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If 5% of the families in large population city do not use gas as a fuel, what will be the probability of selecting 10 families in a random sample of 100 families who do not use gas as a fuel? [Given that e−5=0.0067\displaystyle e^{-5} = 0.0067]

Options

A0.038
BZero
C0.018
D0.048
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Correct Answer

✅ Option c — 0.018

All Options:

  • A0.038
  • BZero
  • C0.018
  • D0.048

Detailed Solution & Explanation

This problem can be solved using the Poisson approximation to the Binomial distribution, as the number of trials n\displaystyle n is large and the probability of success p\displaystyle p is small.
Let:
- Sample size n=100\displaystyle n = 100.
- Probability of not using gas p=5%=0.05\displaystyle p = 5\% = 0.05.
- The parameter λ\displaystyle \lambda of the Poisson distribution is:
λ=np=100×0.05=5\lambda = n p = 100 \times 0.05 = 5
We want to find the probability of selecting exactly x=10\displaystyle x = 10 families who do not use gas. The Poisson probability formula is:
P(X=x)=e−λλxx!P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} Substituting λ=5\displaystyle \lambda = 5 and x=10\displaystyle x = 10:
P(X=10)=e−551010!P(X = 10) = \frac{e^{-5} 5^{10}}{10!} Given that e−5=0.0067\displaystyle e^{-5} = 0.0067, let us calculate 51010!\displaystyle \frac{5^{10}}{10!}:
510=9,765,6255^{10} = 9,765,625 10!=3,628,80010! = 3,628,800 51010!≈2.69113\frac{5^{10}}{10!} \approx 2.69113
Now substitute these back into the probability equation:
P(X=10)≈0.0067×2.69113≈0.01803≈0.018P(X = 10) \approx 0.0067 \times 2.69113 \approx 0.01803 \approx 0.018
Hence, **Option C** 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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