Question

The probability that a non-leap year contains 53 Sundays is

• A. \( \frac{1}{7} \)
• B. \( \frac{1}{9} \)
• C. \( \frac{2}{7} \)
• D. \( \frac{1}{5} \)

Hint:

For a non-leap year, there are always 52 Sundays, and the extra day can be any of the 7 days. The probability of it being a Sunday is \( \frac{1}{7} \).

Answer 1

The Correct Option is A

Step 1: A non-leap year has 365 days, and since a week contains 7 days, we divide 365 by 7: \[ 365 \div 7 = 52 \text{ weeks with 1 extra day} \] This means the year consists of 52 complete weeks and 1 extra day.
Step 2: Since there are 52 complete weeks, there are 52 Sundays in the non-leap year. The extra day can be any one of the 7 days of the week, including Sunday.
Step 3: To have 53 Sundays in the year, the extra day must be a Sunday. Since the extra day is equally likely to be any of the 7 days, the probability that it is a Sunday is: \[ \frac{1}{7} \] Thus, the probability that a non-leap year contains 53 Sundays is \( \frac{1}{7} \).