ProbabilityMCQMTP Nov 20Question 2799 of 295
All Questions

What is the probability of getting neither total of 7 nor 11 when the pair of dice is tossed?

Options

A79\displaystyle \frac{7}{9}
B29\displaystyle \frac{2}{9}
C39\displaystyle \frac{3}{9}
D49\displaystyle \frac{4}{9}
For any discrepancies in this question, email contact@cadada.in

Correct Answer

Option a79\displaystyle \frac{7}{9}

All Options:

  • A79\displaystyle \frac{7}{9}
  • B29\displaystyle \frac{2}{9}
  • C39\displaystyle \frac{3}{9}
  • D49\displaystyle \frac{4}{9}

Ad

Detailed Solution & Explanation

When a pair of fair six-sided dice is tossed, the total number of possible outcomes is 6×6=36\displaystyle 6 \times 6 = 36. 1. Let's find the outcomes that sum to 7 or 11: - Sum of 7: {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} (6 outcomes) - Sum of 11: {(5, 6), (6, 5)} (2 outcomes) Total outcomes with sum 7 or 11 = 6+2=8\displaystyle 6 + 2 = 8. 2. The probability of getting a sum of 7 or 11 is: P(7 or 11)=836=29P(7 \text{ or } 11) = \frac{8}{36} = \frac{2}{9} 3. The probability of getting neither sum is the complement: P(neither 7 nor 11)=1P(7 or 11)=129=79P(\text{neither } 7 \text{ nor } 11) = 1 - P(7 \text{ or } 11) = 1 - \frac{2}{9} = \frac{7}{9} This corresponds to Option A. 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.

Key Concepts to Understand

More Questions from Probability

Ready to Master Probability?

Practice all 295 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.

Start Practicing — It's Free