Sample Space Diagrams

This lesson focuses on constructing and using sample space diagrams to list all outcomes for combined events. Sample space diagrams are a key tool in the Edexcel GCSE Mathematics (1MA1) probability topic and are regularly tested across Foundation and Higher papers.

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What Is a Sample Space Diagram?

A sample space diagram is a systematic way of listing all possible outcomes when two events are combined. It is usually drawn as a grid (table) with one event along the top and the other down the side.

The total number of outcomes = (number of outcomes for event 1) × (number of outcomes for event 2).

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Two Dice — Adding Scores

Worked Example 1

Two fair six-sided dice are rolled. Their scores are added together. Draw a sample space diagram and use it to find:

(a) P(total = 9) (b) P(total is at least 10) (c) P(total is a prime number)

Sample Space Diagram:

+	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 outcomes = 36

Solution:

(a) Outcomes giving total 9: (3,6), (4,5), (5,4), (6,3) = 4 outcomes P(total = 9) = 4/36 = 1/9

(b) Outcomes giving total ≥ 10: total 10 has (4,6), (5,5), (6,4) = 3; total 11 has (5,6), (6,5) = 2; total 12 has (6,6) = 1. That is 3 + 2 + 1 = 6 outcomes. P(total ≥ 10) = 6/36 = 1/6

(c) The possible totals range from 2 to 12. The prime numbers in this range are: 2, 3, 5, 7, 11.

• Total 2: 1 outcome
• Total 3: 2 outcomes
• Total 5: 4 outcomes
• Total 7: 6 outcomes
• Total 11: 2 outcomes
• Sum = 1 + 2 + 4 + 6 + 2 = 15 outcomes

P(total is prime) = 15/36 = 5/12

Edexcel Exam Tip: You are almost always asked to draw the sample space diagram. The diagram itself earns marks — do not skip it and try to work out probabilities in your head.

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Two Dice — Multiplying Scores

Worked Example 2

Two fair six-sided dice are rolled and the scores are multiplied. Draw a sample space diagram and find:

(a) P(product = 6) (b) P(product is odd)

Sample Space Diagram:

×	1	2	3	4	5	6
1	1	2	3	4	5	6
2	2	4	6	8	10	12
3	3	6	9	12	15	18
4	4	8	12	16	20	24
5	5	10	15	20	25	30
6	6	12	18	24	30	36

Solution:

(a) Products equal to 6: (1,6), (2,3), (3,2), (6,1) = 4 outcomes P(product = 6) = 4/36 = 1/9

(b) The product is odd only when both dice show odd numbers. The odd numbers are 1, 3, 5. Odd × odd pairs: 3 choices for first dice × 3 choices for second dice = 9 outcomes P(product is odd) = 9/36 = 1/4

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Two Dice — Finding the Difference

Worked Example 3

Two fair six-sided dice are rolled. The smaller score is subtracted from the larger (or the difference is 0 if they are equal). Find P(difference = 0).

Sample Space Diagram (showing |a − b|):

| |a−b| | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | 1 | 0 | 1 | 2 | 3 | 4 | 5 | | 2 | 1 | 0 | 1 | 2 | 3 | 4 | | 3 | 2 | 1 | 0 | 1 | 2 | 3 | | 4 | 3 | 2 | 1 | 0 | 1 | 2 | | 5 | 4 | 3 | 2 | 1 | 0 | 1 | | 6 | 5 | 4 | 3 | 2 | 1 | 0 |

Outcomes with difference = 0: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) = 6 outcomes

P(difference = 0) = 6/36 = 1/6

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Spinners

Worked Example 4

Spinner A has sections labelled 1, 2, 3. Spinner B has sections labelled 1, 2, 3, 4. Both spinners are fair. The two spinners are spun and the scores are added.

(a) Draw a sample space diagram. (b) Find P(total = 5). (c) Find P(total > 4).

Solution:

+	1	2	3	4
1	2	3	4	5
2	3	4	5	6
3	4	5	6	7

Total outcomes = 3 × 4 = 12

(b) Outcomes giving total 5: (1,4), (2,3), (3,2) = 3 outcomes P(total = 5) = 3/12 = 1/4

(c) Outcomes with total > 4: 5 appears 3 times, 6 appears 2 times, 7 appears 1 time = 6 outcomes P(total > 4) = 6/12 = 1/2

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Coins

Worked Example 5

Three fair coins are tossed. List the sample space and find P(exactly two heads).

Sample Space (systematic listing):

Coin 1	Coin 2	Coin 3	Outcome
H	H	H	HHH
H	H	T	HHT
H	T	H	HTH
H	T	T	HTT
T	H	H	THH
T	H	T	THT
T	T	H	TTH
T	T	T	TTT

Total outcomes = 2 × 2 × 2 = 8

Outcomes with exactly 2 heads: HHT, HTH, THH = 3 outcomes

P(exactly 2 heads) = 3/8 = 3/8

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Systematic Listing Strategies

When the number of outcomes is small enough, a systematic list is effective. Use these strategies:

1. Fix the first item and list all possibilities for the remaining items.
2. Work alphabetically or numerically to avoid missing outcomes.
3. Use a tree-like structure (first → second → third) for three or more events.
4. Count using the multiplication principle first to know how many outcomes to expect.

Worked Example 6

A 3-digit code is formed using the digits 1, 2, 3 with no digit repeated. List all codes.

Solution: Fix the first digit and list systematically.

• 1, 2, 3
• 1, 3, 2
• 2, 1, 3
• 2, 3, 1
• 3, 1, 2
• 3, 2, 1

Total = 3 × 2 × 1 = 6 codes

If a code is chosen at random, P(code starts with 2) = 2/6 = 1/3.

Worked Example 7 — A Card and a Coin

A card is drawn at random from a standard 52-card pack and a fair coin is tossed.

(a) How many equally likely outcomes are there in total? (b) Find P(heart ∩ tails). (c) Find P(face card ∩ heads), where face cards are Jack, Queen and King.

Solution:

(a) Total outcomes = 52 × 2 = 104

(b) Favourable: 13 hearts × 1 (tails) = 13. P(heart ∩ tails) = 13/104 = 1/8