Following question is given in my book:
What is the probability that a leap year has $53$ Sundays or $53$ Mondays?
My understanding:
I understand that in a leap year there are $366$ days i.e. $52$ complete weeks and $2$ extra days.
All the possibilities of these extra days are:
Sunday Monday
Monday Tuesday
Tuesday Wednesday
Wednesday Thursday
Thursday Friday
Friday Saturday
Saturday Sunday
The question asks us to find the probability of $53$ Sundays or $53$ Mondays which includes the following cases.
- Saturday Sunday
- Sunday Monday
- Monday Tuesday
Here Sunday and Monday occur in two cases each and they occur simultaneously in one case.
Let occurrence of Sunday be denotes by $A$ and that of Monday be $B$, the probability is given by,
$$\begin{aligned}P(A \cup B) &= P(A) + P(B) – P(A\cap B)\\& = \frac27 + \frac 27 – \frac17 \\& = \frac 37\end{aligned}$$
I'm not sure about my work and logic. I wish if someone could check my work and tell me the mistakes if any.
Best Answer
The answer depends on what you have assumed about the calendar you are using. If your book has not explained how the calendar works, I would say the question is not well posed, because the calendar has features that cannot be predicted by pure mathematics; they depend on detailed knowledge about the way people have agreed to adjust the civil calendar to the Earth's rotation and its orbit around the Sun.
In the Julian calendar, in which there is a leap year every four years without exception, the calendar will repeat the exact same sequence of days of the week on January $1$ every $28$ years, and in this sequence each weekday occurs on January $1$ of a leap year exactly once. According to that calendar, if we were to check the number of Sundays and the number of Mondays in the upcoming $n$ leap years, starting from any given date, the frequency of years with either $53$ Sundays or $53$ Mondays would approach $3/7$ as $n$ increased, which is a reasonable model of "probability" for a calendar problem like this.
But the civil calendar used almost everywhere in the world today is the Gregorian calendar, in which not every year whose year number is divisible by $4$ is a leap year. A year number divisible by $100$ is a leap year only if it is divisible by $400$. In the next $400$ years, $2100$, $2200$, and $2300$ will not be leap years.
There are therefore just $97$ leap years every $400$ years on average, for a total of $400\times 365 + 97 = 146097$ days every $400$ years. Since $146097 = 20871 \times 7$, the $400$-year cycle of leap years is also a $400$-year cycle of weeks, that is, the same sequence of month/day-of-month/day-of-week triplets repeats exactly every $400$ years. January 1, 2000 was a Saturday and January 1, 2400 will be a Saturday.
Within this cycle of $400$ years, every year the year starts one day later in the week, except that if the year is a leap year then the next year starts two days later in the week. By starting at Saturday, January 1, 2000 and tabulating the day of the week on January 1 for each year of the cycle, you can develop a table from which you can count you many times a leap year starts on each day of the week. (You can also use various techniques to count these days more efficiently, but that doesn't change the result.)
The result is that in every $400$ years,
$15$ leap years start on Sunday,
$13$ leap years start on Monday,
$14$ leap years start on Tuesday,
$14$ leap years start on Wednesday,
$13$ leap years start on Thursday,
$15$ leap years start on Friday, and
$13$ leap years start on Saturday.
That's a total of $97$ leap years, and you want the years that start on Saturday, Sunday, or Monday. The probability according to the Gregorian calendar therefore is
$$ \frac{13+15+13}{97} = \frac{41}{97}, $$
which is slightly less than $3/7$.
If we wanted to be even more pedantic, we might note that even though the Gregorian calendar does a much better job of synchronizing with the orbit of the Earth than the Julian calendar did, it is not perfect and the civil calendar might be modified again in the far future. But that's beyond anything we can assign an exact probability to.
The final conclusion is that your work is fine under certain simple assumptions (namely, a leap year every $4$ years without exception, which is what most of us alive today can expect to see in their lifetimes). Whether that is a good or bad answer according to the author of the book depends on what assumptions the author made, although if they were assuming $97$ leap years every $400$ years they really should have said something about that.