Solution

The number of orders arriving at the website per minute is modelled by a $\mathrm{Poisson}(2)$ distribution (since we have an average of $2$ “successes” within a specified time limit.)

(A) Recall that for a Poisson distribution $\lambda$ is the expectation or average rate of successes (arrivals of orders). In this example we have been told that the average number of orders per minute is 2. We therefore have $\lambda = 2$.

(B) The probability that the website crashes between $10:30$am and $10:31$am morning is the same as the probability of the website crashing in any one minute interval.

Let $X$ denote the number of orders in a one-minute interval. We know that the website crashes if the number of orders exceeds 5 in one minute. We thus have:

$\mathrm{P}$(website crashes between $10:30$am and $10:31$am) $=\mathrm{P}(X \gt 5)$, where:

\begin{align} \mathrm{P}(X \gt 5) &=1-\mathrm{P}(X \leq 5)\\ &=1-(\mathrm{P}(X =0) + \mathrm{P}(X =1) + \mathrm{P}(X =2) + \mathrm{P}(X =3) + \mathrm{P}(X =4))\\ &=1-\left(\dfrac{~2^0 \times e^{-2}~}{0!} + \dfrac{~2^1 \times e^{-2}~}{1!} + \dfrac{~2^2 \times e^{-2}~}{2!} + \dfrac{~2^3 \times e^{-2}~}{3!} + \dfrac{~2^4 \times e^{-2}~}{4!}\right)\\ &=0.095 \text{ (to 3 d.p.)}.\\ \end{align}

Note: in this example we were able to sum the probabilities to obtain $\mathrm{P}(X \leq 5)$: i.e.

\[\mathrm{P}(X \leq 5)=\mathrm{P}(X =0) + \mathrm{P}(X =1) + \mathrm{P}(X =2) + \mathrm{P}(X =3) + \mathrm{P}(X =4)\]

because the Poisson is a discrete probability distribution.

(C) We have:

\begin{align} \mathrm{P}(X= \lambda)&=\mathrm{P}(X =2)\\ &=\dfrac{~2^2 \times e^{-2}~}{2!}\\ &=0.27 \text{ (to 2 d.p.)}. \end{align}

(D) Using the formula for the variance of a Poisson distribution we have:

\begin{align} \mathrm{Var(X)}&=\lambda\\ &=2. \end{align}

Solution

(A) Let $X$ denote the number of calls received by the call centre in a minute.

The number of calls received by the call centre per minute follows an average rate of $7$. Therefore $X\sim\mathrm{Po}(7)$ and so:

\begin{align} \mathrm{P}(X=4) &= \dfrac{~7^4 \times e^{-7}~}{4!}\\ &=0.091\text{(3 d.p.)}\\ \end{align}

(B) Now we need to calculate $\mathrm{P}(X=8)$.

\begin{align} \mathrm{P}(X=8) &= \dfrac{~7^8 \times e^{-7}~}{8!}\\ &=0.130\text{ (3 d.p.)}\\ \end{align}

(C) Now we need to calculate $\mathrm{P}(X=0)$. \begin{align} \mathrm{P}(X=0) &= \dfrac{~7^0\times e^{-7}~}{0!}\\ &=e^{-7}\\ &=0.001\text{ (3 d.p.)}\\ \end{align}

(D)

\begin{align} \mathrm{P}(X\leq3) &= \mathrm{P}(Y=0) + \mathrm{P}(Y=1) + \mathrm{P}(Y=2) + \mathrm{P}(Y=3)\\ &= \dfrac{~7^0 \times e^{-7}~}{0!} + \dfrac{~7^1 \times e^{-7}~}{1!} + \dfrac{~7^2 \times e^{-7}~}{2!} + \dfrac{~7^3 \times e^{-7}~}{3!}\\ &= 0.001+0.006+0.022+0.052\\ &= 0.082 \text{ (3 d.p.)}\\ \end{align}

(E) The expected number of calls per minute is the same as the rate for the Poisson Distribution, $7$.

Note: There are also Poisson tables available which can tell us the cumulative probabilities, which is useful for when $r$ is relatively large, for example, $P(X \leq 12)$.

Solution

Let $X$ denote the number of fires in the city in the next decade. We wish to find $\mathrm{P}(X\leq1)$.

First we shall calculate this using the binomial distribution.

\begin{align} \mathrm{P}(X\leq1) &= \mathrm{P}(X = 0) + \mathrm{P}(X = 1)\\ &= {}^{5000}C_0 \times 0.001^{0} \times 0.999^{5000} + {}^{5000}C_1 \times 0.001^{1} \times 0.999^{4999}\\ &= 0.00672+0.03364\\ &= 0.040\text{ (3 d.p.)}\\ \end{align}

Now we calculate using the Poisson approximation to this Binomial distribution. Here, $\lambda = \mathrm{E}[X] = n \times p = 5000 \times 0.001 = 5$ so $X\sim\mathrm{Po}(5)$.

\begin{align} \mathrm{P}(X\leq1) &= \mathrm{P}(X=0) + \mathrm{P}(X=1)\\ &= \dfrac{~5^0 e ^{-5}~}{0!} + \dfrac{~5^1 e ^{-5}~}{1!}\\ &= 0.00674 + 0.03369\\ &= 0.040\text{ (3 d.p.)}. \end{align}

As we would expect, the two methods give the same result (to 3 d.p), which shows that in this instance the Poisson distribution is a good approximation to the Binomial distribution .

An insurance company could use this approximation to calculate how likely it is a business will need to claim for fire damage or the expected number of claims likely to be made.

Solution

Now we have seen how to calculate probabilities with the Binomial distribution, we can go back to solving the Business example from earlier.

The manager needs to know how likely it is that the performance rates will drop to a poor standard. He needs to know the probability that there will be five or more absences caused by the Norovirus.

Let $X$ be the number of people who catch the Norovirus. We know from above that there are $10$ people in the office who could catch the Norovirus with probability, $p = 0.6$. So we know that $X$ ~ $\mathrm{Bin}(10,0.6).$

\begin{align} \mathrm{P}(X \geq 5) &= 1 - P(X \leq 4)\\ &= 1 - \big( \mathrm{P}(X = 0) + \mathrm{P}(X = 1) + \mathrm{P}(X = 2) + \mathrm{P}(X = 3) + \mathrm{P}(X = 4) \big)\\ &= 1 - \big( ({}^{10}C_0 \times 0.6^0 \times 0.4^{(10-0)}) + ({}^{10}C_1 \times 0.6^1 \times 0.4^{(10-1)}) + ({}^{10}C_2 \times 0.6^2 \times 0.4^{(10-2)})\\ &\quad + \; ({}^{10}C_3 \times 0.6^3 \times 0.4^{(10-3)}) + ({}^{10}C_4 \times 0.6^4 \times 0.4^{(10-4)}) \big)\\ &= 1 - \big( (0.4^10) + (10 \times 0.6 \times 0.4^9) + (45 \times 0.6^2 \times 0.4^8) + (120 \times 0.6^3 \times 0.4^7)\\ &\quad + (210 \times 0.6^4 \times 0.4^6)\big)\\ &= 1 - \big(0.0001 + 0.0016 + 0.0106 + 0.0425 + 0.1115)\big)\text{ (4 d.p.)}\\ &= 1 - 0.1662\text{ (4 d.p.)}\\ &= 0.8338 \text{ (4 d.p.)}\\ \end{align}

There is a high probability that more than $5$ people will be absent due to the illness. Therefore, it would be wise for the manager to request for help from other branches. Here, the expected number of people who catch the virus is $n \times p$, which is $10 \times 0.6 = 6$. This also suggests that the manager should request help for his department.