What is the probability that a leap year will have 53 Sundays?

Consider the number of complete weeks in a leap year and the scenarios in which a leap year could have 53 Sundays.

The probability of an event E occurring is the ratio of the number of cases in its favour to the total number of equally likely cases.

$P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{\text{ no. of cases favourable to event }E}{\text{ Total no. of cases }}$

A leap year has 366 days $=52\times 7+2$ days

That means there will be 52 Sunday/Monday/Saturdays plus 2 additional days. Now these 2 additional days would be combination of any of the successive two days.

i)Sunday + Monday

ii)Monday + Tuesday

iii)Tuesday + Wednesday

iv) Wednesday + Thursday

v)Thursday + Friday

vi)Friday + Saturday

vii)Saturday + Sunday

Now the probability of two Sundays will be (i) & (vii) =2 outcomes out of above 7 outcomes. So the probability of having two additional Sundays =2 / 7 .