What is the probability that a leap year selected at random, will have either 53 Sundays or 53 Saturdays?

$\frac{3}{7}$

We know that a Leap year has 366 Days that means 52 weeks and 2 extra days.

The sample space for these $2$ extra days is given below

S = {(Sunday, Monday), (Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday))}

There are 7 numbers of cases.

Thus, possible events when a leap year has either 1 extra Sunday or Saturday = {(Sunday, Monday), (Friday, Saturday), (Saturday, Sunday))} = 3 cases

Required Probability = 3/7

The correct answer is Option (1) → $\frac{3}{7}$