A packet of 10 electronic components is known to include 2 defectives. If a sample of 4 components is selected at random from the packet, what is the probability that the sample does not contain more than 1 defective?

Option c: 13/15. Solution: Probability that the Sample Contains At Most 1 Defective Total electronic components = 10 - Defective components = 2 - Good components = 8 Sample size = 4 The event "not more than 1 defective" means the sample contains either 0 defectives or exactly 1 defective. Step 1: Calculate total ways to choose 4 components Total ways = (10 choose 4) = 10 × 9 × 8 × 7/4 × 3 × 2 × 1 = 210 Step 2: Calculate ways to choose 0 defectives (all good) Ways for 0 defectives = (8 choose 4) × (2 choose 0) = 70 × 1 = 70 Step 3: Calculate ways to choose exactly 1 defective Ways for 1 defective = (8 choose 3) × (2 choose 1) = 56 × 2 = 112 Step 4: Total favorable ways and probability Favorable ways = 70 + 112 = 182 P = 182/210 = 13/15 Hence, Option C is the correct answer.

ProbabilityMCQMTP Oct 21Question 3385 of 187

All Questions

A packet of $\displaystyle 10$ electronic components is known to include $\displaystyle 2$ defectives. If a sample of $\displaystyle 4$ components is selected at random from the packet, what is the probability that the sample does not contain more than $\displaystyle 1$ defective?

Options

\displaystyle \frac{1}{3}

\displaystyle \frac{2}{3}

\displaystyle \frac{13}{15}

\displaystyle \frac{3}{15}

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Correct Answer

✅ Option c — $\displaystyle \frac{13}{15}$

All Options:

A $\displaystyle \frac{1}{3}$
B $\displaystyle \frac{2}{3}$
C $\displaystyle \frac{13}{15}$
D $\displaystyle \frac{3}{15}$

Detailed Solution & Explanation

**Probability that the Sample Contains At Most 1 Defective** Total electronic components = 10 - Defective components = 2 - Good components = 8 Sample size = 4 The event "not more than 1 defective" means the sample contains either 0 defectives or exactly 1 defective. **Step 1: Calculate total ways to choose 4 components**

\text{Total ways} = \binom{10}{4} = \frac{10 \times 9 \times 8 \times 7}{4 \times 3 \times 2 \times 1} = 210

**Step 2: Calculate ways to choose 0 defectives (all good)**

\text{Ways for 0 defectives} = \binom{8}{4} \times \binom{2}{0} = 70 \times 1 = 70

**Step 3: Calculate ways to choose exactly 1 defective**

\text{Ways for 1 defective} = \binom{8}{3} \times \binom{2}{1} = 56 \times 2 = 112

**Step 4: Total favorable ways and probability**

\text{Favorable ways} = 70 + 112 = 182

P = \frac{182}{210} = \frac{13}{15}

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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A packet of 10\displaystyle 1010 electronic components is known to include 2\displaystyle 22 defectives. If a sample of 4\displaystyle 44 components is selected at random from the packet, what is the probability that the sample does not contain more than 1\displaystyle 11 defective?