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Problem of the Week
Problem D and Solution
A Big Leap


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:

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

An alternative format for the flowchart


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

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 \(\frac{97}{400} = 0.2425\).