Skip to contentSkip to search

Probability

CIMA Free Mock Exam
CIMA-BA2 · Chapter 14
  1. Probability

A probability measures how likely an outcome is, on a scale from 0 (impossible) to 1 (certain). It can be written as a fraction, a decimal or a percentage – it is simply the proportion of times the outcome is expected to happen.

This chapter explains the difference between risk and uncertainty, calculates simple and joint probabilities (the addition and multiplication rules), and then uses probabilities to calculate expected values – the weighted average outcome – to help make decisions. It ends with the limitations of expected values.

1 Introduction

YouTube video

In this chapter we will explain what we mean by probabilities, and look at a range of different calculations of probabilities that could be required of you in the exam.

Risk and uncertainty

Most business decisions are about the future, and the future is rarely known for certain. A new product may sell well or badly; next month’s demand may be high or low. Management accountants describe this in two ways.

Meaning

Example

Risk

There are several possible outcomes, and the probability of each outcome can be measured or estimated – usually from past experience or research.

A company has recorded weekly complaints for a year (Example 3), so it can estimate the probability of 0, 1, 2 or 3 complaints next week.

Uncertainty

The possible outcomes, or their probabilities, cannot be measured reliably – there is no past data to go on.

A completely new product in a new market: management can guess at demand, but cannot say how likely each level is.

Where risk can be measured, probabilities can be put on the outcomes and used to calculate an expected value (section 4). The more uncertain a situation is, the less reliable those probabilities – and the decisions based on them – become (section 5). In BA2 you will be asked to work with probabilities and expected values; you will not be asked to apply special techniques for decisions under uncertainty.

2 Simple probabilities

The probability of an event occurring is the likelihood or the chance that it will occur.

So, for example, if we toss a coin then there are two possible outcomes - that it falls as a head or as a tail - and both outcomes have an equal chance of occurring. Therefore the probability of it being a head - one of the two outcomes - is 1 in 2, which can be expressed as a probability of 1/2, or as 0.5, or as 50%.

Where every outcome is equally likely:

Probability of an event = number of outcomes that give the event ÷ total number of possible outcomes

A probability is always between 0 and 1. A probability of 0 means the event cannot happen; 1 means it is certain. If a list covers every possible outcome, their probabilities add up to 1 – one of them is certain to happen.

If we toss a 6-sided die, what is the probability of getting:

  1. a six

  2. a one or a two

  3. a number greater than 3

Show answerHide answer

Outcomes that give the event

Probability

(a) a six

6 (1 out of 6)

1/6

(b) a one or a two

1 or 2 (2 out of 6)

2/6 (= 1/3)

(c) a number greater than 3

4, 5 or 6 (3 out of 6)

3/6 (= 1/2)

Parts (b) and (c) can also be done with the addition rule (below): for (b), 1/6 + 1/6 = 2/6; for (c), 1/6 + 1/6 + 1/6 = 3/6.

The addition rule – ‘either … or …’

When we want the probability that either one event or another happens, we add the probabilities.

  • Mutually exclusive events cannot happen at the same time (a single die cannot show a 1 and a 2). Then P(A or B) = P(A) + P(B).

  • Events that can happen together (a card can be an ace and a diamond). Adding the two probabilities would count the overlap twice, so it is subtracted once: P(A or B) = P(A) + P(B) − P(A and B).

A standard pack of 52 playing cards has four suits – hearts, diamonds, spades and clubs – each of 13 cards: ace, 2 to 10, jack, queen and king.

If we pick one card from a pack of 52 playing cards, what is the probability of that card being:

  1. the ace of diamonds

  2. an ace

  3. a heart

  4. an ace or a diamond

Show answerHide answer

Working

Probability

(a) the ace of diamonds

1 card out of 52

1/52

(b) an ace

4 aces out of 52

4/52 (= 1/13)

(c) a heart

13 hearts out of 52

13/52 (= 1/4)

(d) an ace or a diamond

4 aces + 13 diamonds − 1 (the ace of diamonds is counted twice) = 16 cards

16/52 (= 4/13)

Using the addition rule for (d): 4/52 + 13/52 − 1/52 = 16/52.

Probabilities, proportions and percentages

In business, probabilities usually come from past results. If an outcome has happened in a certain proportion of past cases, that proportion is used as its probability. A probability of 0.25 is the same as a proportion of 1/4 and a percentage of 25% – it means the outcome is expected to happen in a quarter of cases.

A company has recorded the number of complaints received per week over the last year, and has produced the following table:

Number of complaints

Frequency (weeks)

0

12

1

16

2

20

3

4

52

What is the probability that in one particular week there are 2 complaints?

Show answerHide answer

There were 2 complaints in 20 of the 52 weeks, so:

Probability = 20/52 (= 5/13) = 0.385, or 38.5%

In a group of 200 potential voters, 150 are male and 50 are female.

60 of the male voters will vote for Party A. 25 of the female voters will not vote for Party A.

If one voter is picked at random:

  1. What is the probability that they are female

  2. What is the probability that they will vote for Party A

  3. If the voter picked will be voting for Party A, then what is the probability that they are female?

Show answerHide answer

(a) 50 of the 200 voters are female: 50/200 = 1/4 (0.25 or 25%).

(b) Set out the whole group in a table first. 60 of the 150 males vote for Party A, so 90 do not; 25 of the 50 females will not vote for Party A, so the other 25 will.

For A

Not for A

Total

Male

60

90

150

Female

25

25

50

Total

85

115

200

Probability of voting for Party A = 85/200 (= 17/40 = 0.425)

(c) We know the voter is one of the 85 who vote for Party A. Of those 85, 25 are female: probability = 25/85 (= 5/17 = 0.294).

Part (c) is a conditional probability – the probability of one event (female) given that another has already happened (voting for Party A). The total we divide by is no longer the whole group of 200, only the 85 for whom the condition is true. Drawing up a table of the whole group, as in the answer, is the safest way to get it right.

3 Joint probabilities

If we want to know the probability of two or more events occurring, then we can calculate it by multiplying the individual probabilities together.

For example, if we toss a die two times, then there are 36 possible outcomes - a 1 and a 1, a 1 and a 2, a 1 and a 3, and so on. So the probability of getting two 6s is 1/36 because two 6s is just one of the 36 possible outcomes. However, we could have got the same answer by saying that the probability of getting a 6 on the first toss is 1/6; the probability of getting a 6 on the second toss is 1/6; and therefore the probability of getting two 6s is 1/6 x 1/6 = 1/36.

This is the multiplication rule: for independent events – where the result of one has no effect on the other, like two tosses of a die – P(A and B) = P(A) × P(B). Use it whenever the question wants one thing and another to happen.

A card is picked at random from a pack of 52 playing cards. It is replaced, and then a second card is picked at random from the same pack.

What is the probability of having picked:

  1. two Kings

  2. two hearts

  3. an Ace and a King

Show answerHide answer

First card

Second card

Probability

(a) two Kings

4/52

4/52

4/52 × 4/52 = 1/169

(b) two hearts

13/52

13/52

13/52 × 13/52 = 1/16

(c) an Ace and a King

8/52 (an Ace or a King)

4/52 (the matching card)

8/52 × 4/52 = 2/169

In (c) the order does not matter, so the first card can be any of the 4 Aces or 4 Kings (8 cards). The second card must then be the other kind – a King if the first was an Ace, an Ace if the first was a King – and there are 4 of those.

In Example 5, we replaced the playing card after the first pick, and therefore there were 52 cards to pick from on both picks. However, if the first card had not been replaced, then it would change the probabilities for the second card.

A card is picked at random from a pack of 52 playing cards. It is NOT replaced, and then a second card is picked at random from the same pack.

What is the probability of having picked:

  1. two Kings

  2. two hearts

  3. an Ace and a King

Show answerHide answer

First card

Second card (51 left)

Probability

(a) two Kings

4/52

3/51 (3 Kings left)

4/52 × 3/51 = 1/221 (0.0045)

(b) two hearts

13/52

12/51 (12 hearts left)

13/52 × 12/51 = 1/17 (0.0588)

(c) an Ace and a King

8/52

4/51 (all 4 of the matching kind are still there)

8/52 × 4/51 = 8/663 (0.0121)

Because the first card was not replaced, the probability for the second card depends on what happened first – the events are dependent, and the second probability is a conditional probability. The multiplication rule still applies, using the conditional probability for the second pick.

Words in the question

Rule

Calculation

‘A or B’, mutually exclusive

Addition rule

P(A) + P(B)

‘A or B’, can happen together

Addition rule, less the overlap

P(A) + P(B) − P(A and B)

‘A and B’, independent

Multiplication rule

P(A) × P(B)

‘A and B’, dependent

Multiplication rule with a conditional probability

P(A) × P(B given A)

Do not confuse mutually exclusive with independent. Mutually exclusive events cannot happen together, so P(A and B) = 0. Independent events can happen together, and P(A and B) = P(A) × P(B).

4 Expected values

YouTube video

Probabilities can be useful in a business situation where there are various possible results that could occur for which we know the probabilities. We can use the probabilities to arrive at an ‘average’ result, which we call the expected value.

Expected value (EV) = the sum of (each possible outcome × its probability)

It is a weighted average: each outcome is weighted by how likely it is. The probabilities used must add up to 1.

A company is considering launching a new product. The demand for the new product is uncertain, but the company has estimated that if demand is high then the revenue will be $500,000 a year; if the demand is medium then the revenue will be $300,000 a year; and if the demand is low then the revenue will be $200,000 a year.

The probabilities of high, medium, and low are 0.2, 0.5 and 0.3 respectively.

What is the expected revenue?

Show answerHide answer

Demand

Revenue
($)

Probability

Revenue ×
probability
($)

High

500,000

0.2

100,000

Medium

300,000

0.5

150,000

Low

200,000

0.3

60,000

Expected revenue

1.0

310,000

($500,000 × 0.2) + ($300,000 × 0.5) + ($200,000 × 0.3) = $310,000 a year. If demand varies from year to year with these probabilities, this is the average revenue per year over many years.

Expected values can be used to help make decisions, in which case it is common to need to produce a table first showing the possible outcomes and their probabilities.

John has a factory capacity of 1,200 units per month.

Units cost him $6 each to make and his normal selling price is $11 each. However, the demand per month is uncertain and is as follows:

Demand (units)

Probability

400

0.2

500

0.3

700

0.4

900

0.1

He has been approached by a customer who is prepared to contract to a fixed quantity per month at a price of $9 per unit. The customer is prepared to sign a contract to purchase 300, 500, 700 or 800 units per month.

The company can vary production levels during the month up to the maximum capacity, but cannot carry forward any unsold units in inventory.

  1. Calculate all possible profits that could result

  2. Determine for what quantity John should sign the contract, using expected values.

Show answerHide answer

(a) Each unit sold under the contract earns $9 − $6 = $3. Each unit sold to normal customers earns $11 − $6 = $5. Capacity is 1,200 units, so after the contract only 1,200 − contract units can be made for normal customers, whatever their demand.

Profit = (contract units × $3) + (normal units sold × $5), where normal units sold = the lower of demand and the capacity left. For example, a 700-unit contract with demand of 900: 700 × $3 = $2,100, plus only 500 normal units × $5 = $2,500, total $4,600.

Contract size

Demand
400u

Demand
500u

Demand
700u

Demand
900u

300u

2,900

3,400

4,400

5,400

500u

3,500

4,000

5,000

5,000

700u

4,100

4,600

4,600

4,600

800u

4,400

4,400

4,400

4,400

(b) Expected value if contract size =

Contract size

Working

Expected value ($)

300 units

(0.2 × 2,900) + (0.3 × 3,400) + (0.4 × 4,400) + (0.1 × 5,400)

3,900

500 units

(0.2 × 3,500) + (0.3 × 4,000) + (0.5 × 5,000)

4,400

700 units

(0.2 × 4,100) + (0.8 × 4,600)

4,500

800 units

4,400 whatever the demand

4,400

Sign the contract for 700 units – it gives the highest expected profit, $4,500 a month.

Where the same profit appears in several columns, their probabilities can be added first (for 700 units, 0.3 + 0.4 + 0.1 = 0.8). Note too that the expected value must be worked out on the profits, not on the average demand: extra demand earns nothing once capacity is full, so the average demand would give the wrong answer.

5 The limitations of expected values

There are two main limitations as to the use of expected values in decision making:

  • it is unlikely that the probabilities can be determined with accuracy, and should the probabilities turn out to be different then the wrong decision may have been made.

  • for a ‘one-off’ decision, the actual outcome is unlikely to coincide with the expected outcome - it may turn out to be better or it may turn out to be worse.

In Example 7 the expected revenue is $310,000 a year, but no single year will actually produce $310,000. If the product turns out to be popular, demand may be high every year and revenue $500,000 every year; if it is unpopular, revenue may be $200,000 every year. The expected value is only a reliable guide when the decision is repeated many times, so that the good and bad outcomes average out. In the same way, John in Example 8 will never earn exactly $4,500 in a month – it is not one of the possible profits.

6 Test your knowledge

Two quick checks before you move on: work through the flashcards to fix this chapter’s terms and definitions, then sit the objective questions for exam-style practice. Both mark themselves and explain the answers as you go.

Practice questions

Probability

7 questions

Answer the questions one at a time. Your progress is saved so you can leave and come back.

Open chapter practice