Problem
of the Week
Problem
D and Solution
A
Big Leap

Problem

Most people think of a year as \(365\) days, however it is actually slightly
more than \(365\) days. To account for
this extra time we use leap years, which are years containing one extra
day.

Mara uses the flowchart shown to determine whether or not a given
year is a leap year. She has concluded the following:

\(2018\) was
not a leap year because \(2018\) is not divisible by \(4\).

\(2016\) was a leap year because
\(2016\) is divisible by \(4\), but not \(100\).

\(2100\) will
not be a leap year because \(2100\) is divisible by \(4\) and \(100\), but not \(400\).

\(2000\) was a leap year because
\(2000\) is divisible by \(4\), \(100\), and \(400\).

If Mara chooses a year greater than \(2000\) at random, what is the probability
that she chooses a leap year?

Start with a year and proceed through the following steps:

Is the year divisible by \(4\)?

If NO, then the year is not a leap
year.

If YES, then go to 2.

Is the year divisible \(100\)?

If NO, then the year is a leap year.

If YES, then go to 3.

Is the year divisible \(400\)?

If NO, then the year is not a leap
year.

If YES, then the year is a leap year.

Solution

The probability of an event occurring is calculated as the number of
favourable outcomes (that is, the number of outcomes where the event
occurs) divided by the total number of possible outcomes. This is an
issue in our problem because the number of years greater than \(2000\) is infinite. However, the cycle of
leap years repeats every \(400\) years.
For example, since \(2044\) is a leap
year, so is \(2444\).

Thus, to determine the probability, we need to count the number of
leap years in a \(400\)-year cycle.
From the flowchart we can determine that leap years are either

multiples of \(4\) that are not
also multiples of \(100\), or

multiples of \(4,~100,\) and
\(400\).

Note that we can simplify the second case to just multiples of \(400\), since any multiple of \(400\) will also be a multiple of \(4\) and \(100\).

The number of multiples of \(4\) in
a \(400\)-year cycle is \(\frac{400}{4}=100\). However, we have
included the multiples of \(100\), so
we need to subtract these multiples. There are \(\frac{400}{100}=4\) multiples of \(100\) in a \(400\)-year cycle. Thus, there are \(100-4=96\) multiplies of \(4\) that are not multiples of \(100\). We now need to add back the the
multiples of \(400\). There is \(\frac{400}{400}=1\) multiple of \(400\) in a \(400\)-year cycle. Thus, there are \(96+1=97\) numbers that are multiples of
\(4\) and are not multiples of \(100\), or that are multiples of \(400\).

Therefore, for every \(400\)-year
cycle, \(97\) of these years will be a
leap year.

Therefore, the probability of Mara choosing a leap year is \(\frac{97}{400} = 0.2425\).