A leap year has 366 days (52 weeks + 2 days). If a year is chosen uniformly at random among leap years, what is the probability that it contains 53 Mondays?

Introduction / Context:A leap year has 366 days, which is 52 full weeks plus 2 extra days. Exactly two weekdays will occur 53 times; the others occur 52 times. We want the chance that Monday is among the 53-times weekdays.

Given Data / Assumptions:

• Leap year: 366 days = 52 weeks + 2 days.
• Start-of-year weekday is equally likely to be any of the 7 weekdays.
• Two consecutive weekdays occur 53 times each; all others occur 52 times.

Concept / Approach:Let the first day of the year be D. Then the two 53-times weekdays are D and the weekday immediately following D. Monday is a 53-times weekday iff D is Monday or Sunday.

Step-by-Step Solution:Favorable start days for “53 Mondays”: Sunday or Monday → 2 possibilities.Total possible start days = 7.Probability = 2/7.

Verification / Alternative check:Compare with ordinary (365-day) years, where exactly one weekday is 53-times; then P(53 Mondays) = 1/7. The leap-year case doubles the favorable possibilities to 2/7.

Why Other Options Are Wrong:5/7 and 3/7 overstate; 1/7 is for ordinary years; “Data inadequate” is incorrect under the uniform start-day assumption.

Common Pitfalls:Forgetting that in leap years, two adjacent weekdays get the extra occurrences.

Final Answer:2/7