If a 4 digit year is choosen randomly, what is the probability that it is not a leap year ?
This problem has come in my exam and i have written like this
I know that the number of four digit year is divisible by 4, so probability of a year which is not a leap year is 3/4. But When i have seen the answer key , answer is probability of a year which is not a leap year is $> 3/4$.
Please clear my doubt, Thank you
On
There are 9000 four-digit years, running from 1000 AD to 9999 AD. Of these, there would be 2250 leap years, except for the century rule that excludes the years 2100, 2200, 2300, 2500, etc. There are 60 such excluded years from 2100 through 9900.
Furthermore depending on which country you're in, one or more of the years 1700 AD, 1800 AD and 1900 AD may also have been non-leap years, so the fraction of 4-digit years from 1000 AD to 9999 AD that are leap years is one of
$$ \frac{2187}{9000}=\frac{243}{1000} \qquad\quad \frac{2188}{9000}=\frac{547}{2550} \qquad\quad \frac{2189}{9000} \qquad\quad \frac{2190}{9000}=\frac{73}{300} $$
On
You can look directly to the number of days in a year.
Depending on the used calendar / year, we get:
$$ \begin{array}{l|r|l} \textrm{calendar / year} & days / year & P\\ \hline \textrm{Gregorian calendar} & 365.2425\color{gray}{0} & 0.2425\\ \textrm{Julian calendar} & 365.25\color{gray}{000}& 0.25\\ \textrm{Hebrew calendar} & 365.2468\color{gray}{0} & 0.2468\\ \textrm{mean tropical year} & 365.24219 & 0.24219 \end{array} $$
$$ \Big\lfloor (n+k+1) \phi \Big\rfloor - \Big\lfloor (n + k ) \phi \Big\rfloor, $$
where $k$ is an offset and $\phi$ is the number of days in a year, depending on the calandar or year.
The probability is then given by
$$ \Big\lfloor \phi \Big\rfloor. $$
As for the question not a leap year, you need to reverse it, thus
$$ 1 - \Big\lfloor \phi \Big\rfloor. $$
HINT:
From the very definition the number of leap years in every $400$ years in $$\left\lfloor\frac{400}4\right\rfloor -3$$