Probability Distributions

Probability

Distributions

DR ALFRED MIFSUD

Probability at work:

Probability distributions;

Analyse discrete random variables

Find and interpret summary measures of probability distributions;

Apply probability in the analysis of business decisions;

Probability distributions:

A probability distribution is very similar to a frequency distribution.

Like a frequency distribution, a probability distribution has a series of categories, but instead of categories of values it has categories of types of outcomes.

Histograms

Histograms portray frequency distributions

Likewise we can use histograms to portray a probability distribution;

Mean and Standard Deviation

Summary measures including mean and standard deviation to summarise distribution of data.

We can use the mean and standard deviation to summarize distributions of probabilities.

Simple Probability distributions

We need the set of data to construct a frequency distribution, we need to identify the set of compound outcomes in order to create a

probability distribution.

We also need the probabilities of the simple outcomes that make up the combinations of outcomes.

Exhibit 1

The sales Director of a large organisation requires all members of sales staff to attend a team building event at an outward bound centre.

Teams of three staff will be selected by drawing names out of a hat. If there are equal numbers of females and males employed in the sales force, what are the chances that a team of three includes zero, one, two, and three females?

Solution 1:

Since there are equal numbers of females and males, the probability that a female is selected is 0.5 and the probability that a male is

selected is also 0.5.

The probability that a team includes no females is the probability that a sequence of three males are selected:

P(MMM) = 0.5*0.5*0.5 = 0.125

Solution 1(cont)

The probability that one female is selected in a team of three is a little more complicated because we have to take into account the fact that the female could be the first or the second or the third person to be selected.

So,

P(1 Female) = P(FMM or MFM or MMF)

Solution 1 (cont)

Because these three sequences are mutually exclusive, according to the addition rule:

P(1 Female) = P(FMM) + P(MFM) + P(MMF)

Since the probability that a female is selected is the same as the probability that a male is selected each of these three ways of getting one female will have the same probability.

Solution 1 (cont)

That is:

P(FMM) = P(MFM) = P(MMF) 0.5 * 0.5 * 0.5 = 0.125 So,

P(1 Female) = 0.125 + 0.125 + 0.125

= 0.375

We get the same answer for the probability that two females are selected :

Solution 1 (Cont)

P(2 Females) = P(FFM or FMF or MFF)

P(FFM) + P(FMF) + P(MFF) 0.125 + 0.125 + 0.125

0.375 Finally

P(3 Females) = 0.5 * 0.5 * 0.5 = 0.125

We can bring these results together and present them in the form of a probability distribution:

Probability distribution:

Number of Females (x) P(x)

0 0.125

1 0.375

2 0.375

3 0.125

Continuation of Exhibit 1:

The probability distribution presents the number of females as a variable, x whose values are represented as x. The variable X is a discrete random variable. It is discrete because it can only take a limited set of values. It is random because the values occur as the result of a random process.

The symbol ‘P(x)’ represents the probability that the variable X takes a particular value, x.

For instance, we can represent the probability that the number of females is one:

P(x = 1) = 0.375

The probability distribution of x, the number of females

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4

1 2 3 4

P(x)

Mean probability distribution:

The mean probability distribution by multiplying each x value by its probability and then adding up the results:

µ = ∑xP(x)

µ = mean of distribution;

The mean of a probability distribution is a population mean because we are dealing with a distribution that represents the probabilities of all possible values of the variable.

Variance and standard deviation:

The variance and standard deviation are obtained by squaring each x value, multiplying the square of it by its probability and adding the results. From this sum we subtract the square of the mean:

σ² = ∑x²P(x) - µ²

The standard deviation, σ by taking the square root of the variance.

Exhibit 2

x P(x) xP(x) x² x²P(x)

0 0.125 0 0 0

1 0.375 0.375 1 0.375

2 0.375 0.375 4 1.500

3 0.375 0.375 9 1.125

1.500 3.000

Calculate the mean and the standard deviation for the probability distribution In Exhibit 1:

Solution 2:

The mean µ is 1.5, the total of the xP(x) column;

The variance σ², is 3 the total of the line x² P(x) column minus the square of the mean:

σ² = 3 - 1.5² = 3 – 2.25 = 0.75 The standard deviation:

σ = σ ² 0.75 = 0,866

Summing up:

The mean of a probability distribution is sometimes referred to as the expected value of the distribution. Unlike the mean of a set of data, which is based on what the observed values of a variable actually were, the mean of a probability distribution tells us what values of the variable are likely or expected to be.

Summing up:

We may need to know the probability that a discrete random variable takes a particular value or a lower value. This is known as a cumulative probability because in order to get it we have to add up or accumulate other probabilities.

Exhibit 3

Calculate a set of cumulative probabilities form the probability distribution in Exhibit 1.

Solution 3:

Suppose we want the probability that x, the number of females is tow or less than two. Another way of saying this is the probability that X is less than or equal to two. We can use the symbol ‘<‘ to represent ‘less than equal to’, so we are looking for P(X<2). It may help you to recognise the symbol if you remember that the small end of the ‘<‘ part is printing at the x and the large end at the 2 , implying that x is smaller than 2.

Solution 3

We can find the cumulative probabilities for each value of X by taking the probability that x takes that value and adding the probability that X takes a lesser value. You can see these cumulative probabilities in the right hand side column of the following table.

Solution 3 (cont)

Number of females (x) P(x) P(X< x)

0 0.125 0.125

1 0.375 0.500

2 0.375 0.875

3 0.125 1.000

Solution 3 (cont)

The cumulative probability that X is zero or less, P (x<0) is the

probability that X is zero, 0.125, plus the probability that X is less than zero. Since it is impossible for X to be less than zero we do not have to add anything to 0.125.

The second cumulative probability that X is one or less, P(X<1) is the probability that X is one, 0.375 plus the probability the X is less than one, in other words that it is zero, 0.125. Adding these two probabilities together gives us 0.5.

Solution 3 (cont)

The third cumulative probability is the probability that X is two or less, P(X<2). We obtain this by adding the probability that X is two, 0.375, to the probability that X is less than two, in other words that it is one or less. This is the previous cumulative probability, 0.5. If we add this to the 0.375 we get 0.875.

Solution 3 (cont)

The fourth and final cumulative probability is the probability that X is three or less. Since we know that X can’t be more than three (there are only three members in a team), it is certain to be three or less, so the cumulative probability is 1. We would get the same result arithmetically if we add the probability that X is three, 0.125, to the cumulative

probability that X is less than three in other words that it is two or less, 0.875

Binomial Distribution:

Binomial distribution are used when you have a binomial structure.

These are situations that arise when you have a series of finite, or limited number of ‘experiments’ or ‘trials’ takes place repeatedly. Each trial has the same two mutually exclusive and collectively exhaustive outcomes, as the bi in the word binomial suggest. These two outcomes are referred to success and failure.

Formula of Binomial Distribution

The symbol X represents the number of successes in a certain number of trials, n. X can be described as a binomial random variable. The probability of success in any one trial is represented by the letter p,

The probability that there are x successes in n trials is:

P(X = x) = n! * p^x(1- p)^n-x x!(n-x)!

Exhibit 4

Refer to exhibit 1

Solution 4

We will begin by identifying the number of trials to insert the binomial equation.

Selecting a team of three involves conducting three trials, so n=3.

The variable X is the number of females selected in a team of three. We need to find the probabilities that X is 0, 1, 2, and 3, so these will be x values.

It will define success as selecting a female, p the probability of success in any one trial is 0.5.

Solution 4 (cond)

We can now put these numbers into the equation, We will start by working out the probability that no females are selected in a team of three, that is X = 0

P(X = 0) = 3 * 0.5⁰ (1 - 0.5)^3-0 0!(3! - 0!)

Solution 4 (cond)

The expression can be simplified considerably. Any number raised to the power of zero is one, so 0.5⁰ = 1

Conveniently zero factorial, 0! Is also one. We can also clear up some of the subtractions.

P(X = 0) = 3!/1(3)! * 1(0.5)³

= (3*2*1/3*2*1) * (0.5 * 0.5 * 0.5)

= 1*0.125 = 0.125

Solution 4

P(X =1) = 3!/ 1!(3 -1)! * 0.5¹ (1- 0.5)^3-1

= 6/1 (2 *1) * 0.5(0.25)

= 3 * 0.125 = 0.375

Binomial distribution:

The binomial distribution is called a discrete probability distribution because it describes the behaviour of some discrete random variables, binomial variables.

These variables concern the number of times things happen in the course of a finite number of trials.

The binomial distribution enables us to analyse situations where a specific number of trials occur. But what if we need to analyse how many things happen over a period of time? For this sort of situation we use another type of discrete probability distribution known as the Poisson distribution

The Poisson Distribution

Some types of business problems involve the analysis of incidents that are unpredictable.

The Poisson Distribution describes the behaviour of variables like the number of calls per hour or the number of stains per square metre. It enables us to find the probability that a specific number of incidents happen over a particular period or area.

Comparing the Binomial and Poisson

Binomial

Two parameters:

1. Number of trials;

2. The probability of success

Poisson

Single parameter;

1. Mean

Exhibit 5

The first aid tent at a music festival can deal with no more than three people requiring treatment in any one hour. The mean number of

people requiring treatment is two per hour. What is the probability that they will be able to deal with all the people requiring treatment in an hour?

Solution 5:

The variable X in this case is the number of people per hour who require treatment. We can use the Poisson distribution to investigate the

problem because it involves a discrete number of occurrences, or incidents over a period of time. The mean of X is 2.

Solution 5:

The first aid facility can deal with three people an hour, so the

probability that there are more people requiring treatment, than they can handle is the probability that X is more than 3, P(X>3).

The appropriate distribution is the Poisson distribution with a mean of 2, The table below demonstrates the following information about the distribution.

Solution 5: Extract of tables

µ = 2.0

P(x) P(X<x)

X = 0 0.135 0.135

X = 1 0.271 0.406

X = 2 0.271 0.677

X = 3 0.180 0.857

X = 4 0.090 0.947

X = 5 0.036 0.983

X = 6 0.012 0.995

X = 7 0.003 0.999

X = 8 0.001 1.000

Solution:

The column headed P(x) provides the probabilities that a specific number of incidents, x, occurs, e.g., the probability of four incidents, P(4), is 0.090. This column headed P(X<x) provides the probability that x or fewer incidents occur, e.g.; the probability that there are four or

fewer incidents, P(X<4) is 0.947.

Solution:

To obtain the probability that more than three people require

treatment at the first aid tent, P(x>3), we can subtract the probabilities that X is 3 or fewer P(X<3) which is the probability that the number of people requiring treatment in an hour can be dealt with from one.

P(X>3) = 1 – P (x> 3) = 1 – 0.857 = 0.143 of 14.3%

Poisson Probability:

P(X = x) = ^{𝑒−𝜇𝜇𝑥}

𝑥!

In this expression you can see the letter e, which represents a mathematical constant know as Euler’s number.

The value of this to four places of decimals, is 2.7183, so we can put this in the formula.

P(X = x) = (2.7183^−𝜇) 𝜇^𝑥)/𝑥!

Poisson Probability:

P(X = 0) = ^{2.7183−𝜇𝜇0}

0!

P(X = 0) = (2.7183⁻²) 𝜇⁰)/1

P(X = 0) = (2.7183⁻²) )/1

= 1/2.7183²

= 0.135

Expectation:

A reference has been made to the mean of a probability distribution as the expected value of the distribution because if can be used to guide to what we could expect or predict that the values will be like.

In fact one of the most important uses of probabilities is to make predictions. The production of predicted or expected values is called expectation.

Expectation 2:

A probability assesses the chance of a certain outcome in general. To use it to make predictions we have to apply it to something specific. If the probability refers to a process that is repeated, we can predict how many times the outcome will occur if the process happens a specific number of times by multiplying the probability by the number of times the process happens.

Exhibit 6

In exhibit 1 the probability that a team of three members of the sales force chosen at random contained one female was 0.375. If there were enough sales staff for 200 teams, how many teams will contain exactly one female?

Solution 6

Selecting these teams is a process that is to be repeated 200 times so that we can work out how many teams include one female by

multiplying the probability that a team contains one female, 0.375, by the number of times the process is repeated, 200.

Expected number of teams with one female = 0.375 * 200 = 75

So we would expect 75 of the teams to contain one female.

Expectation 3:

The result of exhibit 6 means that if 200 teams were to be selected at random many times over, in the long run we would expect that the average number of teams that contain one female will be 75.

Expectation also allows us to predict incidence over time. To do this you multiply the probability that a certain number of incidence occur over a period of time by the time you want your prediction to cover.

Exhibit 7

In exhibit 5 we found the probability that the number of people requiring treatment at the first aid tent was more than three in any hour was 0.143. If the music festival lasts 10 hours, in how many hours will there be more than three people seeking treatment?

Solution 7

We can obtain the answer by multiplying the probability of more than three people requiring treatment in an hour by the number of the music festival lasts.

Expected hours with more than three people requiring treatment = 0.143 * 10 =1.43

Since the first aid tent can only treat three people an hour, we would expect them to have to summon help during 1 or 2 hours of the music festival.

Expected Monetary Values (EMV)

In some situations the outcomes could be associated with a specific financial results.

In these cases the probabilities can be applied to the monetary consequences of the outcomes to produce a prediction of the amount of money that the process will generate. These types of prediction are called expected monetary values.

Exhibit 8

A rail operating company incurs extra costs, if its long distance trains are late. Passengers are given a voucher to put towards the cost of a future journey if the delay is between 30 minutes and 2 hours. If the train is more than 2 hours late the company refunds the cost of the ticket for every passenger. The cost of issuing vouchers costs the company €5000.

The cost of refunding all the fares cost the company €60,000.

The probability that a train is between 30 minutes and 2 hours late is 10% and the probability a train is more than 2 hours late is 2%. What is the EMV of the operating company’s extra costs?

Solution 8:

To answer this we need to take the probability of each of the three of the three possible outcomes (less than 30 minutes late, 30 minutes to 2 hours late, more than 2 hours late) and multiply them by their

respective costs (€0, €5000 and €60,000) The EMV is the sum of these results:

EMV = (0.88 * 0) + (0.1 * 5000) + (0.2 * 60000)

= 0 + 500 + 1200 = 1700

The company can therefore expect that extra costs will amount to

€1700 per journey.

Decision Trees

EMVs play an important part in models that can be used to analyse business situations involving a number of stages of outcomes and decisions. These models are called decision trees.

Exhibit 9

A retail grocery company is interested in opening a new supermarket in a disused cinema on the outskirts of a major town. Because of local opposition to the proposals there is only a 65% chance that planning permission will be granted. If the planning permission is obtained there is an 80% chance that a major competitor will build a rival establishment close enough to seriously affect the prospects of the supermarket.

Financial planners at the company say that the venture would make a profit of €20million in its first 5 years of operation if the rival establishment is not built and a loss of €4 million over the 5 years if the rival establishment is built.

Exhibit 9 (2)

Assuming that the company takes all investment decisions on the basis of financial prospects over the first 5 years of any venture, should they proceed with their plan?

Solution 9

One needs to distinguish between the decisions the company can take and the outcomes they could face. The decisions are to open the supermarket or not to open the supermarket. The outcomes are whether they get planning permission or not and whether the rival establishment is built or not.

Although granting planning permission and building the rival establishment are decisions they are decisions taken by people outside the company. As far as the company is concerned they are outcomes and they do not have direct control over them.

Solution 9

“O” for opening the supermarket and “No O” for not opening it;

“PP” for planning permission and “No PP” for no planning permission;

“RE” for a rival establishment and “No RE” for no rival establishment.

Solution 9

Monetary result:

“PP”

0.65

RE 0.8

-€4,000,000

“O” No RE

0.2

€20,000,000

“No PP”

0.35

€0

“No O” €0

Solution 9

EMV

= (0.35 * 0) + (0.65 * 0.8 * (-4,000,000)) + (0.65 *0.2 * 20,000,000)

=0 + (-2,080,000) + 2,600,000

= 520,000

Since the EMV of opening the supermarket, €520,000 is greater than the EMV of not opening it €0 they should open the supermarket.