Question

The probability that a non-leap year selected at random will have 53 Sundays or 53 Saturdays is:

• A. \( \frac{1}{7} \)
• B. \( \frac{2}{7} \)
• C. \( 1 \)
• D. \( \frac{2}{365} \)

Hint:

Focus on the remainder when the number of days in a year is divided by 7. This remainder determines the number of days that occur 53 times.

Answer 1

The Correct Option is B

Step 1: Calculate days and extra days in a non-leap year.
\nA non-leap year has 365 days, equivalent to \( 52 \) weeks and \( 1 \) extra day (\( 365 = 52 \times 7 + 1 \)).\n\n
Step 2: Define possible outcomes for the extra day.
\nThe extra day can be any of the 7 weekdays: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}. Each has a probability of \( \frac{1}{7} \).\n\n
Step 3: Determine the conditions for 53 Sundays or 53 Saturdays.
\n 53 Sundays occur if the extra day is a Sunday.
\n 53 Saturdays occur if the extra day is a Saturday.
\n\n
Step 4: Determine the probability of each event.
\n\[\nP(53 \text{ Sundays}) = P(\text{extra day is Sunday}) = \frac{1}{7}\n\]\n\[\nP(53 \text{ Saturdays}) = P(\text{extra day is Saturday}) = \frac{1}{7}\n\]\n\n
Step 5: Calculate the probability of 53 Sundays or 53 Saturdays.
\nThese are mutually exclusive events, so
\n\[\nP(53 \text{ Sundays or } 53 \text{ Saturdays}) = P(53 \text{ Sundays}) + P(53 \text{ Saturdays}) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7}\n\]