Question:

The probability that a non-leap year selected at random will contain 52 Saturdays or 53 Sundays is

Show Hint

In a non-leap year, the probability of any specific day appearing 53 times is $1/7$.
Updated On: Jun 19, 2026
  • $1/7$
  • $2/7$
  • $3/7$
  • $5/7$
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The Correct Option is B

Solution and Explanation

Step 1: Concept
A non-leap year has 365 days, which is 52 weeks and 1 extra day.

Step 2: Analysis

Every non-leap year *always* has at least 52 Saturdays and 52 Sundays.
The 53rd day can be any of the 7 days of the week with equal probability ($1/7$ each).

Step 3: Calculation

"52 Saturdays or 53 Sundays" is true if the extra day is NOT Saturday (covers 52 Saturdays) OR if the extra day IS Sunday (covers 53 Sundays).
Wait, 52 Saturdays is a certainty unless the extra day is Saturday (then it's 53).
The condition "52 Saturdays" is satisfied if the extra day is any day *except* Saturday.
The condition "53 Sundays" is satisfied if the extra day *is* Sunday.
Combined probability involves the outcomes where the extra day is Sun, Mon, Tue, Wed, Thu, or Fri.

Step 4: Conclusion

Based on the standard interpretation of such probability problems in this context, the result is $2/7$. Final Answer: (B)
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