Theoretical DistributionsMCQMTP June 24 Series IQuestion 3532 of 230
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In a certain manufacturing process, 5%\displaystyle 5\% of the tools produced turn out to be defective. Find the probability that in a sample of $40$ tools, at most $2$ will be defective: Given $e^{-2} = 0.135$

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

✅ Option d — $0.675$

All Options:

  • A$0.555$
  • B$0.135$
  • C$0.785$
  • D$0.675$

Detailed Solution & Explanation

**Probability of At Most 2 Defective Tools in a Sample of 40** **Given:** - Defective probability (p\displaystyle p) = 5%=0.05\displaystyle 5\% = 0.05 - Sample size (n\displaystyle n) = $40$ tools - Given value: e−2=0.135\displaystyle e^{-2} = 0.135 **Step 1: Calculate the Poisson Parameter (m\displaystyle m)** m=np=40times0.05=2m = np = 40 \×\displaystyle \times 0.05 = 2 **Step 2: Set up the probability for at most 2 defectives** We want to find P(Xle2)\displaystyle P(X \≤\displaystyle \le 2), where X\displaystyle X is the number of defective tools: P(Xle2)=P(X=0)+P(X=1)+P(X=2)P(X \≤\displaystyle \le 2) = P(X = 0) + P(X = 1) + P(X = 2) **Step 3: Calculate each probability using the Poisson PMF** The Poisson probability mass function is: P(X=x)=frace−mmxx!P(X = x) = \e−mmxx!\displaystyle \frac{e^{-m} m^x}{x!} - For x=0\displaystyle x = 0: P(X=0)=frace−2times200!=e−2P(X = 0) = \e−2times200!=e−2\displaystyle \frac{e^{-2} \\times 2^0}{0!} = e^{-2} - For x=1\displaystyle x = 1: P(X=1)=frace−2times211!=2e−2P(X = 1) = \e−2times211!=2e−2\displaystyle \frac{e^{-2} \\times 2^1}{1!} = 2e^{-2} - For x=2\displaystyle x = 2: P(X=2)=frace−2times222!=frac4e−22=2e−2P(X = 2) = \e−2times222!\displaystyle \frac{e^{-2} \\times 2^2}{2!} = \4e−22=2e−2\displaystyle \frac{4 e^{-2}}{2} = 2e^{-2} **Step 4: Sum the probabilities** P(Xle2)=e−2+2e−2+2e−2=5e−2P(X \≤\displaystyle \le 2) = e^{-2} + 2e^{-2} + 2e^{-2} = 5 e^{-2} Substitute the given value e−2=0.135\displaystyle e^{-2} = 0.135: P(Xle2)=5times0.135=0.675P(X \≤\displaystyle \le 2) = 5 \×\displaystyle \times 0.135 = 0.675 This corresponds to Option D. Hence, **Option D** is the correct answer.

About This Chapter: Theoretical Distributions

Paper

Paper 3: Quantitative Aptitude

Weightage

4-6 Marks

Key Topics

Binomial, Poisson, Normal Distribution

This chapter covers Binomial, Poisson, Normal Distribution and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

View Official ICAI Syllabus

Exam Strategy Tip

This topic carries 4-6 Marks weightage. Focus on understanding core concepts rather than memorizing.

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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