Calculating Probabilities

This lesson covers calculating probabilities for single events, listing outcomes systematically, using sample spaces, and applying the complementary event rule. These skills are essential for the Edexcel GCSE Mathematics (1MA1) specification and appear frequently on all three exam papers.

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Single Event Probabilities

For any single event with equally likely outcomes:

P(event) = number of favourable outcomes / total number of outcomes

Worked Example 1

A standard pack of 52 playing cards is shuffled and one card is drawn at random. Find the probability that the card is:

(a) a heart (b) a king (c) the queen of spades

Solution:

(a) There are 13 hearts in a pack of 52 cards. P(heart) = 13/52 = 1/4

(b) There are 4 kings in a pack of 52 cards. P(king) = 4/52 = 1/13

(c) There is exactly 1 queen of spades. P(queen of spades) = 1/52 = 1/52

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Listing Outcomes Systematically

When working with probability, it is essential to list all possible outcomes in a systematic (organised) way so that none are missed or counted twice.

Worked Example 2

A café offers a meal deal: choose one main from {burger, wrap, salad} and one drink from {cola, juice, water}. List all possible meal combinations.

Solution:

Main	Drink	Combination
Burger	Cola	Burger + Cola
Burger	Juice	Burger + Juice
Burger	Water	Burger + Water
Wrap	Cola	Wrap + Cola
Wrap	Juice	Wrap + Juice
Wrap	Water	Wrap + Water
Salad	Cola	Salad + Cola
Salad	Juice	Salad + Juice
Salad	Water	Salad + Water

There are 9 possible combinations (3 × 3 = 9).

If one meal deal is chosen at random, P(wrap and juice) = 1/9.

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Sample Spaces

A sample space is the set of all possible outcomes of an experiment. It is often denoted by the symbol S or listed using curly brackets {}.

Worked Example 3

A fair six-sided dice is rolled once. Write down the sample space.

Solution: S = {1, 2, 3, 4, 5, 6}

The sample space has 6 elements.

Worked Example 4

Two fair coins are tossed. List the sample space.

Solution: Using H for heads and T for tails:

S = {HH, HT, TH, TT}

The sample space has 4 equally likely outcomes.

Important: HT and TH are different outcomes (the first coin shows heads and the second shows tails, or vice versa). A common mistake is to list only 3 outcomes: HH, HT, TT.

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Sample Space Diagrams for Two Events

When two events are combined, a sample space diagram (also called a two-way table or possibility space) helps you list every outcome.

Worked Example 5

Two fair six-sided dice are rolled and their scores are added together. Draw a sample space diagram and find the probability of getting a total of 7.

Solution:

+	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Total number of outcomes = 6 × 6 = 36

The outcomes that give a total of 7 are: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) — that is 6 outcomes.

P(total of 7) = 6/36 = 1/6

Edexcel Exam Tip: Sample space diagrams are a favourite in Edexcel papers. Always draw the full table — do not try to count outcomes in your head. The marks are awarded for showing the diagram.

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Complementary Events — P(not A)

The complement of an event A (written A' or "not A") is the event that A does not happen.

P(not A) = 1 − $P(A)$P(A)

This is because event A either happens or it does not, and the two probabilities must sum to 1.

Worked Example 6

The probability that it rains tomorrow is 0.35. Find the probability that it does not rain tomorrow.

Solution: P(not rain) = 1 − 0.35 = 0.65

Worked Example 7

A bag contains red, blue and green counters. The probability of picking a red counter is 2/5 and the probability of picking a blue counter is 1/3. Find the probability of picking a green counter.

Solution: P(green) = 1 − P(red) − P(blue) = 1 − 2/5 − 1/3

Find a common denominator (15): = 15/15 − 6/15 − 5/15 = 4/15

Worked Example 8

The probability of a biased dice landing on 6 is 0.2. The dice is rolled 150 times. Estimate the number of times it will land on a number that is not 6.

Solution: P(not 6) = 1 − 0.2 = 0.8 Expected number of "not 6" = 0.8 × 150 = 120

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The Multiplication Principle

When you can list outcomes for two separate stages, the total number of outcomes is found by multiplying:

Total outcomes = outcomes for stage 1 × outcomes for stage 2

Worked Example 9

A padlock has 3 dials. Each dial shows a digit from 0 to 9. How many different 3-digit codes are possible?

Solution:

• Dial 1: 10 choices (0–9)
• Dial 2: 10 choices (0–9)
• Dial 3: 10 choices (0–9)
• Total codes = 10 × 10 × 10 = 1000

If the code is chosen at random, P(getting the correct code) = 1/1000.

Worked Example 10

A fair spinner has 8 equal sections numbered 1 to 8. Find: (a) P(prime number) (b) P(factor of 8) (c) P(multiple of 3)

Solution:

(a) Primes in 1–8: 2, 3, 5, 7 — that is 4 numbers. P(prime) = 4/8 = 1/2

(b) Factors of 8: 1, 2, 4, 8 — that is 4 numbers. P(factor of 8) = 4/8 = 1/2

(c) Multiples of 3 in 1–8: 3, 6 — that is 2 numbers. P(multiple of 3) = 2/8 = 1/4

Worked Example 11

A menu offers 4 starters, 6 mains and 3 desserts. A customer orders one of each course at random. How many different three-course meals are possible? If one meal is picked at random, find the probability it consists of a specific starter, main and dessert combination.

Solution:

By the multiplication principle: Total meals = 4 × 6 × 3 = 72

For one specific combination, there is only 1 favourable outcome out of 72: P(specific combination) = 1/72

Worked Example 12