Find the probability that a leap year chosen at random will have 53 fridays .

Concept: A common year has 365 days = 52 weeks + 1 extra (“odd”) day. A leap year has 366 days = 52 weeks + 2 odd days. Because these odd days come right after 52 complete weeks, they are always two CONSECUTIVE days of the week, and which pair occurs depends only on which day the year starts on — giving 7 equally likely pairs in total.

1. List the 7 possible pairs of consecutive odd days, one for each possible starting day of the year: (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun).

2. For the year to contain 53 Fridays instead of the usual 52, Friday must be one of the two odd days.

3. Count how many of the 7 pairs contain Friday: the Thursday–Friday pair and the Friday–Saturday pair — exactly two pairs.

4. Since all 7 pairs are equally likely, the required probability = (pairs containing Friday) / (total pairs) = 2/7.

Cross-check: By the same listing, every day of the week appears in exactly two of the seven consecutive pairs (once as the earlier day, once as the later day). This symmetry confirms the counting method is consistent and not a one-off coincidence for Friday.

Result: the probability that a randomly chosen leap year has 53 Fridays is 2/7.