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The probability of success of three students in CA Foundation examination are 1/5,1/4\displaystyle 1/5, 1/4 and 1/3\displaystyle 1/3 respectively. Find the probability that at least two students will get success.

Select Your Answer (Keyboard: A, B, C, D)
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Correct Answer

✅ Option c — 1/6\displaystyle 1/6

All Options:

  • A3/4\displaystyle 3/4
  • B2/5\displaystyle 2/5
  • C1/6\displaystyle 1/6
  • D1/5\displaystyle 1/5

Detailed Solution & Explanation

**Probability of Success of At Least Two Students** Let the events of success for the three students be A,B,\displaystyle A, B, and C\displaystyle C. Given: - P(A)=15⇒P(Aˉ)=45\displaystyle P(A) = 15\displaystyle \frac{1}{5} \Rightarrow P(Aˉ\displaystyle \bar{A}) = 45\displaystyle \frac{4}{5} - P(B)=14⇒P(Bˉ)=34\displaystyle P(B) = 14\displaystyle \frac{1}{4} \Rightarrow P(Bˉ\displaystyle \bar{B}) = 34\displaystyle \frac{3}{4} - P(C)=13⇒P(Cˉ)=23\displaystyle P(C) = 13\displaystyle \frac{1}{3} \Rightarrow P(Cˉ\displaystyle \bar{C}) = 23\displaystyle \frac{2}{3} The event "at least two students succeed" consists of the following mutually exclusive cases: 1. Exactly two students succeed. 2. All three students succeed. **Step 1: Probability of exactly two students succeeding** - Only A\displaystyle A and B\displaystyle B succeed: P(A∩B∩Cˉ)=P(A)P(B)P(Cˉ)=15×14×23=260P(A \cap B \cap Cˉ\displaystyle \bar{C}) = P(A) P(B) P(Cˉ\displaystyle \bar{C}) = 15\displaystyle \frac{1}{5} ×\displaystyle \times 14\displaystyle \frac{1}{4} ×\displaystyle \times 23=260\displaystyle \frac{2}{3} = \frac{2}{60} - Only A\displaystyle A and C\displaystyle C succeed: P(A∩Bˉ∩C)=P(A)P(Bˉ)P(C)=15×34×13=360P(A \cap Bˉ\displaystyle \bar{B} \cap C) = P(A) P(Bˉ\displaystyle \bar{B}) P(C) = 15\displaystyle \frac{1}{5} ×\displaystyle \times 34\displaystyle \frac{3}{4} ×\displaystyle \times 13=360\displaystyle \frac{1}{3} = \frac{3}{60} - Only B\displaystyle B and C\displaystyle C succeed: P(Aˉ∩B∩C)=P(Aˉ)P(B)P(C)=45×14×13=460P(Aˉ\displaystyle \bar{A} \cap B \cap C) = P(Aˉ\displaystyle \bar{A}) P(B) P(C) = 45\displaystyle \frac{4}{5} ×\displaystyle \times 14\displaystyle \frac{1}{4} ×\displaystyle \times 13=460\displaystyle \frac{1}{3} = \frac{4}{60} Sum of probabilities of exactly two succeeding: P(exactly 2)=260+360+460=960P(exactly 2\displaystyle \text{exactly 2}) = 260+360+460=960\displaystyle \frac{2}{60} + \frac{3}{60} + \frac{4}{60} = \frac{9}{60} **Step 2: Probability of all three students succeeding** P(A∩B∩C)=P(A)P(B)P(C)=15×14×13=160P(A \cap B \cap C) = P(A) P(B) P(C) = 15\displaystyle \frac{1}{5} ×\displaystyle \times 14\displaystyle \frac{1}{4} ×\displaystyle \times 13=160\displaystyle \frac{1}{3} = \frac{1}{60} **Step 3: Probability of at least two succeeding** P(at least 2)=P(exactly 2)+P(exactly 3)=960+160=1060=16P(at least 2\displaystyle \text{at least 2}) = P(exactly 2\displaystyle \text{exactly 2}) + P(exactly 3\displaystyle \text{exactly 3}) = 960+160=1060=16\displaystyle \frac{9}{60} + \frac{1}{60} = \frac{10}{60} = \frac{1}{6} This matches Option C. (Note: The official answer key incorrectly lists Option A, but the mathematically derived answer is 16\displaystyle 16\displaystyle \frac{1}{6}). 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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