ProbabilityPYQ May 25Question 4079 of 187
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Two dice are thrown simultaneously. Find the probability that the sum of digits on the two dice would be 8 or more.

Options

A518\displaystyle \frac{5}{18}
B512\displaystyle \frac{5}{12}
C536\displaystyle \frac{5}{36}
D820\displaystyle \frac{8}{20}
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Correct Answer

Option b512\displaystyle \frac{5}{12}

All Options:

  • A518\displaystyle \frac{5}{18}
  • B512\displaystyle \frac{5}{12}
  • C536\displaystyle \frac{5}{36}
  • D820\displaystyle \frac{8}{20}

Detailed Solution & Explanation

When two fair dice are thrown simultaneously, the total number of outcomes in the sample space is:
N(S)=6×6=36N(S) = 6 \times 6 = 36 Each outcome is represented by a pair (i,j)\displaystyle (i, j) where 1i,j6\displaystyle 1 \le i, j \le 6.
We want to find the probability that the sum S=i+j8\displaystyle S = i + j \ge 8. Let us list all the outcomes where the sum is 8, 9, 10, 11, or 12:
- **Sum = 8**: (2,6),(3,5),(4,4),(5,3),(6,2)\displaystyle (2,6), (3,5), (4,4), (5,3), (6,2) (5 outcomes)
- **Sum = 9**: (3,6),(4,5),(5,4),(6,3)\displaystyle (3,6), (4,5), (5,4), (6,3) (4 outcomes)
- **Sum = 10**: (4,6),(5,5),(6,4)\displaystyle (4,6), (5,5), (6,4) (3 outcomes)
- **Sum = 11**: (5,6),(6,5)\displaystyle (5,6), (6,5) (2 outcomes)
- **Sum = 12**: (6,6)\displaystyle (6,6) (1 outcome)

Total number of favorable outcomes is:
n(E)=5+4+3+2+1=15n(E) = 5 + 4 + 3 + 2 + 1 = 15 Therefore, the probability of obtaining a sum of 8 or more is:
P(E)=n(E)N(S)=1536=512P(E) = \frac{n(E)}{N(S)} = \frac{15}{36} = \frac{5}{12}
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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