In this lecture we will consider the basic concept of probability which is of such fundamental importance in statistics. We will calculate probabilities from symmetry and relative frequency for both single and combined events and use expected values in order to make logical decisions. We will only use contingency tables for combined probabilities in this lecture so make sure you read Chapter 4 of Business Statistics for descriptions of tree diagrams. Make sure you complete your tutorial, if not finished in class time. The concept of probability was initially associated with gambling. Its value was later realised in field of business, initially in calculations for shipping insurance. Other areas of business: Quality control: Sampling schemes based on probability, e.g. acceptance sampling. Sample design for sample surveys. In decision making, e.g. the use of decision trees. Stock control, e.g. probable demand for stock before delivery of next order.
Probability can be calculated from various sources: simple probability calculated from symmetry; simple probability calculated from frequencies; conditional probability calculated from contingency tables; conditional probability calculated from tree diagrams; etc.
Probability is measure of uncertainty Probability measures the extent to which an event is likely to occur and can be calculated from the ratio of favourable outcomes to the whole number of all possible outcomes. (An outcome is the result of an trial, i.e. the result of rolling a die might be a six. An event may be a set of outcomes, 'an even number', or a single outcome, 'a six'.) There are two main ways of assessing the probability of a single outcome occurring: from the symmetry of a situation to give an intuitive expectation, from past experience using relative frequencies to calculate present or future probability.
Value of Probability Probability always has a value between 0, impossibility, and 1, certainty. Probability of 0.0 0.1 0.5 0.8 1.0 corresponds to ‘impossible’ corresponds to ‘extremely unlikely’ corresponds to ‘evens chance’ corresponds to ‘very likely’ corresponds to ‘certainty’
Probability from symmetry We can define the probability of an event A taking place as:
Pevent A Number of equally likely outcomes in which A occurs Total number of equally likely outcomes
Invoking symmetry there is no need to actually do any experiments. This is the basis for the theory of probability which was developed for use in the gaming situation. – 'a priori' probability. Examples 1. Tossing 1 coin: Possibilities: Head, Tail.
P(a head) = 1/2 = 0.5
P(a tail) = 1/2 = 0.5
Tossing 2 coins: Possibilities: Head Head, Head Tail, Tail Head, Tail Tail. P(2 heads) = P(2 tails) = P(1 head and 1 tail) = Note that the sum of all the possible probabilities is always one. This makes sense as one of them must occur and P(certainty) = 1. Rolling a single die Sample space: 1, 2, 3, 4, 5, 6. P(6) = P(anything but 6) = P(an even number) = P(an odd number) = Again each pair of probabilities sums to 1.
Rolling a pair of dice (all possibilities shown below - known as total sample space) 1,1 2,1 3,1 4,1 5,1 6,1 P(double six) = P(more than 8) = 1,2 2,2 3,2 4,2 5,2 6,2 1,3 2,3 3,3 4,3 5,3 6,3 1,4 2,4 3,4 4,4 5,4 6,4 1,5 2,5 3,5 4,5 5,5 6,5 1,6 2,6 3,6 4,6 5,6 6,6 P(less than 4) = P(an even total) =
P(any double) = P(at least one six) =
Probability from Frequency An alternative way to look at Probability would be to carry out an experiment to determine the proportion of favourable outcomes in the long run. P(Event E) = Number of times E occurs Number of possible occurrences
This is the frequency definition of Probability. The relative frequency of an event is simply the proportion of the possible times it has occurred. As the number of trials increases the relative frequency demonstrates a long term tendency to settle down to a constant value. This constant value is the probability of the event. But how long is this 'long term'? Table of Relative Frequencies from one experiment of flipping a coin: Total number of throws 1 2 3 4 5 6 7 8 Number of heads 1 2 2 2 3 3 3 (4) etc. Proportion of heads 1.00 1.00 0.67 0.50 0.60 0.50 0.43 (0.57) etc.
Result H H T T H T T (if H) etc.
In this experiment the ‘long term’ would be the time necessary for the relative frequency to settle down to two decimal places.
Proportion of heads
Value Proportion of Heads
.5 .4 1 3 5 7 9 11 13 15 17 19 21 23 25 27 29
NOTE: The outcome of an individual trial is either 1 or 0 heads. It is the long-term tendency which gives the estimated probability. Also: n must be a large number - the larger n is, the better the estimate of the probability.
Contingency tables Another alternative is that probabilities can be estimated from contingency tables. Relative frequency is used as before, this time from past experience. Contingency tables display all possible combined outcomes and the frequency with which each of them has happened in the past. These frequencies are used to calculate future probabilities. Example 5 A new supermarket is to be built in Bradfield. In order to estimate the requirements of the local community, a survey was carried out with a similar community in a neighbouring area. Part of the results of this are summarised in the table below:
Expenditure on drink Mode of travel On foot By bus By car Total Suppose we put all the till invoices into a drum and thoroughly mix them up. If we close our eyes and take out one invoice, we have selected one customer at random. We first complete all the row and column totals. From the 'sub-totals' we can now calculate, for example: P(customer spends at least £20) = 67/250 = 0.268 P(customer will travel by car) = From the ‘cells’ we can calculate, for example: P(customer spends over £20 and travels by car) = 42/250 = 0.168 P(customer arrives on foot and spends no money on drink) = P(customer spends over £20 or travels by car) = (40 + 20 + 10 + 30 + 25)/250 = 0.50 P(customer arrives on foot or spends no money on drink) = None 40 30 25 1p and under £20 20 35 33 At least £20 10 15 42 Total
These last two values are probably more easily calculated by adding the row and column totals and then subtracting the intersecting cell which has been included twice: P(customer spends over £20 or travels by car) = (70 + 95 - 40)/250 = 0.50 P(customer arrives on foot or spends no money on drink) =
Sometimes we need to select more than one row or column: P(a customer will spend less than £20) = (95 + 88)/250 = 0.732 P(a customer will not travel by car) = P(a customer will not travel by car and will spend less than £20) = P(a customer will not travel by car or will spend less than £20) =
In the examples above all the customers have been under consideration without any condition being applied which might exclude any of them from the overall ratio. Often not all are included as some condition applies.
If the probability of the outcome of a second event depends upon the outcome of a previous event then the second is conditional on the result of the first. This does not imply a time sequence but simply that we are asked to find the probability of an event given additional information - an extra condition, i.e. if we know that the customer travelled by car all the other customers who did not are excluded from the calculations. If we need P(a customer spends at least £20, if it is known that he/she travelled by car) We eliminate from the choice all those who did not arrive in a car, i.e. we are only interested in the third row of the table. A short hand method of writing 'if it is know that . .' or 'given that . . .' is '|'. P(a customer spends at least £20, | he/she travelled by car) = 42/100 = 0.420 P(a customer came by car | he spent at least £20 on drink) = 42/67 = 0.627 Note that P(spending £20, | he/she travelled by car) P(coming by car | he spent £20 on drink)
P(a customer spends at least £20, | he/she travelled by bus) = P(a customer spends less than £20, | he/she did not travel by car) = P(a customer came on foot, | he/she spent at least £20 on drink) =
Many variations on the theme of probability have been covered in this lecture but they all boil down to answering the question 'how many out of how many?'
A commonly used method in decision making problems is the consideration of expected values. The expected value for each decision is calculated and the option with the maximum or minimum value (dependent upon the situation) is selected. The expected value of each decision is defined by: E(x) = px where x is the value associated with each out outcome, E(x) is the expected value of the event x and p is the probability of it happening. If you bought 5 out of 1000 raffle tickets for a prize of £25 the expected value of your winnings would be 0.005 x £25 = £0.125. If there was also a £10 prize, you would also expect to win 0.005 x £10 = £0.05 giving £0.175 in total. (Strictly speaking there are only 999 tickets left but the difference is negligible.) Clearly this is only a theoretical value as in reality you would win either £25, £10 or £0. Example 6: Two independent operations A and B are started simultaneously. The times for the operations are uncertain, with the probabilities given below: Operation A Operation B Duration (days) (x) Probability (p) Duration (days) (x) Probability (p) 1 0.0 1 0.1 2 0.5 2 0.2 3 0.3 3 0.5 4 0.2 4 0.2 Determine whether A or B has the shorter expected completion time. Operation A E(x) = px = (1 x 0.0) + (2 x 0.5) + (3 x 0.3) + (4 x 0.2) = 2.7 days Operation B E(x) = px = (1 x 0.1) + (2 x 0.2) + (3 x 0.5) + (4 x 0.2) = 2.8 days Hence Operation A has the shorter completion time. Example 7 A marketing manager is considering whether it would be more profitable to distribute his company's product on a national or on a more regional basis. Given the following data, what decision should be made? National distribution Net profit Prob. that £m (x) demand is met (p) 4.0 0.50 2.0 0.25 0.5 0.25 Regional distribution Level of Net Prob. that Demand profit demand is £m (x) met (p) High 2.5 0.50 Medium 2.0 0.25 Low 1.2 0.25
Level of Demand High Medium Low
Expected profits = px