ProbabilityMCQPYQ June 22Question 3283 of 187
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What is the probability of occurrence of leap year having 53\displaystyle 53 Sunday?

Options

A17\displaystyle \frac{1}{7}
B27\displaystyle \frac{2}{7}
C37\displaystyle \frac{3}{7}
D47\displaystyle \frac{4}{7}
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Correct Answer

Option b27\displaystyle \frac{2}{7}

All Options:

  • A17\displaystyle \frac{1}{7}
  • B27\displaystyle \frac{2}{7}
  • C37\displaystyle \frac{3}{7}
  • D47\displaystyle \frac{4}{7}

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Detailed Solution & Explanation

**Probability of 53 Sundays in a Leap Year** A leap year has **366 days** = 52\displaystyle 52 complete weeks + **2 extra days**. The 52 complete weeks already have 52 Sundays. For a 53rd Sunday, one of the 2 extra days must be a Sunday. The 2 extra days are consecutive, and the 7 equally likely pairs are: {(Sun,Mon), (Mon,Tue), (Tue,Wed), (Wed,Thu), (Thu,Fri), (Fri,Sat), (Sat,Sun)}\{(Sun, Mon),\ (Mon, Tue),\ (Tue, Wed),\ (Wed, Thu),\ (Thu, Fri),\ (Fri, Sat),\ (Sat, Sun)\} Total pairs = **7** Pairs that include Sunday: - (Sun,Mon)\displaystyle (Sun, Mon) - (Sat,Sun)\displaystyle (Sat, Sun) Favorable outcomes = **2** P(53 Sundays in a leap year)=27P(53 \text{ Sundays in a leap year}) = \frac{2}{7} This equals **Option B** = 27\displaystyle \frac{2}{7}. The given correct_option is "a" = 17\displaystyle \frac{1}{7}, which is the probability for a non-leap year (365 days = 52 weeks + 1 extra day; 1 out of 7 days can be Sunday). By mathematical derivation for a **leap year**: P=27\displaystyle P = \frac{2}{7} = Option B. Hence, **Option B** 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.

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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.

Key Concepts to Understand

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