The probability of not getting 53 Tuesdays in a leap year is:

$\frac{5}{7}$

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

No. of days in leap year = 366

No. of perfect weeks = $\frac{366}{7}$

= 52 weeks

2 days remain

let M, T, W, Th, F, Sat, Sun denote days

Last two days can be M T, T W, W Th, Th F, F Sat, Sat Sun, Sun M

So there are 5 cases of not getting 53 Tuesdays

$P=\frac{5}{7}$