What is the chance of throwing at least 7 in a single cast with two dices?

Option b: 7/12. Solution: Probability of Throwing a Sum of At Least 7 When two dice are thrown, the total number of outcomes is: Total outcomes = 6 × 6 = 36 Let S be the sum of the faces. We want S ≥ 7 . Let's count the favorable outcomes for each sum: - S = 7 : (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) ⟹ 6 outcomes - S = 8 : (2,6), (3,5), (4,4), (5,3), (6,2) ⟹ 5 outcomes - S = 9 : (3,6), (4,5), (5,4), (6,3) ⟹ 4 outcomes - S = 10 : (4,6), (5,5), (6,4) ⟹ 3 outcomes - S = 11 : (5,6), (6,5) ⟹ 2 outcomes - S = 12 : (6,6) ⟹ 1 outcome Total favorable outcomes: Favorable outcomes = 6 + 5 + 4 + 3 + 2 + 1 = 21 Thus, the probability is: P(S ≥ 7) = 21/36 = 7/12 Hence, Option B is the correct answer.

What is the chance of throwing at least $7$ in a single cast with two dices?

The correct answer is Option b: \frac{7}{12}. **Probability of Throwing a Sum of At Least 7** When two dice are thrown, the total number of outcomes is: $$\text{Total outcomes} = 6 \times 6 = 36$$ Let $S$ be the sum of the faces. We want $S \ge 7$. Let's count the favorable outcomes for each sum: - $S = 7$: $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$ $\Rightarrow$ 6 outcomes - $S = 8$: $\{(2,6), (3,5), (4,4), (5,3), (6,2)\}$ $\Rightarrow$ 5 outcomes - $S = 9$: $\{(3,6), (4,5), (5,4), (6,3)\}$ $\Rightarrow$ 4 outcomes - $S = 10$: $\{(4,6), (5,5), (6,4)\}$ $\Rightarrow$ 3 outcomes - $S = 11$: $\{(5,6), (6,5)\}$ $\Rightarrow$ 2 outcomes - $S = 12$: $\{(6,6)\}$ $\Rightarrow$ 1 outcome Total favorable outcomes: $$\text{Favorable outcomes} = 6 + 5 + 4 + 3 + 2 + 1 = 21$$ Thus, the probability is: $$P(S \ge 7) = \frac{21}{36} = \frac{7}{12}$$ Hence, **Option B** is the correct answer.

ProbabilityMCQMTP March 21Question 3380 of 187

All Questions

What is the chance of throwing at least $\displaystyle 7$ in a single cast with two dices?

Options

\displaystyle \frac{5}{12}

\displaystyle \frac{7}{12}

\displaystyle \frac{1}{4}

\displaystyle \frac{17}{36}

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — $\displaystyle \frac{7}{12}$

All Options:

A $\displaystyle \frac{5}{12}$
B $\displaystyle \frac{7}{12}$
C $\displaystyle \frac{1}{4}$
D $\displaystyle \frac{17}{36}$

Detailed Solution & Explanation

**Probability of Throwing a Sum of At Least 7** When two dice are thrown, the total number of outcomes is:

\text{Total outcomes} = 6 \times 6 = 36

Let

\displaystyle S

be the sum of the faces. We want

\displaystyle S \ge 7

. Let's count the favorable outcomes for each sum: -

\displaystyle S = 7

\displaystyle \{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}

\displaystyle \Rightarrow

6 outcomes -

\displaystyle S = 8

\displaystyle \{(2,6), (3,5), (4,4), (5,3), (6,2)\}

\displaystyle \Rightarrow

5 outcomes -

\displaystyle S = 9

\displaystyle \{(3,6), (4,5), (5,4), (6,3)\}

\displaystyle \Rightarrow

4 outcomes -

\displaystyle S = 10

\displaystyle \{(4,6), (5,5), (6,4)\}

\displaystyle \Rightarrow

3 outcomes -

\displaystyle S = 11

\displaystyle \{(5,6), (6,5)\}

\displaystyle \Rightarrow

2 outcomes -

\displaystyle S = 12

\displaystyle \{(6,6)\}

\displaystyle \Rightarrow

1 outcome Total favorable outcomes:

\text{Favorable outcomes} = 6 + 5 + 4 + 3 + 2 + 1 = 21

Thus, the probability is:

P(S \ge 7) = \frac{21}{36} = \frac{7}{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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