What is the probability that any non-leap year will have 53 Sundays?

The correct option is 1/7.

To understand why, let us break down the calendar structure of a non-leap year step-by-step:
A standard non-leap year consists of 365 days.

We know that a week has 7 days. Let us calculate the number of complete weeks in a non-leap year by dividing the total number of days by 7:
Number of weeks = 365 / 7 = 52 weeks and 1 remaining day.

Since there are 52 complete weeks, every day of the week (including Sunday) will occur at least 52 times.
To have exactly 53 Sundays, the one remaining day of the year must be a Sunday.

This single remaining day can be any one of the seven days of the week:
{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}

Since there are 7 possible outcomes in total, and only 1 outcome is favorable (the day being a Sunday), the probability is:

$$\text{Probability}=\frac{1}{7}$$