Find the probability of getting $53$ Sundays in a non-leap year.
Step-by-Step Solution
Key Concept: Non-leap year $= 365$ days $= 52$ weeks $+ 1$ extra day.<br>Extra day can be any of the 7 days. Favourable $= 1$ (Sunday). $P = 1/7$.
1 extra day can be any day of the week (7 outcomes). [1.0 Mark]
$P(\text{Sunday}) = 1/7$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Identifying 1 extra day out of 7: 1.0 Mark
Evaluating probability $= 1/7$: 1.0 Mark
Correct Answer: