ProbabilityMCQMTP Nov 19Question 2797 of 295
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Three balls are drawn at random from a bag containing 6 blue and 4 red balls. What is the chance that 2 balls are blue and 1 is red?

Options

A14\displaystyle \frac{1}{4}
B34\displaystyle \frac{3}{4}
C44\displaystyle \frac{4}{4}
DNone of these
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Correct Answer

✅ Option d — None of these

All Options:

  • A14\displaystyle \frac{1}{4}
  • B34\displaystyle \frac{3}{4}
  • C44\displaystyle \frac{4}{4}
  • DNone of these

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Detailed Solution & Explanation

We are given a bag containing 6 blue and 4 red balls (total of 10 balls). Three balls are drawn at random. 1. Total number of ways to draw 3 balls from 10 is: (103)=10×9×83×2×1=120\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120 2. We want to find the number of ways to draw exactly 2 blue balls and 1 red ball: - Ways to select 2 blue balls from 6: (62)=15\displaystyle \binom{6}{2} = 15 - Ways to select 1 red ball from 4: (41)=4\displaystyle \binom{4}{1} = 4 - Favorable outcomes = 15×4=60\displaystyle 15 \times 4 = 60 3. The probability is: P(2 Blue and 1 Red)=60120=12P(\text{2 Blue and 1 Red}) = \frac{60}{120} = \frac{1}{2} Since 12\displaystyle \frac{1}{2} is not among Option A (14\displaystyle \frac{1}{4}), Option B (34\displaystyle \frac{3}{4}), or Option C (44\displaystyle \frac{4}{4}), the correct choice is Option D (None of these). 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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