Question

The manager of the hospital blood bank in Lancaster receives daily requests for a rare blood type from two hospitals, one in Lancaster and one in Morecambe. Requests for this blood are known to occur at random, at different rates from the two hospitals. The request rates per day from the two hospitals are given in the table below.

	Lancaster	Morecambe
Request Rate (per day)	0.4	0.7

1. What is the probability distribution of the number of requests for the blood in a 1-week period (i.e. 7 days) from Lancaster? [There is no need to calculate any probabilities.] Justify your answer.

2. What is the probability of exactly 4 requests from Lancaster in a 1-week period? [Do not use tables and show your working].

3. What is the probability of more than 4 requests from Lancaster in a 1-week period? [You may use tables, but show your working.]

4. What is the probability of more than 4 requests from Lancaster occurring in exactly three out of eight 1-week periods. [Do not use tables and show your working]. Justify your answer.

5. Given that requests from Lancaster and Morecambe are independent, what is the probability of a total of exactly six requests occurring in a 1-week period across both Lancaster and Morecambe? Justify your answer carefully.

Part (b)        

When banks are lending money to companies there is always a chance that the company will become bankrupt and the bank will lose the money that it has lent. To better understand the risks they are facing banks often study historical data. In one such study the company performance was categorised each year as poor (P), good (G) or excellent (E), and the following probabilities of becoming bankrupt in the following year were obtained:

P(company bankrupts next year| performance is poor) = 0.25

P(company bankrupts next year| performance is good) = 0.10

P(company bankrupts next year| performance is excellent) = 0.05

The current performance of companies is distributed 10% ‘poor’, 60% ‘good’ and 30% ‘excellent’.

1. Assuming that the historic pattern of performance continues, what is the probability that a company that bankrupts next year is one that had an ‘excellent’ performance this year, i.e.

P(current performance is excellent | company bankrupts next year)?

2. Describe briefly two ways in which a credit crisis might impact on the calculation you performed to answer part (i).

Answer 1

Solution

Back –up Theory

If an event has infinite number of possibilities of occurrence, but the actual probability of occurrence is very low,

then the number of times the event occurs is Poisson distributed………………………………...………………………….. (1)

If a random variable X ~ Poisson (λ), i.e., X has Poisson Distribution with mean λ then

probability mass function (pmf) of X is given by P(X = x) = e ^{– λ}.λ^x/(x!) ……………...………………………………….……..(2)

where x = 0, 1, 2, ……. , ∞

This probability can also be obtained by using Excel Function, Statistical, POISSON …………………………………….. (2a)

Now, to work out the solution,

Let

X = number of requests for the blood in a 1-week period (i.e. 7 days) from Lancaster

Y = number of requests for the blood in a 1-week period (i.e. 7 days) from Morecambe

Since number of requests for the blood from hospitals can be, at least conceivably, infinite, but the

actual rate of request is very low, vide (1), both X and Y are Poisson variables.

Further, given rate per day, rate per week = (7 x rate per day). Thus,

X ~ Poisson (2.8) ………………….…………………………………………………………………….. (3a)

Y ~ Poisson (4.9) ………………….…………………………………………………………………….. (3b)

Part (i)

Probability distribution of the number of requests for the blood in a 1-week period (i.e. 7 days) from Lancaster is Poisson (4.9) [vide (3b)] Answer 1

Justification

Since number of requests for the blood from hospitals can be, at least conceivably, infinite, but the

actual rate of request is very low, vide (1) the distribution is Poisson.

Further, given rate per day, rate per week = (7 x rate per day). Answer 2

Part (ii)

Probability of exactly 4 requests from Lancaster in a 1-week period

= P(X = 4)

= e ^{– 2.8}.(2.8)⁴/(4!) [vide (2) and (3a)]

= 0.1557

Answer 3

Part (iii)

Probability of more than 4 requests from Lancaster in a 1-week period

= P(X > 4)

= 1 – Σ(x = 0 to 4)P(X = x)

= 1 – 0.8447 [using PoissonTables or alternatively vide (2a)]

= 0.1553 Answer 4

Part (iv)

Back-up Theory

If X ~ B(n, p). i.e., X has Binomial Distribution with parameters n and p, where n = number of trials and

p = probability of one success, then, probability mass function (pmf) of X is given by

p(x) = P(X = x) = (ⁿC_x)(p^x)(1 - p)^{n – x}, x = 0, 1, 2, ……. , n ………….........................................................................………..(4)

[This probability can also be directly obtained using Excel Function: Statistical, BINOMDIST].............................………….(4a)

Let Z = number of weeks out of 8 when more than 4 requests from Lancaster occurred.

Then,

Z ~ B(8, 0.1553) [vide justification below and Answer 4] …………………………………………..................................……… (5)

So, probability of more than 4 requests from Lancaster occurring in exactly three out of eight 1-week periods

= P(X = 3)

= (⁸C₃)(0.1553³)(0.8447)⁵ [vide (4)]

= 35 x 0.0037 x 0.43

= 0.0564 Answer 4

Justification

If an experiment is dichotomous, i.e., has only two possible outcomes, say success and failure, which are mutually exclusive and collectively exhaustive, n repetitions of the experiment are identical, outcomes are independent and probability of a success, say p, is constant, then the random variable X, that represents the number of successes of the experiment, is a binomial variable and it is represented as: X ~ B(n, p)

DONE