For the Question "Find the probability that a leap year has 53 Sundays".
The Solution goes :
For 53 Sundays, we proceed as: $\frac{366}{7} = 52.28$; So we can be sure that there are 52 Sundays, now the only question is of the 1 Sunday. So $7\cdot 52 = 364$. Therefore the remaining 2 days (366-364) decide the probability of being a Sunday. So we consider pairings of 2 consecutive days –
- Sunday-Monday
- Monday-Tuesday
- Tuesday-Wednesday
- Wednesday-Thursday
- Thursday-Friday
- Friday-Saturday
- Saturday-Sunday
So out of the 7 pairs, we have 2 pairs that include a Sunday. Therefore the probability of having 53 Sundays in a Leap Year is $\frac{2}{7}$.
Then I was wondering, what if we ask "Probability that a leap year has 52 Sundays", Considering the above solution for 53 Sundays, the answer to the 52 Sundays problem should be 1. (Since $\frac{366}{7} = 52.28$) So we have complete 52 weeks and hence always 52 Sundays, therefore 1.
I found some online forums, where the answer to $52$ Sundays Problem is not 1. Hence my confusion. Please help.
Best Answer
By the way, the exact probability that a leap year has 53 Sundays is a bit higher than 2/7: The distribution of days of the week repeats exactly every 400 years: January 1st, 2000, was a Saturday, as will be January 1st, 2400. Within these 400 years, there are 97 leap years, 28 of which start on a Saturday or Sunday; so the probability is $28/97\approx 0.28866$ rather than $2/7\approx 0.28571$.