Discussion
The probability of a leap year selected at random contain 53 Sunday is:
Answer: C
A leap year has 366 days, which is
\[ 366 = 52 \times 7 + 2 \]
So it holds 52 complete weeks - and therefore 52 Sundays for certain - plus two extra days. A 53rd Sunday appears only if one of those two is a Sunday.
The two leftover days are one of seven equally likely consecutive pairs:
\[ \{\text{Sun,Mon}\},\ \{\text{Mon,Tue}\},\ \{\text{Tue,Wed}\},\ \{\text{Wed,Thu}\},\ \{\text{Thu,Fri}\},\ \{\text{Fri,Sat}\},\ \{\text{Sat,Sun}\} \]
Two of the seven contain a Sunday, so
\[ P = \frac{2}{7} \]
In a non-leap year there is only one leftover day and the probability drops to \( \tfrac{1}{7} \).
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