Theoretical DistributionsMCQPYQ Jun 23Question 3500 of 230
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Between 9 AM and 10 AM, the average number of phone calls per minute coming into the switchboard of a company is 4. Find the probability that in one particular minute there will be either 2 phone calls or no phone calls (given e−4=0.018316\displaystyle e^{-4} = 0.018316)

Options

A0.156
B0.165
C0.149
D0.194
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Correct Answer

✅ Option b — 0.165

All Options:

  • A0.156
  • B0.165
  • C0.149
  • D0.194

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Detailed Solution & Explanation

Let X\displaystyle X represent the number of phone calls in a minute. X\displaystyle X follows a Poisson distribution with parameter m=4\displaystyle m = 4. The probability mass function is: P(X=k)=e−44kk!P(X = k) = \frac{e^{-4} 4^k}{k!} We need to find the probability that there are either 2 phone calls or no phone calls: P(X=2 or X=0)=P(X=2)+P(X=0)P(X = 2 \text{ or } X = 0) = P(X = 2) + P(X = 0) P(X=0)=e−4400!=e−4P(X = 0) = \frac{e^{-4} 4^0}{0!} = e^{-4} P(X=2)=e−4422!=8e−4P(X = 2) = \frac{e^{-4} 4^2}{2!} = 8 e^{-4} Adding these two probabilities: P(X=2)+P(X=0)=8e−4+e−4=9e−4P(X = 2) + P(X = 0) = 8 e^{-4} + e^{-4} = 9 e^{-4} Given e−4=0.018316\displaystyle e^{-4} = 0.018316: 9e−4=9×0.018316=0.164844≈0.1659 e^{-4} = 9 \times 0.018316 = 0.164844 \approx 0.165 Hence, **Option B** 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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