ProbabilityMCQPYQ Dec 23Question 3290 of 187

A box contain 20 electrical bulbs out of which 4 are defective. Two bulbs are chosen at random from this box. The probability that at least one of them is defective.

Options

\displaystyle \frac{7}{19}

\displaystyle \frac{4}{19}

\displaystyle \frac{12}{19}

\displaystyle \frac{15}{19}

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Correct Answer

✅ Option a — $\displaystyle \frac{7}{19}$

All Options:

A $\displaystyle \frac{7}{19}$
B $\displaystyle \frac{4}{19}$
C $\displaystyle \frac{12}{19}$
D $\displaystyle \frac{15}{19}$

Detailed Solution & Explanation

**At Least One Defective Bulb from 2 Chosen** Total bulbs = 20 (4 defective, 16 non-defective) Total ways to choose 2 from 20:

\binom{20}{2} = \frac{20 \times 19}{2} = 190

**Using complement** — P(at least one defective) = 1 − P(none defective): P(none defective) =

\displaystyle \frac{\binom{16}{2}}{\binom{20}{2}} = \frac{\frac{16 \times 15}{2}}{190} = \frac{120}{190} = \frac{12}{19}

P(\text{at least one defective}) = 1 - \frac{12}{19} = \frac{7}{19}

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

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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