<p>Find the probability that a leap year will have 53 Fridays or 53 Saturdays.</p>
Step-by-Step Solution
Key Concept: A leap year has 366 days = 52 weeks + 2 days, so exactly one day of the week appears 53 times (the day on which the year starts or the extra 2 days fall). Find the probability that this extra day is Friday or Saturday.
<p><strong>Step 1:</strong> A leap year has 366 days = 52 weeks + 2 days.</p><p><strong>Step 2:</strong> Since 366 = 52(7) + 2, exactly 2 days of the week appear 53 times in a leap year. These must be consecutive days determined by which day the year starts on.</p><p><strong>Step 3:</strong> There are 7 equally likely possibilities for the starting day (Mon, Tue, Wed, Thu, Fri, Sat, Sun).</p><p><strong>Step 4:</strong> Count favorable outcomes where Friday or Saturday appears 53 times:</p><ul><li>If year starts on <strong>Friday</strong>: 53 Fridays and 53 Saturdays ✓</li><li>If year starts on <strong>Saturday</strong>: 53 Saturdays and 53 Sundays ✓</li><li>If year starts on <strong>Thursday</strong>: 53 Thursdays and 53 Fridays ✓</li></ul><p><strong>Step 5:</strong> Favorable outcomes = 3 (Thursday, Friday, Saturday starts)</p><p><strong>Step 6:</strong> Probability = $\dfrac{3}{7}$</p><p>∴ Answer: $\dfrac{3}{7}$</p>
Correct Answer: \(\dfrac{3}{7}\)