<p>The probability that a year chosen at random has 53 Sundays, is</p>
<p>(a) \(\frac{1}{7}\)</p>
<p>(b) \(\frac{2}{7}\)</p>
<p>(c) \(\frac{3}{7}\)</p>
<p>(d) \(\frac{4}{7}\)</p>
Step-by-Step Solution
Key Concept: A non-leap year has 365 days = 52 weeks + 1 day, so exactly one day appears 53 times. For 53 Sundays to occur, that extra day must be a Sunday. Since any day of the week is equally likely to be that extra day, the probability is 1/7.
<p><strong>Step 1: Analyze the structure of a year</strong></p><p>A non-leap year has 365 days.</p><p>365 = 7 × 52 + 1</p><p>This means a non-leap year contains exactly 52 complete weeks plus 1 extra day.</p><p><strong>Step 2: Determine which days appear 53 times</strong></p><p>In 52 complete weeks, every day of the week (Monday through Sunday) appears exactly 52 times.</p><p>The 1 extra day means exactly one day of the week will appear 53 times in that year.</p><p><strong>Step 3: Identify the condition for 53 Sundays</strong></p><p>For a year to have 53 Sundays, Sunday must be that extra day occurring beyond the 52 complete weeks.</p><p><strong>Step 4: Calculate the probability</strong></p><p>The extra day can be any of the 7 days of the week with equal likelihood.</p><p>Therefore, the probability that the extra day is Sunday = $\frac{1}{7}$</p><p><strong>Step 5: Consider all years (assumption)</strong></p><p>For a random year chosen from non-leap years, this probability applies directly.</p><p>∴ <strong>Answer: A</strong></p>
Correct Answer: A