Q.The mortality rate for a certain disease is 0.007. Using Poisson distribution, calculate the probability for 2 deaths in a group of 400 people
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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 Poisson is often called the "law of small numbers" — it works beautifully when the event is rare but the opportunity for it to happen is large. For example, the number of typos on a page: each word has a tiny chance of being a typo, but there are many words.
The Precise Statement
Let X be the number of events occurring in a fixed interval. Let λ (lambda) be the average number of events in that interval. Then X follows a Poisson distribution with parameter λ, written as:
X∼Poisson(λ)
The probability of observing exactly k events (where k=0,1,2,…) is:
P(X=k)=k!e−λλk
Here:
- e≈2.71828 is Euler's number
- λ is the average rate (must be positive)
- k! is factorial: k!=k×(k−1)×(k−2)×⋯×1, and 0!=1
What This Formula Says
Let's test it with the call centre example (λ=5 calls per hour):
- Probability of exactly 0 calls: P(0)=e−5⋅50/0!=e−5≈0.0067 — very unlikely
- Probability of exactly 5 calls: P(5)=e−5⋅55/120≈0.175 — the most likely outcome
- Probability of exactly 10 calls: P(10)=e−5⋅510/10!≈0.018 — quite rare
The distribution is unimodal (one peak) and skewed right when λ is small, becoming more symmetric as λ grows.
Two Key Properties
The Poisson distribution has a remarkable feature: its mean and variance are equal.
For X∼Poisson(λ):
- Mean: E[X]=λ
- Variance: Var(X)=λ
- Standard deviation: λ
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 …
Since 400 is large and the mortality rate 0.007 is small, the number of deaths can be modelled by a Poisson distribution with λ=np=400×0.007=2.8. …
With mortality rate 0.007 over 400 people, λ=np=2.8; then P(X=2)=2!e−2.8(2.8)2≈0.2384.
λ=np, P(X=k)=k!e−λλk
where n=400, p=0.007 (mortality rate).
Steps
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Poisson parameter: λ=np=400×0.007=2.8.
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Required: P(X=2)=2!e−2.8(2.8)2.
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Compute powers: (2.8)2=7.84 and 2!=2.
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Use e−2.8=0.060810.
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Substitute: P(X=2)=20.060810×7.84=0.060810×3.92.
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Compute: P(X=2)=0.238376≈0.2384. …
- CBSE 2025Set 465/S/WXYZ/41 markMCQQ.If the variance of a Poisson distribution is 2, then P(X=2) is : (A) 4e2 (B) 2e2 (C) e22 (D) e24
›Reveal solutionSolution
Poisson variance =λ=2, so P(X=2)=2!e−222=e22.
Poisson: P(X=x)=x!e−λλx, where both mean and variance equal λ.
- Variance =λ=2 (mean and variance coincide for Poisson).
- Substitute x=2: P(X=2)=2!e−2⋅22. …
- CBSE 2024Set 465/RQPS/41 markMCQQ.If for a Poisson variate X, P(X=k)=P(X=k+1), then the variance of X is : (A) k−1 (B) k (C) k+1 (D) k+2
›Reveal solutionSolution
The equal-probability condition forces λ=k+1, and Poisson variance =λ=k+1.
Poisson: P(X=r)=r!λre−λ, with mean = variance =λ.
- P(X=k)=P(X=k+1)⇒k!λke−λ=(k+1)!λk+1e−λ. …
- CBSE 2024Set 465/RQPS/41 markMCQQ.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)=3e8512. (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.
›Reveal solutionSolution
λ=np=8 (A true) and P(X=4)=3e8512 (R true), but R does not justify why the mean is 8 — so (B).
Poisson approximation to binomial: λ=np; P(X=r)=r!λre−λ.
- Mean: λ=np=200×0.04=8, so Assertion (A) is true.
- Compute P(X=4)=4!λ4e−λ=2484e−8=244096e−8.
- Simplify: 244096=3512, so P(X=4)=3512e−8=3e8512 — Reason (R) is true. …
- CBSE 2023Set 465/EF1GH/41 markMCQQ.If X is a Poisson variable such that P(X=1)=2P(X=2), then P(X=0) is :(a) e(b) e1(c) 1(d) e2
›Reveal solutionSolution
The condition P(X=1)=2P(X=2) forces the Poisson mean λ=1, so P(X=0)=e−1=e1.
Poisson: P(X=x)=x!e−λλx, where λ>0 is the mean.
- P(X=1)=e−λλ and P(X=2)=2e−λλ2.
- Apply P(X=1)=2P(X=2): e−λλ=2⋅2e−λλ2=e−λλ2. …
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