CIMA - C3 Fundamentals Of Business Mathematics

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180 6: Probability ⏐ Part C Probability

Answe

The EV of demand is as follows.

Demand

Probability

Expected value

Units

5,000

0.3

1,500

6,000

0.6

3,600

8,000

0.1

800

EV of demand

5,900

The EV of the variable cost per unit is as follows.

Variable costs

Probability

Expected value

3.00

0.1

0.30

3.50

0.3

1.05

4.00

0.5

2.00

4.50

0.1

0.45

EV of unit variable costs

3.80

Sales 5,900 units × $6.00

35,400

Less: variable costs 5,900 units × $3.80

22,420

Contribution

12,980

Less: fixed costs

10,000

Expected profit

2,980

4 Expectation and decision making

4.1 Decision making

Probability and expectation should be seen as an aid to decision making.

The concepts of probability and expected value are vital in business decision making. The expected values for

single events can offer a helpful guide for management decisions.

• A project with a positive EV should be accepted

• A project with a negative EV should be rejected

Another decision rule involving expected values that you are likely to come across is the choice of an option or

alternative which has the

highest EV of profit (or the lowest EV of cost).

Choosing the option with the highest EV of profit is a decision rule that has both merits and drawbacks, as the

following simple example will show.

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Part C Probability ⏐ 6: Probability 181

4.2 Example: The expected value criterion

Suppose that there are two mutually exclusive projects with the following possible profits.

Project A Project B

Probability Profit Probability Profit/(loss)

$ $

0.8 5,000 0.1 (2,000)

0.2 6,000 0.2 5,000

0.6 7,000

0.1 8,000

Required

Determine which project should be chosen.

Solution

The EV of profit for each project is as follows.

(a) Project A (0.8

× 5,000) + (0.2 × 6,000) = 5,200

(b) Project B (0.1

× (2,000)) + (0.2 × 5,000) + (0.6 × 7,000) + (0.1 × 8,000) = 5,800

Project B has a higher EV of profit. This means that on the balance of probabilities, it could offer a better return

than A, and so is arguably a better choice.

On the other hand, the minimum return from project A would be $5,000 whereas with B there is a 0.1 chance of a

loss of $2,000. So project A might be a safer choice.

Question

Expected values

A company is deciding whether to invest in a project. There are three possible outcomes of the investment:

Outcome Profit/(Loss)

$'000

Optimistic 19.2

Most likely 12.5

Pessimistic (6.7)

There is a 30% chance of the optimistic outcome, and a 60% chance of the most likely outcome arising. The

expected value of profit from the project is

A $7,500 C $13,930

B $12,590 D $25,000

Answe

B Since the probabilities must total 100%, the probability of the pessimistic outcome = 100% – 60% – 30% =

10%.

Outcome

Profit/(Loss)

Probability

Expected value

Optimistic

19,200

0.3

5,760

Most likely

12,500

0.6

7,500

Pessimistic

(6,700)

0.1

(670

)

1.0

12,590

182 6: Probability ⏐ Part C Probability

If you selected option A, you calculated the expected value of the most likely outcome instead of the entire

project.

If you selected option C, you forgot to treat the 6,700 as a loss, ie as a negative value.

If you selected option D, you forgot to take into account the probabilities of the various outcomes arising.

4.3 Payoff tables

Decisions have to be taken about a wide variety of matters (capital investment, controls on production, project

scheduling and so on) and under a wide variety of conditions from

virtual certainty to complete uncertainty.

There are, however, certain common factors in many business decisions.

(a) When a decision has to be made, there will be a range of possible

actions.

(b) Each action will have certain consequences, or payoffs (for example, profits, costs, time).

circumstances (for example, high demand or

low demand), which may or may not be known when the decision is taken. Frequently each

circumstance will be assigned a probability of occurrence. The circumstances are

not dependent on

the action taken.

For a decision with these elements, a

payoff table can be prepared.

A payoff table is simply a table with rows for circumstances and columns for actions (or vice versa), and the

payoffs in the cells of the table.

For example, a decision on the level of advertising expenditure to be undertaken given different states of the economy,

would have payoffs in $

'000 of profit after advertising expenditure as follows.

Actions: expenditure

High Medium Low

Circumstances:

Boom +50 +30 +15

the state of the Stable +20 +25 +5

economy Recession 0 –10 –35

4.4 Example: Payoff table

A cinema has to decide how many programmes to print for a premiere of a film. From previous experience of

similar events, it is expected that the probability of sales will be as follows.

Number of programmes demanded Probability of demand

250 0.1

500 0.2

750 0.4

1,000 0.1

1,250 0.2

The best print quotation received is $2,000 plus 20 pence per copy. Advertising revenue from advertisements

placed in the programme totals $2,500. Programmes are sold for $2 each. Unsold programmes are worthless.

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Part C Probability ⏐ 6: Probability 183

Required

(a) Construct a payoff table.

(b) Find the most profitable number of programmes to print.

Solution

(a) Actions: print levels

250 500 750 1,000 1,250

250 (p = 0.1) 950 900 850 800 750

Circumstances: 500 (p = 0.2) 950 1,400 1,350 1,300 1,250

demand levels 750 (p = 0.4) 950 1,400 1,850 1,800 1,750

1,000 (p = 0.1) 950 1,400 1,850 2,300 2,250

1,250 (p = 0.2) 950 1,400 1,850 2,300 2,750

These figures are calculated as the profit under each set of circumstances. For example, if the cinema

produces 1,000 programmes and 1,000 are demanded, the profit is calculated as follows.

Total revenue = advertising revenue + sale of programmes

= $2,500 + $(1,000

× 2)

= $4,500

Total costs = $2,000 + $(0.20 × 1,000)

= $2,000 + $200

= $2,200

Profit = total revenue – total costs = $4,500 – $2,200 = $2,300

Similarly, if the cinema produces 750 programmes, but only 500 are demanded, the profit is calculated as

follows.

Total revenue = $2,500 + $(500 × 2)

= $2,500 + $1,000 = $3,500

Total costs = $2,000 + $(0.20 × 750)

= $2,000 + $150

= $2,150

Profit = total revenue – total costs = $3,500 – $2,150 = $1,350

Note that whatever the print level, the maximum profit that can be earned is determined by the demand.

This means that when 250 programmes are printed, the profit is $950 when demand is 250. Profit is also

$950 when demand is 500, 750, 1,000 or 1,250.

(b) The expected profits from each of the possible print levels are as follows.

Print 250

Expected profit = $((950 × 0.1) + (950 × 0.2) + (950 × 0.4) + (950 × 0.1) + (950 × 0.2)) = $950

Print 500

Expected profit = $((900 × 0.1) + (1,400 × (0.2 + 0.4 + 0.1 + 0.2))) = $1,350

Print 750

Expected profit = $((850 × 0.1) + (1,350 × 0.2) + (1,850 × 0.7)) = $1,650

Print 1,000

Expected profit = $((800 × 0.1) + (1,300 × 0.2) + (1,800 × 0.4) + (2,300 × 0.3)) = $1,750

184 6: Probability ⏐ Part C Probability

Print 1,250

Expected profit = $((750 × 0.1) + (1,250 × 0.2) + (1,750 × 0.4) + (2,250 × 0.1) + (2,750 × 0.2)) = $1,800

1,250 programmes should therefore be printed in order to

maximise expected profit.

E(X) = Expected value = Probability

× Payoff

Question

Payoff tables

In a restaurant there is a 30% chance of five apple pies being ordered a day and a 70% chance of ten being

ordered. Each apple pie sells for $2. It costs $1 to make an apple pie. Using a payoff table, decide how many apple

pies the restaurant should prepare each day, bearing in mind that unsold apple pies must be thrown away at the

end of each day.

Answe

Prepared

Five Ten

Demand

Five (P = 0.3) 5 0

Ten (P = 0.7) 5 10

Prepare five, profit = ($5

× 0.3) + ($5 × 0.7) = $5

Prepare ten, profit = ($0

× 0.3) + ($10 × 0.7) = $7

Ten pies should be prepared.

4.5 Limitations of expected values

Evaluating decisions by using expected values have a number of limitations.

(a) The

probabilities used when calculating expected values are likely to be estimates. They may

therefore be

unreliable or inaccurate.

(b) Expected values are

long-term averages and may not be suitable for use in situations involving one-

off decisions

. They may therefore be useful as a guide to decision making.

attitudes to risk of the people involved in the decision-making

process. They do not, therefore, take into account all of the factors involved in the decision.

(d) The time value of money may not be taken into account: $100 now is worth more than $100 in ten

years' time. We shall study the time value of money in Section F of this Study Text.

Limitations of methods can be tested in a computer based assessment so make sure you know what they are.

4.6 Risk and uncertainty

Probability is used to help to calculate risk in decision making.

Assessment

formula

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Assessment

focus point

Part C Probability ⏐ 6: Probability 185

Risk involves situations or events which may or may not occur, but whose probability of occurrence can be

calculated statistically and the frequency predicted.

Uncertainty involves situations or events whose outcome cannot be predicted with statistical confidence.

Do not underestimate the importance of probability in the

Business Mathematics assessment – this topic accounts

for 15% of the syllabus. The key to being able to answer probability questions is lots of practice.

Chapter Roundup

• Probability is a measure of likelihood and can be stated as a percentage, a ratio, or more usually as a number from 0

to 1.

• The simple addition law for two mutually exclusive events, A and B is as follows.

P(A or B) = P(A) + P(B)

• Mutually exclusive outcomes are outcomes where the occurrence of one of the outcomes excludes the possibility

of any of the others happening.

• The simple multiplication law for two independent events A and B, is as follows.

P(A and B) = P(A) P(B)

• Independent events are events where the outcome of one event in no way affects the outcome of the other events.

• The general rule of addition for two events, A and B, which are not mutually exclusive, is as follows.

P(A or B) = P(A) + P(B) – P(A and B)

• The general rule of multiplication for two dependent events, A and B, is as follows.

P(A and B) = P(A) × P(B/A)

= P(B) × P(A/B)

• Dependent or conditional events are events where the outcome of one event depends on the outcome of the

others.

• Contingency tables can be useful for dealing with conditional probability.

• An expected value (or EV) is a weighted average, based on probabilities. The expected value for a single event can

offer a helpful guide for management decisions: a project with a positive EV should be accepted and a project

with a negative EV should be rejected.

• Probability and expectation should be seen as an aid to decision making.

• A payoff table is simply a table with rows for circumstances and columns for actions (or vice versa), and the

payoffs in the cells of the table.

• Probability is used to help to calculate risk in decision making.

Assessment

focus point

186 6: Probability ⏐ Part C Probability

Quick Quiz

1 Complete the following equations

(a) P

()

X = 1 –

(b) Simple addition/OR law

P(A or B or C) =

where A, B and C are

P(A and B) =

where A and B are

(d) General rule of addition

P(A or B) =

where A and B are

(e) General rule of multiplication

P(A and B) =

where A and B are

2 1 Mutually exclusive outcomes

2 Independent events

3 Conditional events

A The occurrence of one of the outcomes excludes the possibility of any

of the others happening

1 2 3

B Events where the outcome of one event depends on the outcome of

the others

1 2 3

C Events where the outcome of one event in no way affects the outcome

of the other events

1 2 3

3 An analysis of 480 working days in a factory shows that on 360 days there were no machine breakdowns.

Assuming that this pattern will continue, what is the probability that there will be a machine breakdown on a

particular day?

A 0%

B 25%

C 35%

D 75%

4 A production director is responsible for overseeing the operations of three factories – North, South and West. He

visits one factory per week. He visits the West factory as often as he visits the North factory, but he visits the South

factory twice as often as he visits the West factory.

What is the probability that in any one week he will visit the North factory?

Part C Probability ⏐ 6: Probability 187

A 0.17

B 0.20

C 0.25

D 0.33

5 A project may result in profits of $15,000 or $20,000, or in a loss of $5,000. The probabilities of each profit are

0.2, 0.5 and 0.3 respectively.

What is the expected profit?

188 6: Probability ⏐ Part C Probability

Answers to Quick Quiz

1 (a) 1 – P(X)

(b) P(A) + P(B) + P(C) Mutually exclusive outcomes

× P(B) Independent events

(d) P(A) + P(B) – P(A and B) Not mutually exclusive outcomes

(e) P(A)

× P(B/A) = P(B) × P(A/B) Dependent events

2 A = 1

B = 3

C = 2

3 B The data tells us that there was a machine breakdown on 120 days (480 – 360) out of a total of 480.

P(machine breakdown) = 120/480

× 100%

= 25%

You should have been able to eliminate option A immediately since a probability of 0% = impossibility.

If you selected option C, you calculated the probability of a machine breakdown as 120 out of a possible

365 days instead of 480 days.

If you selected option D, you incorrectly calculated the probability that there was

not a machine breakdown

on any particular day.

Factory Ratio of visits

North 1

South 2

West

Pr(visiting North factory) = 1/4 = 0.25

If you didn't select the correct option, make sure that you are clear about how the correct answer has been arrived

at. Remember to look at the

ratio of visits since no actual numbers of visits are given.

EV = (15,000 × 0.2) + (20,000 × 0.5) + (– 5,000 × 0.3)

= 3,000 + 10,000 – 1,500

= 11,500

Now try the questions below from the Exam Question Bank

Question numbers Pages

49-57 335-338

11,500

189

Topic list Syllabus references

1 Probability distributions C, (vi), (6)

2 The normal distribution C, (vi), (6)

3 The standard normal distribution C, (vi), (6)

4 Using the normal distribution to calculate

probabilities

C, (vi), (6)

5 The Pareto distribution and '80:20 rule' C, (vii), (6)

Distributions

Introduction

This chapter will build on many of the statistical topics already covered. We saw in Chapter

4b that a frequency distribution of continuous data can be drawn as a symmetrical bell-

shaped curve called the normal distribution. This can be linked with the calculation of

probabilities studied in Chapter 6 and is a useful business decision making tool. We will also

be looking at the Pareto Distribution which demonstrates that 80% of value is concentrated

in 20% of items.