a. What is the probability that a randomly chosen leap year has 53 Fridays?
b. What is the probability of having 53 Fridays in a leap year
of the current century?
Rem: Gregorian calendar, years from 1600 till 2099
D1 answer:
A leap year has 53 Fridays if it begins on a Thursday or Friday, since 366/7 is 52 with a remainder of 2. Since any day of the week is equally likely to begin the year, the answer is 2/7.
D3 answer:
In the long haul in the Gregorian calendar, not every starting day of the week is equally likely. The Gregorian calendar has a cycle of 400 years. Most years ending in 00 are not leap years despite being a multiple of 4; only if divisible by 400, such as in 2000, are they leap years. So 400 years has 365*400 + 100 - 3 = 146097 days, which is a multiple of 7, so each 400-year cycle begins on the same day of the week, so the same irregularities in one 400-year cycle keep repeating every cycle.
So 2000 was a leap year beginning on a Saturday. 2100, 2200, and 2300 will not be leap years.
Three regular years plus a leap year make up 1461 days, which is 2 days less than an integral number of weeks. If there are four ordinary years in a row, that is 3 days fewer than an integral number of weeks.
Rather than do the messy work,
clearvars,clc
dow='SmTwtfs'
ctgood=0; ctall=0;
for y=2000:4:2099
if y~=2100 && y~=2200 && y~=2300
t=datetime(y,1,1);
d=day(t,'dayofweek');
fprintf('%4d %s ',y, dow(d))
if mod(y,100)==96 || mod(y,100)==48
fprintf('\n')
end
ctall=ctall+1;
if contains('tf',dow(d))
ctgood=ctgood+1;
end
end
end
disp([ctgood ctall])
finds the beginning days of the week for all the leap years in one group of 400 consecutive years:
2000 s 2004 t 2008 T 2012 S 2016 f 2020 w 2024 m 2028 s 2032 t 2036 T 2040 S 2044 f 2048 w
2052 m 2056 s 2060 t 2064 T 2068 S 2072 f 2076 w 2080 m 2084 s 2088 t 2092 T 2096 S
7 25
where Thursday and Friday are lower-case t and f. (Tuesday is upper-case T; Sunday is upper-cased also.)
and reports
7 begin on Thursday or Friday, out of 25 leap years.
>> 7/25
ans =
0.28
compared to
>> 2/7
ans =
0.285714285714286
>>
so the probability, if the curren century is taken as being 2000 - 2099, is 7/25, which is slightly lower than 2/7.
However if we consider the current century to be 2001 - 2100:
2004 t 2008 T 2012 S 2016 f 2020 w 2024 m 2028 s 2032 t 2036 T 2040 S 2044 f 2048 w
2052 m 2056 s 2060 t 2064 T 2068 S 2072 f 2076 w 2080 m 2084 s 2088 t 2092 T 2096 S
7 24
and 7/24 = 0.291666666666667, somewhat higher than 2/7.
And for one Gregorian cycle of 400 years, 1600 through 2099
1600 s 1604 t 1608 T 1612 S 1616 f 1620 w 1624 m 1628 s 1632 t 1636 T 1640 S 1644 f 1648 w
1652 m 1656 s 1660 t 1664 T 1668 S 1672 f 1676 w 1680 m 1684 s 1688 t 1692 T 1696 S
1704 T 1708 S 1712 f 1716 w 1720 m 1724 s 1728 t 1732 T 1736 S 1740 f 1744 w 1748 m
1752 s 1756 t 1760 T 1764 S 1768 f 1772 w 1776 m 1780 s 1784 t 1788 T 1792 S 1796 f
1804 S 1808 f 1812 w 1816 m 1820 s 1824 t 1828 T 1832 S 1836 f 1840 w 1844 m 1848 s
1852 t 1856 T 1860 S 1864 f 1868 w 1872 m 1876 s 1880 t 1884 T 1888 S 1892 f 1896 w
1904 f 1908 w 1912 m 1916 s 1920 t 1924 T 1928 S 1932 f 1936 w 1940 m 1944 s 1948 t
1952 T 1956 S 1960 f 1964 w 1968 m 1972 s 1976 t 1980 T 1984 S 1988 f 1992 w 1996 m
2000 s 2004 t 2008 T 2012 S 2016 f 2020 w 2024 m 2028 s 2032 t 2036 T 2040 S 2044 f 2048 w
2052 m 2056 s 2060 t 2064 T 2068 S 2072 f 2076 w 2080 m 2084 s 2088 t 2092 T 2096 S
35 122
35/122 = 0.286885245901639
D2: But we won't see any missing leap year due to the Gregorian system in our lifetime (for most of us).
D4: Over the long, long haul, due mostly to the lengthening of the day (slowing of the earth's rotation), and less so due to the small inaccuracy of the Gregorian correction itself, it makes sense to go back to the original calculation of 2/7.
|
|
Posted by Charlie
on 2024-02-28 08:17:17 |
solution
