Question

What is the probability that a leap year has $53$ Tuesdays and $53$ Mondays?

Answer 1

A leap year has $366$ days. Now $364$ is divisible by $7$ and therefore there will be two excess weekdays in a leap year.

The two excess weekdays can be (Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday). So, the sample space S has $7$ pairs of excess weekdays. i.e. $n\left(S\right)=7$

Now we want the desired event E to have $53$ Mondays and $53$ Tuesdays. E consists of only one pair in S which is (Monday, Tuesday). So, $n\left(E\right)=1$

Hence, the probability that a leap year will contain $53$ Mondays and $53$ Tuesdays $=\frac{n\left(E\right)}{n\left(S\right)}=\frac{1}{7}$