The probability that a non-leap year selected at random will contain 53 Sundays is
\(\dfrac{1}{7}\)
\(\dfrac{2}{7}\)
\(\dfrac{3}{7}\)
\(\dfrac{5}{7}\)
Step 1: A non-leap year has 365 days.
Step 2: Divide 365 days by 7 (since 1 week = 7 days).
\(365 = 52 \times 7 + 1\)
This means: 52 full weeks and 1 extra day.
Step 3: In 52 full weeks, there are exactly 52 Sundays (one in each week).
Step 4: Whether we get a 53rd Sunday depends on what that extra day is.
Step 5: The extra day can be any one of the 7 days: Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, or Saturday.
Step 6: Out of these 7 possibilities, only 1 case (when the extra day is Sunday) gives us 53 Sundays.
Step 7: Therefore, the probability is:
\(\dfrac{\text{Favourable outcomes}}{\text{Total outcomes}} = \dfrac{1}{7}\)
Final Answer: Option (A) \(\dfrac{1}{7}\)