Q.In a town, 10 accidents take place in the span of 50 days. Assuming that the number of accidents follows Poisson distribution, find the probability that there will be 3 or more accidents on a day.
(Given that $e^{-0.2} = 0.8187$)
Solution:
Here, $m = \square$ and $X - P(m)$ with parameter $m$.
The p.m.f. X is:
$P(X = x) = \frac{e^{-m} \cdot m^x}{x!}, x = 0, 1, 2, \ldots$
$P(X \geq 3) = 1 - P(X < 3)$
$= 1 - [\square + \square + \square]$
$= 1 - \left[\frac{e^{-0.2}(0.2)^0}{0!} + \frac{e^{-0.2}(0.2)^1}{1!} + \frac{e^{-0.2}(0.2)^2}{2!}\right]$
$= 1 - [0.8187(1 + 0.2 + 0.02)]$
$= 1 - \square$
$= \square$

Telephone calls, misprints and bacteria-per-volume are standard Poisson counts (rare, independent events at a constant rate); customer arrivals at a petrol station vary with peak/lean demand, so it is the odd one out.

The Poisson distribution models the number of times a rare event occurs in a fixed interval of time or space, under the assumptions that events are independent and the average rate $\lambda$ is constant.

• (a) bacteria in a cubic foot of soil — a count in a fixed volume of space: a standard Poisson example.
• (b) telephone calls in a ten-minute interval — a count in a fixed time interval at a steady rate: a classic Poisson example.
• (c) misprints per page — a count in a fixed unit (a page): the textbook Poisson example. …

Poisson, $\lambda=np=1$; $P(X=3)=\dfrac{e^{-1}1^3}{3!}=0.0613$.

With $n=100$ toys and defect rate $p=0.01$, the expected number of defectives is $\lambda=np=100\times 0.01=1$. For large $n$ and small $p$ the Binomial is approximated by the Poisson distribution:

…

The Poisson distribution is the one whose mean equals its variance, both being $\lambda$.

For a Poisson distribution with parameter $\lambda$:

$\text{Mean}=\lambda,\qquad \text{Variance}=\lambda.$ …

$P(X=1)=P(X=2)$ gives $m = \dfrac{m^2}{2}$, so $m = 2$, not $1$ — the statement is False.

For $X\sim P(m)$, the Poisson probability is

$P(X=x) = \dfrac{e^{-m} m^{x}}{x!}$.

Apply the given condition $P(X=1) = P(X=2)$:

$e^{-m} m = \dfrac{e^{-m} m^{2}}{2}$.

Divide both sides by $e^{-m} m$ (with $m\neq 0$):

…

$P(\text{at most 2 defective}) = 2.5\,e^{-1}$.

Defectives are rare ($p = 0.02$) over many switches ($n = 50$), so use the Poisson approximation with mean

$\lambda = np = 50 \times 0.02 = 1.$

The Poisson probability is $P(X = r) = \dfrac{e^{-\lambda}\lambda^{r}}{r!}$. "At most two" means $r = 0, 1, 2$:

…

Small $p$ and large $n$ give a Poisson approximation with $\lambda = np = 1$. Then $P(X \le 2) = P(0) + P(1) + P(2) = e^{-1}(1 + 1 + 0.5) = 2.5e^{-1}$.

Step 1 — Parameters. $n = 50$, $p = 0.02$, so $\lambda = np = 50 \times 0.02 = 1$. Since $p$ is small and $n$ large, use the Poisson distribution $P(X = r) = \dfrac{e^{-\lambda}\lambda^{r}}{r!}$.

Step 2 — At most two defective. …

Equating $P(X=2)$ and $P(X=3)$ for a Poisson distribution gives $\lambda=3$.

1. The Poisson probability mass function is $P(X=k)=\dfrac{e^{-\lambda}\lambda^k}{k!}$.
2. So $P(X=2)=\dfrac{e^{-\lambda}\lambda^2}{2!}=\dfrac{e^{-\lambda}\lambda^2}{2}$ and $P(X=3)=\dfrac{e^{-\lambda}\lambda^3}{3!}=\dfrac{e^{-\lambda}\lambda^3}{6}$.
3. Setting $P(X=2)=P(X=3)$: $\dfrac{e^{-\lambda}\lambda^2}{2}=\dfrac{e^{-\lambda}\lambda^3}{6}$.
4. Cancel $e^{-\lambda}$ (never zero) and multiply both sides by $6$: $3\lambda^2=\lambda^3$. …

Since $P(X=0)=e^{-\lambda}=k$ gives $\lambda=\log(1/k)$, and the Poisson variance equals $\lambda$, the variance is $\log(1/k)$.

1. The Poisson pmf is $P(X=x)=\dfrac{e^{-\lambda}\lambda^x}{x!}$.
2. At $x=0$: $P(X=0)=e^{-\lambda}$.
3. Given $P(X=0)=k$, so $e^{-\lambda}=k$. …