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

OR

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

Q: (a) For a Poisson distribution, if mean (m) = 1, then find $P(r = 1)$. OR (b) Find the mean and standard deviation of the Binomial distribution $B\left(4, \frac{1}{3}\right)$.

**Part (a):** For Poisson with $m = 1$, $P(r) = e^{-m} m^r / r!$ - $P(r = 1) = e^{-1} \cdot 1^1 / 1!$ = **$e^{-1} \approx 0.3679$**. **Part (b):** For $B(4, 1/3)$: $n = 4, p = 1/3, q = 2/3$. - Mean = $np = 4 \times 1/3$ = **$4/3 \approx 1.33$**. - Variance = $npq = 4 \times (1/3)(2/3) = 8/9$. - Standard deviation = $\sqrt{8/9}$ = **$(2\sqrt{2})/3 \approx 0.943$**.

Probability Distributions

Poisson and Binomial Distributions

Official Answer

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

$P(r = 1) = e^{-1} \cdot 1^1 / 1!$ = $e^{-1} \approx 0.3679$ .

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

Mean = $np = 4 \times 1/3$ = $4/3 \approx 1.33$ .
Variance = $npq = 4 \times (1/3)(2/3) = 8/9$ .
Standard deviation = $\sqrt{8/9}$ = $(2\sqrt{2})/3 \approx 0.943$ .

Poisson distributionbinomial distributionmean npvariance npqstandard deviationprobability mass functione to the power minus m

Marking Scheme

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

Hint

Poisson: $P(r)=e^{-m} m^r / r!$ . Binomial: mean = np, variance = 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.

Analysis & Explanation

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

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

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

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

Common Mistakes

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

Interesting Facts

The Poisson distribution is the limiting case of the binomial when n is large and p is small with $np = m$ 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'.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

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 $B(4,1/3)$ the variance is $npq = 8/9$ , so the standard deviation is $\sqrt{8/9} = (2\sqrt{2})/3 \approx 0.943$ , 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.