Two dice are thrown simultaneously. The probability of a total score of 5 from the out comes of dice is.

Option c: 1/9. Solution: Probability of Sum = 5 with Two Dice Total outcomes = 6 × 6 = 36 Favorable outcomes where sum = 5: - (1, 4) : 1+4=5 ✓ - (2, 3) : 2+3=5 ✓ - (3, 2) : 3+2=5 ✓ - (4, 1) : 4+1=5 ✓ Number of favorable outcomes = 4 P(sum=5) = 4/36 = 1/9 Wait — 4/36 = 1/9 , which corresponds to Option C. But the correctoption given is "a" ( 1/18 ). Let me re-verify: 4/36 = 1/9 . The correct answer by calculation is 1/9 = Option C. However, the given correctoption is "a" = 1/18 . Since we must derive from first principles: P(sum=5) = 4/36 = 1/9 This matches Option C. Hence, Option C is the correct answer.

Two dice are thrown simultaneously. The probability of a total score of $5$ from the out comes of dice is.

The correct answer is Option c: \frac{1}{9}. **Probability of Sum = 5 with Two Dice** Total outcomes = $6 \times 6 = 36$ Favorable outcomes where sum = 5: - $(1, 4)$: $1+4=5$ ✓ - $(2, 3)$: $2+3=5$ ✓ - $(3, 2)$: $3+2=5$ ✓ - $(4, 1)$: $4+1=5$ ✓ Number of favorable outcomes = **4** $$P(\text{sum}=5) = \frac{4}{36} = \frac{1}{9}$$ Wait — $\frac{4}{36} = \frac{1}{9}$, which corresponds to Option C. But the correct_option given is "a" ($\frac{1}{18}$). Let me re-verify: $\frac{4}{36} = \frac{1}{9}$. The correct answer by calculation is $\frac{1}{9}$ = Option C. However, the given correct_option is "a" = $\frac{1}{18}$. Since we must derive from first principles: $$P(\text{sum}=5) = \frac{4}{36} = \frac{1}{9}$$ This matches **Option C**. Hence, **Option C** is the correct answer.

ProbabilityMCQPYQ Jan. 21Question 3279 of 187

All Questions

Two dice are thrown simultaneously. The probability of a total score of $\displaystyle 5$ from the out comes of dice is.

Options

\displaystyle \frac{1}{18}

\displaystyle \frac{1}{12}

\displaystyle \frac{1}{9}

\displaystyle \frac{2}{5}

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Correct Answer

✅ Option c — $\displaystyle \frac{1}{9}$

All Options:

A $\displaystyle \frac{1}{18}$
B $\displaystyle \frac{1}{12}$
C $\displaystyle \frac{1}{9}$
D $\displaystyle \frac{2}{5}$

Detailed Solution & Explanation

**Probability of Sum = 5 with Two Dice** Total outcomes =

\displaystyle 6 \times 6 = 36

Favorable outcomes where sum = 5: -

\displaystyle (1, 4)

\displaystyle 1+4=5

✓ -

\displaystyle (2, 3)

\displaystyle 2+3=5

✓ -

\displaystyle (3, 2)

\displaystyle 3+2=5

✓ -

\displaystyle (4, 1)

\displaystyle 4+1=5

✓ Number of favorable outcomes = **4**

P(\text{sum}=5) = \frac{4}{36} = \frac{1}{9}

Wait —

\displaystyle \frac{4}{36} = \frac{1}{9}

, which corresponds to Option C. But the correct_option given is "a" (

\displaystyle \frac{1}{18}

). Let me re-verify:

\displaystyle \frac{4}{36} = \frac{1}{9}

. The correct answer by calculation is

\displaystyle \frac{1}{9}

= Option C. However, the given correct_option is "a" =

\displaystyle \frac{1}{18}

. Since we must derive from first principles:

P(\text{sum}=5) = \frac{4}{36} = \frac{1}{9}

This matches **Option C**. 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.

Key Concepts to Understand

Probability

A numerical measure of the likelihood of an event occurring, expressed as a number between 0 (impossible) and 1 (certain). Formula: P(A) = Number of favorable outcomes / Total number of equally likely outcomes.

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Two dice are thrown simultaneously. The probability of a total score of 5\displaystyle 55 from the out comes of dice is.