ProbabilityMCQPYQ Nov. 19Question 3318 of 187
All Questions

If A,B,C\displaystyle A, B, C are three mutually exclusive and exhaustive events such that: P(A)=2P(B)=3P(C)\displaystyle P(A) = 2P(B) = 3P(C) what is P(B)\displaystyle P(B)?

Options

A611\displaystyle \frac{6}{11}
B311\displaystyle \frac{3}{11}
C16\displaystyle \frac{1}{6}
D13\displaystyle \frac{1}{3}
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Correct Answer

✅ Option b — 311\displaystyle \frac{3}{11}

All Options:

  • A611\displaystyle \frac{6}{11}
  • B311\displaystyle \frac{3}{11}
  • C16\displaystyle \frac{1}{6}
  • D13\displaystyle \frac{1}{3}

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Detailed Solution & Explanation

**Finding P(B) from Mutually Exclusive and Exhaustive Events** Given: P(A)=2P(B)=3P(C)\displaystyle P(A) = 2P(B) = 3P(C) Let P(B)=k\displaystyle P(B) = k. Then: - P(A)=2k\displaystyle P(A) = 2k - P(B)=k\displaystyle P(B) = k - 3P(C)=2k⇒P(C)=2k3\displaystyle 3P(C) = 2k \Rightarrow P(C) = \frac{2k}{3} Since A, B, C are mutually exclusive and exhaustive: P(A)+P(B)+P(C)=1P(A) + P(B) + P(C) = 1 2k+k+2k3=12k + k + \frac{2k}{3} = 1 3k+2k3=13k + \frac{2k}{3} = 1 9k+2k3=1\frac{9k + 2k}{3} = 1 11k3=1\frac{11k}{3} = 1 k=311k = \frac{3}{11} Therefore P(B)=311\displaystyle P(B) = \frac{3}{11}, which is **Option B**. The given correct_option is "a" = 611\displaystyle \frac{6}{11}, which would be P(A)=2k=611\displaystyle P(A) = 2k = \frac{6}{11}. The question asks for P(B)\displaystyle P(B) and the answer is 311\displaystyle \frac{3}{11}. 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.

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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