Theoretical DistributionsMCQPYQ Dec 21Question 3494 of 230
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The average number of advertisements per page appearing in a newspaper is 3. What is the probability that in a particular page zero number of advertisements are there?

Options

Ae−3\displaystyle e^{-3}
Be−1\displaystyle e^{-1}
Ce−2\displaystyle e^{-2}
De−4\displaystyle e^{-4}
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Correct Answer

✅ Option a — e−3\displaystyle e^{-3}

All Options:

  • Ae−3\displaystyle e^{-3}
  • Be−1\displaystyle e^{-1}
  • Ce−2\displaystyle e^{-2}
  • De−4\displaystyle e^{-4}

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Detailed Solution & Explanation

Let X\displaystyle X denote the number of advertisements per page. Since it represents a rare event over a continuous space (page), it follows a Poisson distribution with parameter m=3\displaystyle m = 3 (the average rate). The probability mass function of the Poisson distribution is: P(X=k)=e−mmkk!P(X = k) = \frac{e^{-m} m^k}{k!} To find the probability that there are zero advertisements on a particular page, we substitute k=0\displaystyle k = 0 and m=3\displaystyle m = 3: P(X=0)=e−3300!=e−3P(X = 0) = \frac{e^{-3} 3^0}{0!} = e^{-3} 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.

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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For a Poisson distribution,

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