Question

What Is The Probability That An Ordinary Year Has $53$ Sundays?

Answer 1

Find the Probability that an Ordinary Year has $53$ Sundays.

An Ordinary year contains $365$ days
$\therefore 52$ complete weeks and one day.
The following are some possibilities for this one day:
{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
Total possibilities are $7$.
Favorable cases will only be $1$.

Let the event of occurring $53$ Sundays be $\mathrm{S}$.

$$\begin{gathered} \mathrm{P}\left(\mathrm{S}\right)=\frac{\mathrm{no}.\mathrm{of}\mathrm{favourable}\mathrm{cases}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{possible}} \\ \mathrm{P}\left(\mathrm{S}\right)=\frac{1}{7} \end{gathered}$$
Hence, the Probability that an Ordinary Year has $53$ Sundays will be $\frac{1}{7}$.