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: Determine the number of days and extra days in a non-leap year.
A non-leap year has 365 days, which is \( 52 \) weeks and \( 1 \) extra day (\( 365 = 52 \times 7 + 1 \)).
Step 2: Identify the possible outcomes for the extra day.
The extra day can be any of the 7 days of the week: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}. Each is equally likely with a probability of \( \frac{1}{7} \).
Step 3: Determine the conditions for 53 Sundays or 53 Saturdays.
For 53 Sundays, the extra day must be a Sunday.
For 53 Saturdays, the extra day must be a Saturday.

Step 4: Calculate the probability of each event.
\[ P(53 \text{ Sundays}) = P(\text{extra day is Sunday}) = \frac{1}{7} \] \[ P(53 \text{ Saturdays}) = P(\text{extra day is Saturday}) = \frac{1}{7} \]
Step 5: Calculate the probability of 53 Sundays or 53 Saturdays.
Since these are mutually exclusive events,
\[ P(53 \text{ Sundays or } 53 \text{ Saturdays}) = P(53 \text{ Sundays}) + P(53 \text{ Saturdays}) = \frac{1}{7} + \frac{1}{7} = \frac{2}{7} \]