Discussion
What is the possibility of having 53 Thursdays in a non-leap year?
Answer: B
A non-leap year has 365 days:
\[ 365 = 52 \times 7 + 1 \]
So there are 52 Thursdays for certain, plus one leftover day. A 53rd Thursday occurs only if that day is a Thursday, and it is equally likely to be any of the seven days.
\[ P = \frac{1}{7} \]
A leap year has two leftover days and the probability rises to \( \tfrac{2}{7} \).
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