Discussion
Find the probability that a leap year selected at random will contain 53 Sundays.
Answer: D
The probability is \(\frac{2}{7}\).
A leap year has 366 days, which is 52 full weeks plus 2 days. The 52 weeks give 52 Sundays for certain; a 53rd Sunday occurs only if one of the 2 extra days is a Sunday.
The two extra days form one of 7 equally likely pairs: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun). Two of these contain a Sunday, so
\[ P = \frac{2}{7} \]
\(\frac{1}{7}\) would be the answer for an ordinary year of 365 days (one extra day). \(\frac{5}{7}\) is the chance of not getting 53 Sundays, and \(\frac{3}{4}\) and \(\frac{4}{7}\) match no count of the pairs.
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