 leap years start on a Friday
13 leap years start on a Saturday
15 leap years start on a Sunday
So the probablility that a leap year chosen uniformly among the leap years in a cycle starts on a Monday or Tuesday (and so contains 53 Tuesdays) is
13+14/97=27/97
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