A leap year is chosen at random. What is the probability that it contains 53 Sundays?

A leap year has $366=52\times 7+2$ days, so 52 Sundays are certain and only the 2 leftover days can supply a 53rd. Those two days form one of 7 equally likely consecutive pairs — (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun) — and a Sunday appears in 2 of them, giving 2/7. The answer 1/7 belongs to an ordinary year, which has just 1 leftover day.