Question

Find the probability that a non-leap year contain $53$ Sundays.

Answer 1

Normal year has $52$ weeks $\left(\frac{365}{7}\right)+1$ extra day, i.e., $52$ Sundays $+1$ day which can be any of the seven days.

So, probability of $53$ Sundays $=$ Probability of that day being a Sunday $=\frac{1}{7}$.

or the year must start on a Sunday, so you have only one favourable chance out of total $7$ possibilities $-$ a probability of $\frac{1}{7}$.