The probability that a non-leap year selected at random will contain $53$ Sundays is:
(a) $\dfrac{1}{7}$
(b) $\dfrac{2}{7}$
(c) $\dfrac{3}{7}$
(d) $\dfrac{5}{7}$
Step-by-Step Solution
Key Concept: A non-leap year has 365 days $= 52 \text{ weeks} + 1 \text{ extra day}$. Favourable outcome $= 1$ (Sunday). Total outcomes $= 7$ days.
365 days $= 52$ full weeks $+$ 1 extra day. [0.5 Mark]
Probability of the 1 extra day being Sunday $= \dfrac{1}{7}$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Identifying 1 extra day in a non-leap year: 0.5 Mark
Calculating probability $= 1/7$: 0.5 Mark
Correct Answer: $\dfrac{1}{7}$