Question 232 marksVery Short AnswerSection B

(a) For a Poisson distribution, if mean (m) = 1, then find P(r=1)P(r = 1).

OR

(b) Find the mean and standard deviation of the Binomial distribution B(4,13)B\left(4, \frac{1}{3}\right).

Probability Distributions
Poisson and Binomial Distributions

Official answer

Part (a): For Poisson with m=1m = 1, P(r)=e−mmr/r!P(r) = e^{-m} m^r / r!

  • P(r=1)=e−1⋅11/1!P(r = 1) = e^{-1} \cdot 1^1 / 1! = e−1≈0.3679e^{-1} \approx 0.3679.

Part (b): For B(4,1/3)B(4, 1/3): n=4,p=1/3,q=2/3n = 4, p = 1/3, q = 2/3.

  • Mean = np=4×1/3np = 4 \times 1/3 = 4/3≈1.334/3 \approx 1.33.
  • Variance = npq=4×(1/3)(2/3)=8/9npq = 4 \times (1/3)(2/3) = 8/9.
  • Standard deviation = 8/9\sqrt{8/9} = (22)/3≈0.943(2\sqrt{2})/3 \approx 0.943.
Answer keywords:Poisson distributionbinomial distributionmean npvariance npqstandard deviationprobability mass functione to the power minus m

Marking scheme

  1. 1.Part (a) — 1 mark: correct Poisson formula with m=1m=1; 1 mark: P(1)=e−1≈0.3679P(1) = e^{-1} \approx 0.3679.
  2. 2.Part (b) — 1 mark: mean = np=4/3np = 4/3; 1 mark: SD=npq=8/9=(22)/3≈0.943\text{SD} = \sqrt{npq} = \sqrt{8/9} = (2\sqrt{2})/3 \approx 0.943.
  3. 3.Accept decimal or exact-surd forms for the standard deviation.

Hint

Poisson: P(r)=e−mmr/r!P(r)=e^{-m} m^r / r!. Binomial: mean = np, variance = npq, SD=npq\text{SD} = \sqrt{npq}.

Quick oral answer

For Poisson with mean 1, P of r equals 1 is e to the power minus 1, about 0.368; for the binomial B of 4 and one third, the mean is np equal to four thirds and the standard deviation is root of npq, which is two root two over three.

Common mistakes

  1. 1.Reporting the binomial variance (8/98/9) as the standard deviation instead of taking its square root.
  2. 2.Using q=1/3q = 1/3 instead of q=2/3q = 2/3 in npq, since q=1−p=2/3q = 1 - p = 2/3 here.
  3. 3.In the Poisson part, mishandling factorials or writing e−1e^{-1} as e1e^1, changing the answer drastically.

Analysis and explanation

This question tests the parameters of the two key discrete distributions in the syllabus.


Part (a) — Poisson: The Poisson probability mass function is P(r)=e−mmr/r!P(r) = e^{-m} m^r / r!, where m is both the mean and the variance. Substituting m=1m = 1 and r=1r = 1 gives e−1e^{-1}, roughly 0.368.


Part (b) — Binomial: For B(n,p)B(n, p), the mean is np and the variance is npq (with q=1−pq = 1 - p). Standard deviation is the square root of the variance. Here np=4/3np = 4/3 and npq=8/9npq = 8/9, so SD=223\text{SD} = \frac{2\sqrt{2}}{3}.


Exam trap: For the binomial, do not report the variance as the standard deviation — the SD is its square root. For Poisson, remember 0!=10! = 1 and 1!=11! = 1, so the denominator is simply 1.


Application: Poisson models rare events (defects, arrivals per hour); binomial models a fixed number of independent success/failure trials.

Interesting facts

The Poisson distribution is the limiting case of the binomial when n is large and p is small with np=mnp = m fixed — the two distributions in this question are deeply connected.

For a Poisson distribution the mean and variance are always equal, a distinctive property used to test whether real data is 'Poisson-like'.

Question details

Q23 of 44 questions

Board
CBSE
Class
Class 12
Subject
Applied Mathematics
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Probability Distributions
Topic
Poisson and Binomial Distributions
Bloom's level
apply
Syllabus topic
Unit: Probability Distributions — Poisson and Binomial distribution parameters
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is the standard deviation not equal to the variance in the binomial part?

Variance measures spread in squared units, while standard deviation is its square root and shares the units of the variable. For B(4,1/3)B(4,1/3) the variance is npq=8/9npq = 8/9, so the standard deviation is 8/9=(22)/3≈0.943\sqrt{8/9} = (2\sqrt{2})/3 \approx 0.943, not 8/9.

How are the Poisson mean and variance related?

In a Poisson distribution the single parameter m equals both the mean and the variance. This equality is a defining feature: if data has a variance close to its mean, a Poisson model is often appropriate.