Theoretical DistributionsMTP Nov 21Question 3513 of 230
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In a certain Poisson frequency distribution, the probability corresponding to two success is half the probability corresponding to three successes. The mean of the distribution is

Options

A6
B12
C3
D2.45
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Correct Answer

✅ Option a — 6

All Options:

  • A6
  • B12
  • C3
  • D2.45

Detailed Solution & Explanation

Let X\displaystyle X follow a Poisson distribution with mean parameter m\displaystyle m. The probability mass function of the Poisson distribution is given by: P(X=k)=e−mmkk!P(X = k) = \frac{e^{-m} m^k}{k!} We are given that the probability corresponding to two successes is half the probability corresponding to three successes: P(X=2)=12P(X=3)P(X = 2) = \frac{1}{2} P(X = 3) Substituting the probability mass function into the relation: e−mm22!=12×e−mm33!\frac{e^{-m} m^2}{2!} = \frac{1}{2} \times \frac{e^{-m} m^3}{3!} e−mm22=12×e−mm36\frac{e^{-m} m^2}{2} = \frac{1}{2} \times \frac{e^{-m} m^3}{6} e−mm22=e−mm312\frac{e^{-m} m^2}{2} = \frac{e^{-m} m^3}{12} Since e−m≠0\displaystyle e^{-m} \neq 0 and m>0\displaystyle m > 0 for a Poisson variate, we can divide both sides by e−mm22\displaystyle \frac{e^{-m} m^2}{2}: 1=m6  ⟹  m=61 = \frac{m}{6} \implies m = 6 Therefore, the mean of the distribution is 6\displaystyle 6, which corresponds to Option A. Note: The source option key indicates Option B (12), but the correct mathematical calculation yields the mean as 6. 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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