ProbabilityPYQ May 25Question 4385 of 187

In a leap year, what is the probability that there will be 53 Sundays?

Options

\displaystyle \frac{53}{365}

\displaystyle \frac{1}{7}

\displaystyle \frac{3}{7}

\displaystyle \frac{2}{7}

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Correct Answer

✅ Option d — $\displaystyle \frac{2}{7}$

All Options:

A $\displaystyle \frac{53}{365}$
B $\displaystyle \frac{1}{7}$
C $\displaystyle \frac{3}{7}$
D $\displaystyle \frac{2}{7}$

Detailed Solution & Explanation

Let us analyze the days in a leap year:
1. A leap year has

\displaystyle 366

days.
2. A week consists of

\displaystyle 7

days. Therefore, the number of complete weeks in a leap year is:

\frac{366}{7} = 52 \text{ weeks and } 2 \text{ remaining days}

3. The 52 complete weeks guarantee that there will be exactly 52 of each day of the week, including 52 Sundays.
4. The occurrence of a 53rd Sunday depends entirely on the remaining 2 consecutive days. The possible combinations for these 2 remaining days are:

S = \{(\text{Monday, Tuesday}), (\text{Tuesday, Wednesday}), (\text{Wednesday, Thursday}),

(\text{Thursday, Friday}), (\text{Friday, Saturday}), (\text{Saturday, Sunday}), (\text{Sunday, Monday})\}

The total number of outcomes in the sample space is

\displaystyle N(S) = 7

.
5. The combinations containing a "Sunday" are:

E = \{(\text{Saturday, Sunday}), (\text{Sunday, Monday})\} \implies n(E) = 2

6. Therefore, the probability that a leap year has 53 Sundays is:

P(E) = \frac{n(E)}{N(S)} = \frac{2}{7}

Hence, **Option D** is the correct answer.

About This Chapter: Probability

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Probability Operations, Expected Value

A logic-heavy chapter dealing with random experiments, events (mutually exclusive, exhaustive), set theory probability, conditional probability, and Bayes' Theorem. It forms the basis for Theoretical Distributions.

View Official ICAI Syllabus

Exam Strategy Tip

Always draw a quick Venn Diagram or tree when faced with 'At least one' or 'Only A but not B' wording. It saves you from double-counting.

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