Question

Find the probability of getting 53 Sundays in a leap year.

Answer 1

A normal year has 52 Mondays, 52 Tuesdays, 52 Wednesdays, 52 Thursdays, 52 Fridays, 52 Saturdays and 52 Sundays + 1 day that could be anything depending upon the year under consideration. In addition to this, a leap year has an extra day which might be a Monday or Tuesday or Wednesday…or Sunday. We've now reduced the question to : what is the probability that in a given pair of consecutive days of the year one of them is a Sunday?Our sample space is S : {Monday-Tuesday, Tuesday-Wednesday, Wednesday-Thursday,…, Sunday-Monday}Number of elements in S = n(S) = 7What we want is a set A (say) that comprises of the elements Saturday-Sunday and Sunday-Monday i.e. A : {Saturday-Sunday, Sunday-Monday}Number of elements in set A = n(A) = 2By definition, probability of occurrence of $A=\frac{n\left(A\right)}{n\left(S\right)}=\frac{2}{7}$ Therefore, probability that a leap year has 53 Sundays is $\frac{2}{7}$. (Note that this is true for any day of the week, not just Sunday)