4.7 Advanced Probability & Distributions

In a small village there are two doctors’ clinics, one owned by Doctor Black and the other owned by Doctor Green. It was noted after each year that 3.5% of Doctor Black’s patients moved to Doctor Green’s clinic and 5% of Doctor Green’s patients moved to Doctor Black’s clinic. All additional losses and gains of patients by the clinics may be ignored.

At the start of a particular year, it was noted that Doctor Black had 2100 patients on their register, compared to Doctor Green’s 3500 patients.

(a) Write down a transition matrix `T` indicating the annual population movement between clinics. 

(b) Find a prediction for the ratio of the number of patients Doctor Black will have, compared to Doctor Green, after two years. 

(c) Find a matrix `P`, with integer elements, such that `T=PDP^-1`, where `D` is a diagonal matrix. 

(d) Hence, show that the long-term transition matrix `T^infty` is given by `T^infty=((10/17,10/17),(7/17,7/17))`.

(e) Hence, or otherwise, determine the expected ratio of the number of patients Doctor Black would have compared to Doctor Green in the long term. 

(a) `T = ((0.965, 0.05), (0.035, 0.95))`

(b) `((0.965, 0.05), (0.035, 0.95))^2 ((2100), (3500)) = ((2294), (3306))`

Hank sets up a bird table in his garden to provide the local birds with some food. Hank notices that a specific bird, a large magpie, visits several times per month and he names him Bill. Hank models the number of times per month that Bill visits his garden as a Poisson distribution with mean 3.1.

(a) Using Hank’s model, find the probability that Bill visits the garden on exactly four occasions during one particular month. 

(b) Over the course of 3 consecutive months, find the probability that Bill visits the garden: 

1. on exactly 12 occasions.
2. during the first and third month only.

(c) Find the probability that over a 12-month period, there will be exactly 3 months when Bill does not visit the garden. 

After the first year, a number of baby magpies start to visit Hank’s garden. It may be assumed that each of these baby magpies visits the garden randomly and independently, and that the number of times each baby magpie visits the garden per month is modelled by a Poisson distribution with mean 2.1.

(d) Determine the least number of magpies required, including Bill, in order that the probability of Hank’s garden having at least 30 magpie visits per month is greater than 0.2. 

(a) `X_1 ~ Po(3.1)`

`P(X_1=4)=0.173 \ (0.173349 ... )`

(b) 

(i) `X_2 ~ Po(3 times 3.1)=Po(9.3)`

`P(X_2=12)=0.0799 \ (0.0798950 ... )`

(ii) `(P(X_1 > 0))^2 times P(X_1=0) = 0.95495^2 times 0.04505`

`= 0.0411 \ (0.0410817 ... )`

(c) `P(X_1=0)=0.04505`

`X_1 ~ B(12,0.04505)`

`= 0.0133 \ (0.013283 ... )`

(d) 

Long term experience shows that if it is sunny on a particular day in Vokram, then the probability that it will be sunny the following day is 0.8. If it is not sunny, then the probability that it will be sunny the following day is 0.3.

The transition matrix `T` is used to model this information, where `T=((0.8,0.3),(0.2,0.7))`.

(a) It is sunny today. Find the probability that it will be sunny in three days’ time. 

(b) Find the eigenvalues and eigenvectors of `T`.

The matrix `T` can be written as a product of three matrices, `PDP^-1`, where `D` is a diagonal matrix. 

(c)

(i) Write down the matrix `P`.

(ii) Write down the matrix `D`. 

(d) Hence find the long-term percentage of sunny days in Vokram. 

(a) finding `T^3` OR use of tree diagram

`T^3 = ((0.65, 0.525), (0.35, 0.475))`

the probability of sunny in three days’ time is 0.65

(b) attempt to find eigenvalues 

`|(0.8 - lambda, 0.3), (0.2, 0.7 - lambda)| = (0.8 - lambda)(0.7 - lambda) - 0.06 = 0`

`(lambda^2 - 1.5lambda + 0.5 = 0)`

`lambda = 1, lambda = 0.5`

attempt to find either eigenvector

`0.8x + 0.3y = x => -0.2x + 0.3y = 0` so an eigenvector is `((3), (2))`

`0.8x + 0.3y = 0.5x => 0.3x + 0.3y = 0` so an eigenvector is `((1), (-1))`

(c) 

(i) `P = ((3, 1), (2, -1))`

(ii) `D = ((1, 0), (0, 0.5))`

(d) `0.5^n -> 0`

`D^n -> ((1, 0), (0, 0))`

`PD^n P^(-1) = ((0.6, 0.6), (0.4, 0.4))`

60%

Loreto is a manager at the Da Vinci health centre. If the mean rate of patients arriving at the health centre exceeds 1.5 per minute then Loreto will employ extra staff. It is assumed that the number of patients arriving in any given time period follows a Poisson distribution.

Loreto performs a hypothesis test to determine whether she should employ extra staff. She finds that 320 patients arrived during a randomly selected 3-hour clinic.

(a)

(i) Write down null and alternative hypotheses for Loreto’s test. 

(ii) Using the data from Loreto’s sample, perform the hypothesis test at a 5% significance level to determine if Loreto should employ extra staff. 

Loreto is also concerned about the average waiting time for patients to see a nurse. The health centre aims for at least 95% of patients to see a nurse in under 20 minutes.

Loreto assumes that the waiting times for patients are independent of each other and decides to perform a hypothesis test at a 10% significance level to determine whether the health centre is meeting its target.

Loreto surveys 150 patients and finds that 11 of them waited more than 20 minutes.

(b)

(i) Write down null and alternative hypotheses for this test. 

(ii) Perform the test, clearly stating the conclusion in context. 

(a)

(i) let `X` be the random variable “number of patients arriving in a minute”, such that `X ~ Po(m)`

`H_0: m=1.5`

`H_1: m > 1.5`

(ii) under `H_0` let `Y` be the number of patients in 3 hours 

`Y ~ Po(270)`

`P(Y >= 320) = 1 - P(Y <= 319) = 0.00166 \ (0.00165874)`

since `0.00166 < 0.05`

(reject `H_0`)

Loreto should employ more staff

(b)

(i) `H_0`: The probability of a patient waiting less than 20 minutes is 0.95 

`H_1`: The probability of a patient waiting less than 20 minutes is less than 0.95

(ii) under `H_0` let `W` be the number of patients waiting more than 20 minutes

`W ~ B(150, 0.05)`

`P(W >= 11) = 0.132 \ (0.132215 ... )`

since `0.132 > 0.1`

(fail to reject `H_0`)

insufficient evidence to suggest they are not meeting their target

A geneticist uses a Markov chain model to investigate changes in a specific gene in a cell as it divides. Every time the cell divides, the gene may mutate between its normal state and other states.

The model is of the form

`((X_(n+1)),(Z_(n+1))) = M((X_n),(Z_n))`

where `X_n` is the probability of the gene being in its normal state after dividing for the `n`th time, and `Z_n` is the probability of it being in another state after dividing for the `n`th time, where `n in NN`.

Matrix `M` is found to be `((0.94, b),(0.06, 0.98))`.

(a)

(i) Write down the value of `b`.

(ii) What does `b` represent in this context? 

(b) Find the eigenvalues of `M`. 

(c) Find the eigenvectors of `M`. 

(d) The gene is in its normal state when `n=0`. Calculate the probability of it being in its normal state

(i) when `n=5`; 

(ii) in the long term. 

(a)

(i) 0.02 

(ii) the probability of mutating from ‘not normal state’ to ‘normal state’ 

(b) `det((0.94 - lambda, 0.02), (0.06, 0.98 - lambda)) = 0`

`(0.94 - lambda)(0.98 - lambda) - 0.0012 = 0`

`lambda^2 - 1.92lambda + 0.92 = 0`

`lambda = 1, 0.92 \ (23/25)`

(c) `((0.94, 0.02), (0.06, 0.98)) ((x), (y)) = ((x), (y))` OR `((0.94, 0.02), (0.06, 0.98)) ((x), (y)) = 0.92 ((x), (y))`

`0.02y - 0.06x = 0` OR `0.02y + 0.02x = 0`

`((1), (3))` and `((1), (-1))`

(d) 

(i) `((0.94, 0.02), (0.06, 0.98))^5 ((1), (0))` OR `((0.744, 0.0852), (0.256, 0.915)) ((1), (0))`

0.744 (0.744311...)

(ii) `((0.25), (0.75))`

0.25

Jenna is a keen book reader. The number of books she reads during one week can be modelled by a Poisson distribution with mean 2.6.

Determine the expected number of weeks in one year, of 52 weeks, during which Jenna reads at least four books.

let `X` be the random variable “number of books Jenna reads per week.”

then `X ~ Po(2.6)`

`P(X >= 4) = 0.264 \ (0.263998 ... )`

`0.263998 ... × 52`

`= 13.7`

The number of marathons that Audrey runs in any given year can be modelled by a Poisson distribution with mean 1.3.

(a) Calculate the probability that Audrey will run at least two marathons in a particular year. 

(b) Find the probability that she will run at least two marathons in exactly four out of the following five years. 

(a) `X ~ Po(1.3)`

`P(X >= 2) = 0.373`

(b) `V ~ B(5, 0.373)`

`P(V=4) = 0.0608`

The number of taxis arriving at Cardiff Central railway station can be modelled by a Poisson distribution. During busy periods of the day, taxis arrive at a mean rate of 5.3 taxis every 10 minutes. Let `T` represent a random 10 minute busy period.

(a)

(i) Find the probability that exactly 4 taxis arrive during `T`. 

(ii) Find the most likely number of taxis that would arrive during `T`. 

(iii) Given that more than 5 taxis arrive during `T`, find the probability that exactly 7 taxis arrive during `T`.

During quiet periods of the day, taxis arrive at a mean rate of 1.3 taxis every 10 minutes.

(b) Find the probability that during a period of 15 minutes, of which the first 10 minutes is busy and the next 5 minutes is quiet, that exactly 2 taxis arrive. 

(a)

(i) `X ~ Po(5.3)`

`P(X = 4) = e^(-5.3) 5.3^4 / (4!) = 0.164`

(ii) listing probabilities (table or graph) 

mode `X=5` (with probability 0.174)

(iii) attempt at conditional probability 

`(P(X = 7)) / (P(X >= 6)) = (0.1163...) / (0.4365...) = 0.267`

(b) recognising a sum of 2 independent Poisson variables e.g. `X + Y` 

`lambda = 5.3 + 1.3/2`

`P(Z = 2) = 0.0461`

The mean number of squirrels in a certain area is known to be 3.2 squirrels per hectare of woodland. Within this area, there is a 56 hectare woodland nature reserve. It is known that there are currently at least 168 squirrels in this reserve.

Assuming the population of squirrels follow a Poisson distribution, calculate the probability that there are more than 190 squirrels in the reserve. 

`X` is number of squirrels in reserve

`X ~ Po(179.2)`

recognizing conditional probability

`P(X > 190 | X >= 168) = (P(X > 190)) / (P(X >= 168)) = (0.19827...) / (0.80817...) = 0.245`

A discrete random variable `X` follows a Poisson distribution `Po(mu)`.

(a) Show that `P(X = x + 1) = mu / (x + 1) xx P(X = x), x in NN`.

(b) Given that `P(X = 2) = 0.241667` and `P(X = 3) = 0.112777`, use part (a) to find the value of `mu`.

(a) `mu / (x + 1) xx P(X = x) = mu / (x + 1) xx (mu^x e^(-mu)) / (x!)`

`= (mu^(x+1) e^(-mu)) / ((x + 1)!)`

`= P(X = x + 1)`

(b) `P(X = 3) = mu / 3 * P(X = 2) \ \ \ (0.112777 = mu/3 xx 0.241667)`

attempting to solve for `mu`

`mu = 1.40`