Quantitative Aptitude · Theoretical Distributions

Poisson Distribution for CA Foundation

Updated 1 October 2026 · Fact-checked

The Poisson distribution gives the probability of r occurrences of a rare event in a fixed interval: P(X = r) = e^(−m) × m^r ÷ r!. Its mean and variance both equal m. To solve a question, find m, pick r, substitute, and use the e^(−m) value given. It approximates the binomial when n is large and p is small.

Understand Poisson Distribution

Some events happen rarely but over many chances. Examples are defects in a roll of cloth, calls to a helpline per minute, or accidents on a stretch of road per day. You cannot count the number of "failures", only the number of occurrences. The Poisson distribution is built for this.

It has one parameter, m (also written λ). It is the average number of occurrences in the interval you are looking at. The random variable X takes values 0, 1, 2, 3, ... with no upper limit. The probability of exactly r occurrences is e^(−m) × m^r ÷ r!, where e ≈ 2.71828.

The key property is that mean = variance = m. So the standard deviation is √m. If a question gives you a mean and a variance that are clearly different, the distribution is not Poisson. If the mean is given, you already have the variance.

The Poisson distribution is a limiting case of the binomial. Take a binomial with a very large n and a very small p, so that np stays moderate. Then the binomial probabilities are close to Poisson probabilities with m = np. This is why a question may give n = 200 and p = 0.01 and ask you to use Poisson. Just set m = np = 2.

Binomial and Poisson differ in what they need. Binomial needs a fixed number of trials n and a probability p. Poisson needs only the average rate m. Binomial X stops at n. Poisson X has no upper limit. In the binomial, the variance npq is less than the mean np. In the Poisson, they are equal.

Key formulas to remember

Poisson probability: P(X = r) = e^(−m) × m^r ÷ r!, for r = 0, 1, 2, ...; m is the average number of occurrences in the interval. Exam questions usually give the value of e^(−m).
Mean and variance: Mean = m; Variance = m; Standard deviation = √m; Mean equals variance. This is the property most used to identify m.
Recurrence relation: P(r + 1) = m ÷ (r + 1) × P(r); Gives each probability from the previous one without computing factorials.
Poisson as limit of binomial: m = np; Use when n is large and p is small. The binomial probabilities are then close to the Poisson ones.
Probability of at least one: P(X ≥ 1) = 1 − e^(−m); Use the complement for 'at least' questions.
Sum of independent Poisson variables: If X ~ Poisson(m₁) and Y ~ Poisson(m₂) are independent, X + Y ~ Poisson(m₁ + m₂); Useful when the interval is extended or two sources are combined.
Mode: Mode = integer part of m. If m is a whole number, there are two modes: m − 1 and m; Check this when the question asks for the most likely value.

How to solve Poisson Distribution questions

Use this method for any Poisson question. Do the work in order and the answer follows.

1Confirm it is Poisson. Look for a rare event, a rate per interval, or n large with p small.
2Find m. If the average rate is given, use it. If n and p are given, m = np. If mean or variance is given, m equals that value.
3Adjust m to the interval asked. If the rate is 3 per hour and the question asks about 2 hours, m = 6.
4Write what is needed in terms of X. For 'at most 2' use P(0) + P(1) + P(2). For 'at least 1' use 1 − P(0).
5Substitute into e^(−m) × m^r ÷ r! for each required r. Use the value of e^(−m) given in the question.
6Add the terms or subtract from 1, then round to the precision of the options.
7Check the answer lies between 0 and 1, then match it to exactly one option.

Quickest way: Factor out e^(−m) and use the recurrence

When to use it: Use this for 'at most', 'at least' or 'less than' questions, and for questions that give a ratio of probabilities.

Find m first. Write down e^(−m) from the given value.
For a cumulative probability, factor out e^(−m). For example, P(X ≤ 2) = e^(−m) × (1 + m + m²/2).
Compute the bracket first, then multiply once by e^(−m). This saves time and avoids rounding slips.
For 'at least', use 1 − e^(−m) × (bracket). Do not add many terms.
If the question says P(X = a) = P(X = b), equate the two formulas. The e^(−m) cancels and you can solve for m quickly.
Eliminate options: a probability above 1 or below 0 is impossible. Also, P(X = 0) = e^(−m) must match the given e^(−m) value, so drop any option that contradicts it.
Skip a question only if m is large, no e^(−m) value is given and many terms are needed.

Common mistakes in Poisson Distribution

Forgetting to change m when the interval changes.
Students copy the rate from the question without checking the time or space unit asked about.
Fix: Scale m with the interval. A rate of 2 per page becomes m = 6 for 3 pages. Write 'm = ...' before using the formula.
Using m = n × q or m = p instead of m = np.
Students mix up binomial and Poisson parameters.
Fix: For a binomial approximation, always take m = np. Here p is the probability of the rare event.
Taking variance as √m or as npq.
Students recall the binomial variance or mix variance with standard deviation.
Fix: In a Poisson distribution, variance = m and standard deviation = √m. If a question gives variance 4, then m = 4.
Computing 0! as 0.
It feels natural that zero factorial is zero, which makes P(0) undefined.
Fix: Remember 0! = 1. So P(0) = e^(−m) exactly.
Treating 'at least one' as P(X = 1).
Students misread the wording of the event.
Fix: 'At least one' means X ≥ 1, so use 1 − P(0). 'Exactly one' means P(1).
Using Poisson when n is small or p is large.
Students apply the approximation to any binomial.
Fix: Use Poisson as an approximation only when n is large and p is small. Otherwise use the binomial formula.

Worked examples

Example 1

The number of defects per metre of a fabric follows a Poisson distribution with mean 2. Given e^(−2) = 0.1353, the probability of exactly 2 defects in one metre is: (A) 0.1353 (B) 0.2706 (C) 0.4060 (D) 0.5413

Show the solution

The mean is 2, so m = 2 and r = 2.
P(X = 2) = e^(−2) × 2² ÷ 2!
= 0.1353 × 4 ÷ 2
= 0.1353 × 2 = 0.2706

Answer: (B) 0.2706

Example 2

For a Poisson distribution, P(X = 1) = P(X = 2). Given e^(−2) = 0.1353, the value of P(X = 0) is: (A) 0.0677 (B) 0.2706 (C) 0.4060 (D) 0.1353

Show the solution

P(1) = e^(−m) × m and P(2) = e^(−m) × m² ÷ 2.
Setting them equal: m = m² ÷ 2. Since m > 0, divide by m to get 1 = m ÷ 2, so m = 2.
P(X = 0) = e^(−m) × m⁰ ÷ 0! = e^(−2).
e^(−2) = 0.1353.

Answer: (D) 0.1353

Example 3

A binomial distribution has n = 100 and p = 0.01. Using the Poisson approximation with e^(−1) = 0.3679, the probability of at least one success is: (A) 0.2642 (B) 0.3679 (C) 0.6321 (D) 0.7358

Show the solution

n is large and p is small, so use Poisson with m = np = 100 × 0.01 = 1.
'At least one' means P(X ≥ 1) = 1 − P(X = 0).
P(X = 0) = e^(−1) = 0.3679.
P(X ≥ 1) = 1 − 0.3679 = 0.6321.

Answer: (C) 0.6321

Exam tips

Read the first line for m. Mean, variance, np or a rate per interval all lead to the same parameter.
Questions usually give the value of e^(−m). Do not try to compute it yourself. Use the given number and keep four decimal places.
Whenever the question says 'at least', check whether 1 − P(0) is faster. It nearly always is.
Memorise this contrast for theory MCQs: binomial has fixed n and variance less than the mean, while Poisson has mean equal to variance and no upper limit on X.
With 0.25 negative marking, skip a question that needs many terms and has no e^(−m) value. Come back if time allows.

Practice questions from Theoretical Distributions

Poisson Distribution: frequently asked questions

What is the Poisson distribution formula for CA Foundation?

P(X = r) = e^(−m) × m^r ÷ r!, where m is the average number of occurrences and r = 0, 1, 2, ... The mean and variance are both m. Questions usually supply e^(−m).

What is the difference between binomial and Poisson distribution?

Binomial has a fixed number of trials n and a success probability p, and X runs from 0 to n. Poisson needs only the average rate m, and X has no upper limit. In the binomial the variance is less than the mean. In the Poisson they are equal.

When can the Poisson distribution be used instead of the binomial?

Use it when n is large and p is small, so that np is moderate. Then put m = np. This is the Poisson distribution as a limiting case of the binomial. If the question tells you to use Poisson, follow that instruction.

How do I find m if only the variance is given?

Since mean equals variance in a Poisson distribution, m equals the given variance. For example, a variance of 3 means m = 3 and the standard deviation is √3.