Question

Find the chance that a non-leap year contains $53$ Saturdays

• A. $\frac{2}{7}$
• B. $\frac{1}{7}$
• C. $\frac{2}{365}$
• D. $\frac{1}{365}$

Answer 1

The correct option is B $\frac{1}{7}$

non-leap year has $365$ days which is divided into $7\times 52=364+1$ days. There will be $52$ fridays, one for each week except when the first day of the year is a Friday. That probability is $P\left(E\right)=\frac{1}{7}$