The probability of not getting 53 Sundays in a leap year is

$\frac{5}{7}$

The correct answer is Option (4) → $\frac{5}{7}$

A leap year has $366$ days.

$366 = 52$ weeks $+ 2$ days.

Hence two consecutive days in the week will occur $53$ times.

Total possible ordered pairs of consecutive days = $7$.

53 Sundays occur only when Sunday is one of these two extra days, i.e., when the year starts on Saturday or Sunday.

Number of favourable cases for getting $53$ Sundays = $2$.

So,

$P(\text{53 Sundays})=\frac{2}{7}$

$P(\text{not getting 53 Sundays})=1-\frac{2}{7}=\frac{5}{7}$

Final answer: $\frac{5}{7}$