ProbabilityMCQMTP Mar 22Question 3409 of 187
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Four unbiased coins are tossed simultaneously. The expected number of heads is:X: | 0 | 1 | 2 | 3 | 4P(X) | 116\displaystyle \frac{1}{16} | 416\displaystyle \frac{4}{16} | 616\displaystyle \frac{6}{16} | 416\displaystyle \frac{4}{16} | 116\displaystyle \frac{1}{16}

Options

A1\displaystyle 1
B2\displaystyle 2
C3\displaystyle 3
D4\displaystyle 4
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Correct Answer

Option a1\displaystyle 1

All Options:

  • A1\displaystyle 1
  • B2\displaystyle 2
  • C3\displaystyle 3
  • D4\displaystyle 4

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Detailed Solution & Explanation

**Probability of 53 Sundays in a Non-Leap Year** A non-leap year contains 365 days, which is equal to 52 weeks and 1 extra day. The 1 extra day can be any of the 7 days of the week: {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}\{\text{Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}\} So, the total number of outcomes is 7. The year will have 53 Sundays if and only if the extra day is a Sunday. So, the number of favorable outcomes is 1. Thus, the probability is: P=17P = \frac{1}{7} Hence, **Option A** 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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