Theoretical DistributionsMCQPYQ June 24Question 3502 of 230
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The mean of Poisson distribution is 4. The probability of two-successes is____.

Options

Ae−48\displaystyle \frac{e^{-4}}{8}
Be48\displaystyle \frac{e^{4}}{8}
C16e4\displaystyle \frac{16}{e^{4}}
D8e4\displaystyle \frac{8}{e^{4}}
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Correct Answer

✅ Option d — 8e4\displaystyle \frac{8}{e^{4}}

All Options:

  • Ae−48\displaystyle \frac{e^{-4}}{8}
  • Be48\displaystyle \frac{e^{4}}{8}
  • C16e4\displaystyle \frac{16}{e^{4}}
  • D8e4\displaystyle \frac{8}{e^{4}}

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Detailed Solution & Explanation

For a Poisson distribution, the mean is equal to the parameter m\displaystyle m. Here, m=4\displaystyle m = 4. The probability mass function is: P(X=k)=e−mmkk!P(X = k) = \frac{e^{-m} m^k}{k!} For two successes (k=2\displaystyle k = 2): P(X=2)=e−4422!=16e−42=8e−4=8e4P(X = 2) = \frac{e^{-4} 4^2}{2!} = \frac{16 e^{-4}}{2} = 8 e^{-4} = \frac{8}{e^4} Note: The source option key indicates Option A, which is mathematically incorrect. The correct mathematical expression is 8e4\displaystyle \frac{8}{e^4}, which 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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