Question

The probability that a non-leap year has $53$ Sundays is

• A. $\frac{2}{7}$
• B. $\frac{1}{7}$
• C. $\frac{3}{7}$
• D. None of these

Answer 1

The correct option is B

$\frac{1}{7}$

The explanation for the correct option

A non-leap year has $365$ days, which means $52$ weeks and $1$ extra day.

That one extra day can be Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday.

Thus, the total number of spaces is $T=7$.

The number of cases for the day is a Sunday is $S=1$.

Hence, the probability of $53$ Sundays can be given by $P\left(S\right)=\frac{S}{T}=\frac{1}{7}$.

Therefore, the probability that a non-leap year has $53$ Sundays is $\frac{1}{7}$.