Question

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

Answer 1

Given: A leap year contains 366 days.

\[ 366\ \text{days} = 52\ \text{weeks} + 2\ \text{extra days} \]

52 weeks contain 52 Mondays and 52 Tuesdays for certain.

The remaining 2 extra consecutive days can be any of the following 7 pairs:

\[ S = {\text{(Sunday, Monday)}, \text{(Monday, Tuesday)}, \text{(Tuesday, Wednesday)}, \text{(Wednesday, Thursday)}, \text{(Thursday, Friday)}, \text{(Friday, Saturday)}, \text{(Saturday, Sunday)}} \]

\[ n(S) = 7 \]

For the year to have both 53 Mondays and 53 Tuesdays, the two extra days must be Monday and Tuesday.

Favourable outcome = `({\text{(Monday, Tuesday)}})`

\[ m = 1 \]

Answer: \[ P(53\ \text{Mondays and } 53\ \text{Tuesdays}) = \dfrac{m}{n(S)} = \dfrac{1}{7} \]