Question

The probability that a leap year will have $53$ Fridays or $53$ Saturdays is

• A. $\frac{2}{7}$
• B. $\frac{3}{7}$
• C. $\frac{4}{7}$
• D. $\frac{1}{7}$

Answer 1

The correct option is B $\frac{3}{7}$

We know that a leap year has 366 days $(i.e.,7\times 52+2)=52$ 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 $7$ cases.

$\therefore n\left(S\right)=7$

Let $E$ be the event that the leap year has $53$ Fridays or $53$ Saturdays.

i.e., $n\left(E\right)=3$

$\therefore p\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}=\frac{3}{7}$

Hence, the probability that a leap year has 53 Fridays or 53 Saturdays is $\frac{3}{7}$.