Step 1: Understanding the Question:
Determine the probability of an "OR" condition regarding specific counts of weekdays in a non-leap year.
Step 2: Key Formula or Approach:
1. Non-leap year $= 365$ days $= 52$ weeks $+ 1$ extra day.
2. Probability $= \frac{\text{favorable cases}}{\text{total cases}}$.
Step 3: Detailed Explanation:
The "extra day" can be any of the $7$ days of the week: {Mon, Tue, Wed, Thu, Fri, Sat, Sun}.
Let $A$ be the event that the year has $52$ Saturdays. This happens if the extra day is NOT Saturday.
Favorable extra days for $A = \{\text{Mon, Tue, Wed, Thu, Fri, Sun}\}$ ($6$ outcomes).
Let $B$ be the event that the year has $53$ Sundays. This happens if the extra day IS Sunday.
Favorable extra day for $B = \{\text{Sun}\}$ ($1$ outcome).
We need $P(A \cup B)$. Since event $B$ (extra day is Sunday) is already a subset of event $A$ (extra day is not Saturday), the union of outcomes is simply the set of extra days for event $A$.
$\{ \text{Mon, Tue, Wed, Thu, Fri, Sun} \} \cup \{ \text{Sun} \} = \{ \text{Mon, Tue, Wed, Thu, Fri, Sun} \}$.
Number of favorable outcomes $= 6$. Total outcomes $= 7$.
Probability $= \frac{6}{7}$.
Step 4: Final Answer:
The probability is $6/7$.