Theoretical DistributionsMTP Dec 2023 Series IQuestion 3973 of 230
All Questions

If 5%\displaystyle 5\% of the electric bulbs manufactured by a company are defective, use Poisson distribution to find the probability that in a sample of 100\displaystyle 100 bulbs, 5\displaystyle 5 bulbs will be defective. [Given : e−5=0.007\displaystyle e^{-5} = 0.007 ]

Options

A0.1823\displaystyle 0.1823
B0.1723\displaystyle 0.1723
C0.1623\displaystyle 0.1623
D0.1923\displaystyle 0.1923
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Correct Answer

✅ Option a — 0.1823\displaystyle 0.1823

All Options:

  • A0.1823\displaystyle 0.1823
  • B0.1723\displaystyle 0.1723
  • C0.1623\displaystyle 0.1623
  • D0.1923\displaystyle 0.1923

Detailed Solution & Explanation

**Probability of Defective Bulbs Using Poisson Distribution** **Given:** - Defective rate (p\displaystyle p) = 5%=0.05\displaystyle 5\% = 0.05 - Sample size (n\displaystyle n) = 100\displaystyle 100 bulbs - Target number of defective bulbs (x\displaystyle x) = 5\displaystyle 5 - Given value: e−5=0.007\displaystyle e^{-5} = 0.007 **Step 1: Calculate the Poisson Parameter (m\displaystyle m)** The parameter m\displaystyle m (mean) of the Poisson distribution is given by: m=np=100times0.05=5m = np = 100 \\times 0.05 = 5 **Step 2: Apply the Poisson Probability Mass Function** The probability of getting exactly x\displaystyle x successes in a Poisson distribution is: P(X=x)=frace−mmxx!P(X = x) = \\frac{e^{-m} m^x}{x!} For x=5\displaystyle x = 5 and m=5\displaystyle m = 5: P(X=5)=frace−5times555!P(X = 5) = \\frac{e^{-5} \\times 5^5}{5!} **Step 3: Simplify and Calculate** We know that: - 55=3125\displaystyle 5^5 = 3125 - 5!=5times4times3times2times1=120\displaystyle 5! = 5 \\times 4 \\times 3 \\times 2 \\times 1 = 120 Substitute these values and e−5=0.007\displaystyle e^{-5} = 0.007 into the formula: P(X=5)=frac0.007times3125120P(X = 5) = \\frac{0.007 \\times 3125}{120} P(X=5)=frac21.875120approx0.18229approx0.1823P(X = 5) = \\frac{21.875}{120} \\approx 0.18229 \\approx 0.1823 **Discrepancy Note:** The mathematical calculation yields 0.1823\displaystyle 0.1823 which corresponds to **Option A**. The textbook answer key has a typographical error, listing **Option D** (0.1923\displaystyle 0.1923) as the correct answer. We have presented the correct mathematical derivation leading to Option A. Hence, **Option A** 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.

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