ProbabilityMCQMTP Nov 20Question 3377 of 187
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A pair of dice rolled. If the sum of the two dice is 9\displaystyle 9, find the prob. that one of the dice showed is 3\displaystyle 3.

Options

A13\displaystyle \frac{1}{3}
B14\displaystyle \frac{1}{4}
C12\displaystyle \frac{1}{2}
D18\displaystyle \frac{1}{8}
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Correct Answer

✅ Option c — 12\displaystyle \frac{1}{2}

All Options:

  • A13\displaystyle \frac{1}{3}
  • B14\displaystyle \frac{1}{4}
  • C12\displaystyle \frac{1}{2}
  • D18\displaystyle \frac{1}{8}

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Detailed Solution & Explanation

**Conditional Probability of Getting a 3 Given the Sum is 9** When a pair of dice is rolled: **Step 1: Identify the given condition (Sum is 9)** Let S\displaystyle S be the event that the sum of the faces is 9. The outcomes that sum to 9 are: S={(3,6),(4,5),(5,4),(6,3)}S = \{(3, 6), (4, 5), (5, 4), (6, 3)\} So, n(S)=4\displaystyle n(S) = 4. **Step 2: Identify the favorable outcomes** Let F\displaystyle F be the event that at least one of the dice shows 3. The outcomes in S\displaystyle S that satisfy this condition are: {(3,6),(6,3)}\{(3, 6), (6, 3)\} So, n(F∩S)=2\displaystyle n(F \cap S) = 2. **Step 3: Calculate the conditional probability** P(F/S)=n(F∩S)n(S)=24=12P(F/S) = \frac{n(F \cap S)}{n(S)} = \frac{2}{4} = \frac{1}{2} Hence, **Option C** 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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