Q.In a binomial distribution, $n = 200$ and $p = 0.04$. Taking Poisson distribution as an approximation to the binomial distribution : Assertion (A) : Mean of Poisson distribution $= 8$. Reason (R) : $P(X = 4) = \dfrac{512}{3e^8}$. (A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A). (B) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A). (C) Assertion (A) is true, but Reason (R) is false. (D) Assertion (A) is false, but Reason (R) is true.

Poisson Distribution: The Art of Counting Rare Events

Imagine you're watching a busy highway from a bridge. Cars pass by at random moments — sometimes two come together, sometimes there's a gap. You want to answer: How many cars will pass in the next minute? That's a counting problem. But if the traffic is light, most minutes will see 0 or 1 car, and occasionally 2 or 3. This is exactly where the Poisson distribution lives.

The Poisson distribution models the number of times a rare event happens in a fixed interval of time (or space, volume, area) when events occur independently at a constant average rate.

The Intuition

Think of a call centre. On average, you get 5 calls per hour. Some hours you get 3, some 7, rarely 12. The Poisson distribution tells you the probability of each possible count — 0 calls, 1 call, 2 calls, and so on — given that average rate.

The key assumptions are:

• Events happen one at a time (no simultaneous arrivals)
• The rate is constant over the interval
• What happens in one interval doesn't affect the next (independence)

The Precise Statement

Let $X$ be the number of events occurring in a fixed interval. Let $\lambda$ (lambda) be the average number of events in that interval. Then $X$ follows a Poisson distribution with parameter $\lambda$, written as:

The probability of observing exactly $k$ events (where $k = 0, 1, 2, \dots$) is:

Here:

• $e \approx 2.71828$ is Euler's number
• $\lambda$ is the average rate (must be positive)
• $k!$ is factorial: $k! = k \times (k-1) \times (k-2) \times \dots \times 1$, and $0! = 1$

What This Formula Says

Let's test it with the call centre example ($\lambda = 5$ calls per hour):

• Probability of exactly 0 calls: $P(0) = e^{-5} \cdot 5^0 / 0! = e^{-5} \approx 0.0067$ — very unlikely
• Probability of exactly 5 calls: $P(5) = e^{-5} \cdot 5^5 / 120 \approx 0.175$ — the most likely outcome
• Probability of exactly 10 calls: $P(10) = e^{-5} \cdot 5^{10} / 10! \approx 0.018$ — quite rare

The distribution is unimodal (one peak) and skewed right when $\lambda$ is small, becoming more symmetric as $\lambda$ grows.

Two Key Properties

The Poisson distribution has a remarkable feature: its mean and variance are equal.

This equality is a quick check: if you're analysing data and the sample mean and variance are very different, the Poisson model probably doesn't fit.

When to Use It — And When Not To

Use Poisson when:

• You're counting occurrences (not measuring continuous quantities)
• Events are independent …