The probability of not getting 53 Sundays in a leap year is

The correct answer is Option (4) → \(\displaystyle\frac{5}{7}\)

A leap year has 366 days:

    \(366 = 52 \times 7 + 2\)

The 52 full weeks give 52 Sundays. The 2 extra days are consecutive days of the week, and they can be any one of 7 equally likely pairs:

    (Sun, Mon), (Mon, Tue), (Tue, Wed), (Wed, Thu), (Thu, Fri), (Fri, Sat), (Sat, Sun)

The year has 53 Sundays only when a Sunday is one of the extra days, which happens in 2 of these 7 pairs:

    \( \displaystyle P(\text{53 Sundays}) = \frac{2}{7} \)

The question asks for the complement:

    \( \displaystyle P(\text{not 53 Sundays}) = 1 - \frac{2}{7} = \frac{5}{7} \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{1}{7}\) — Incorrect. Only 2 of the 7 pairs contain a Sunday, so the required probability is \(\frac{5}{7}\), not \(\frac{1}{7}\).

• Option (2) \(\displaystyle\frac{2}{7}\) — Incorrect. This is the probability of getting 53 Sundays, not of not getting them.

• Option (3) \(\displaystyle\frac{3}{7}\) — Incorrect. 5 of the 7 pairs contain no Sunday, so the probability is \(\frac{5}{7}\).