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Probability Quiz

Grade 4 • 4V • 10 Questions (36 Marks)

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Score: 0 / 10
Probability — Notes & Worked Examples Show / Hide

Explanation:

Probabilities must always be written as a fraction, decimal, or percentage between 0 and 1 (never write a probability as a ratio like 1 : 4 or in words). For mutually exclusive events (events that cannot happen at the same time), all probabilities in a table add up to 1, and P(not A) = 1 − P(A).

To find the expected number of successes in N trials, use the formula: Expected Frequency = Probability × Number of Trials.

Example 1: A biased spinner lands on Red, Blue, Green, or Yellow. P(Red) = 0.25, P(Blue) = 0.15, and P(Green) is twice P(Yellow). Find P(Green).

Solution:

Remaining probability = 1 − (0.25 + 0.15) = 0.60
Let P(Yellow) = x and P(Green) = 2x:
x + 2x = 0.60 → 3x = 0.60 → x = 0.20
P(Green) = 2 × 0.20 = 0.40 → Answer: 0.4

Example 2: The spinner in Example 1 is spun 300 times. Work out an estimate for the number of times it lands on Red.

Solution:

Expected frequency = P(Red) × Number of spins
= 0.25 × 300
= 75
Answer: 75 times
Question 1
[2 marks]
ColourRedBluePinkGoldProbability0.30.20.2
The table shows the probabilities of a spinner landing on different colours
Find the missing probability
Worked Solution & Answer
Probability = successful outcomestotal possible outcomes (all mutually exclusive outcomes add up to 1).
0.3
ColourRedBluePinkGoldProbability0.30.20.20.3
Answer: 0.3
Question 2
[2 marks]
ShapeStarCubeDiscGemProbability0.10.3
Work out
A biased dice has 4 shapes.
The probabilities are shown below.
The probabilities of landing on a Cube and a Gem are the same.
Complete the table.
Worked Solution & Answer
Probability = successful outcomestotal possible outcomes (all mutually exclusive outcomes add up to 1).
0.30, 0.30
ShapeStarCubeDiscGemProbability0.10.300.30.30
Answer: 0.30, 0.30
Question 3
[2 marks]
There are 3 cards in Box **A** and 3 cards in Box **B**.
There is a number on each card.
Box A345Box B923
Ryan takes at random a card from Box **A** and a card from Box **B**.
He adds together the numbers on the two cards to get a total score.
Work out the probability that the total score is an odd number.
Worked Solution & Answer
Sums: 3+9=12, 3+2=5 (odd), 3+3=6, 4+9=13 (odd), 4+2=6, 4+3=7 (odd), 5+9=14, 5+2=7 (odd), 5+3=8 → 4 odd out of 9 = 4/9 M1 A1
Answer: 49
Question Paper (PDF) • Mark Scheme (PDF)
Question 4 AQA Nov 2018 1F Q21
[3 marks]
Here are five number cards.
1712231516
Two of the five cards are picked at random.
Work out the probability that the total of the two numbers is **more than 30**
Answer
Worked Solution & Answer
Total pairs from 5 cards = 10: M1
17+12 = 29
17+23 = 40 (yes)
17+15 = 32 (yes)
17+16 = 33 (yes)
12+23 = 35 (yes)
12+15 = 27 M1
12+16 = 28
23+15 = 38 (yes)
23+16 = 39 (yes)
15+16 = 31 (yes)
7 pairs out of 10 have total > 30
Probability = 710 = 0.7 A1
Answer: Answer 7/10
Question Paper (PDF) • Mark Scheme (PDF)
Question 5 OCR Nov 2022 2F Q19 / 5H Q3
[3 marks]
Azmi has a fair spinner numbered 2, 5 and 8.
258
(a) Complete the table to show all of the possible totals. [1 mark]
First spinSecondspinTotal258247571081316
(b) Find the probability that the total is a square number. [2 marks]
Worked Solution & Answer
(a)
2 + 8 = 10
5 + 8 = 13
8 + 2 = 10. B1
Answer (a): Missing totals: 10 (row 2, col 8), 13 (row 5, col 8), 10 (row 8, col 2)
(b)
The 9 equally likely totals are 4, 7, 10, 7, 10, 13, 10, 13, 16 M1
The square numbers are 4 and 16 (2 outcomes out of 9) ⇒ probability = 2/9. A1
Answer (b): 29
Question Paper (PDF) • Mark Scheme (PDF)
Question 6 IGCSE Nov 2024 1F Q14
[4 marks]
Kate has a fair triangular spinner and a fair square spinner.
The triangular spinner can land on 2 or 4 or 6
The square spinner can land on 1 or 2 or 3 or 4
Kate spins each spinner once.
She adds the number the triangular spinner lands on to the number the square spinner lands on to get her score.
Square spinner1234Triangularspinner23447610
(a) Complete the table to show all possible scores.
Four of the scores have been done for you. [2 marks]
(b) Find the probability that her score is 9 or less. [2 marks]
Worked Solution & Answer
(a)
Row 2: 3, 4, 5, 6 M1
Row 4: 5, 6, 7, 8
Row 6: 7, 8, 9, 10 A1
Answer (a): Row 2: 3, 4, 5, 6; Row 4: 5, 6, 7, 8; Row 6: 7, 8, 9, 10
(b)
11 outcomes out of 12 are 9 or less (only 6 + 4 = 10 is greater than 9), so P = 11/12 M1 A1
Answer (b): 1112
Question Paper (PDF) • Mark Scheme (PDF)
Question 7 AQA Jun 2023 3H Q12
[4 marks]
Here is a **biased** spinner.
RedGreen
(a) Ali, Ben and Cary want to know the probability of spinning red on the biased spinner.
They each spin it and count how many times it lands on red and divide by the total number of spins.
**Ali says:** I spun red the most times **Ben says:** I spun the spinner the most times **Cary says:** My relative frequency of red is 0.25
Who had the best estimate for the probability of spinning red?
Give a reason for your answer. [1 mark]
(b) Dev spins the spinner 80 times.
He says,
“My relative frequency of red is 0.185”
Give a reason why his relative frequency must be wrong. [1 mark]
(c) Elena spins the spinner 125 times.
The relative frequency of red is 0.32
Work out how many times the spinner landed on **green**. [2 marks]
Answer
Worked Solution & Answer
(a)
Ben did the most trials C1
Answer (a): Ben — because he spun the spinner the greatest number of times (largest sample size)
(b)
0.185 × 80 = 14.8, and the number of times it lands on red must be an integer C1
Answer (b): 80 × 0.185 = 14.8, which is not a whole number of spins
(c)
Red = 0.32 × 125 = 40 M1
⇒ Green = 125 – 40 = 85 A1
Answer (c): Answer 85
Question Paper (PDF) • Mark Scheme (PDF)
Question 8 AQA Jun 2023 2F Q18 / 2H Q7
[5 marks]
In a game,
• an ordinary fair six-sided dice is rolled
• the fair spinner shown is spun.
2x3xx2
The score is the dice number **substituted** into the spinner expression.
(a) Complete the table to show all of the possible scores.
@@TABULAR0@@ [2 marks]
(b) A player wins the game if their score is 10 or more.
Work out the probability that they win the game. [1 mark]
Answer
(c) The game is played 711 times.
Estimate the number of games that are won. [2 marks]
Answer
Worked Solution & Answer
(a)
Substitute x = 1, 2, 3, 4, 5, 6 into 2x, 3x and x² M1 A1
Answer (a): Row 2x: 2, 4, 6, 8, 10, 12; Row 3x: 3, 6, 9, 12, 15, 18; Row x²: 1, 4, 9, 16, 25, 36
(b)
Scores ⩾ 10 are 10, 12 (in 2x), 12, 15, 18 (in 3x), and 16, 25, 36 (in x²) → 8 out of 18 = 8/18 (or 4/9) B1
Answer (b): Answer 818
(c)
711 × 8/18 = 711 × 4/9 = 316 M1 A1
Answer (c): Answer 316
Question Paper (PDF) • Mark Scheme (PDF)
Question 9 AQA Jun 2018 1F Q13
[5 marks]
In a game, two fair spinners are spun.
If the numbers the arrows land on are different, the score is the higher number.
If the numbers the arrows land on are the same, the score is 0
Spinner A1246Spinner B2325
(a) Complete the table to show the possible scores.
@@TABULAR0@@ [2 marks]
(b) Write down the probability that the score is an odd number. [1 mark]
Answer
(c) The same game is played using spinners C and D.
The numbers on C are shown: 1, 4, 6, 7.
The table shows some of the possible scores.
@@TABULAR0@@
Write the missing numbers on spinner D. [2 marks]
Answer
Worked Solution & Answer
(a)
Scores comparing Spinner A and Spinner B: M1
Row 1: 2, 2, 3, 5
Row 2: 2, 0, 3, 5
Row 3: 4, 4, 4, 5
Row 4: 6, 6, 6, 6 A1
Answer (a): Row 1: 2, 2, 3, 5; Row 2: 2, 0, 3, 5; Row 3: 4, 4, 4, 5; Row 4: 6, 6, 6, 6
(b)
There are 16 possible outcomes in total
Odd scores in table: row 1 has 3 and 5 (2), row 2 has 3 and 5 (2), row 3 has 5 (1), row 4 has 0 (0)
Total odd scores = 2 + 2 + 1 + 0 = 5
Probability = 516 B1
Answer (b): Answer 516
(c)
When C = 4, score = 0 means column 2 is 4. M1
When C = 1, score = 4 means column 1 is 4.
When C = 7, score = 0 means column 3 is 7.
When C = 7, score = 8 means column 4 is 8.
Therefore the numbers on spinner D are 4, 4, 7, 8. A1
Answer (c): Answer 4, 4, 7, 8
Question Paper (PDF) • Mark Scheme (PDF)
Question 10 AQA Nov 2024 1F Q9
[6 marks]
A number is picked at random from the first three positive odd numbers.
A number is picked at random from the first four prime numbers.
The two numbers are **multiplied** to get a score.
(a) Complete the table. [4 marks]
primeodd×2371515
(b) What is the probability that the score is a square number?
Give your answer as a fraction. [2 marks]
Answer
Worked Solution & Answer
(a)
Odd numbers: 1, 3, 5. Prime numbers: 2, 3, 5, 7. Multiply each pair M3 A1
primeodd×2371515532357691521102535
Answer (a): Headers 3 and 5; cells 2,3,5,7 / 6,9,15,21 / 10,15,25,35
(b)
Out of 12 outcomes, the square numbers are 9 and 25 (2 outcomes), giving 212 (or 16) M1 A1
Answer (b): Answer 212
Question Paper (PDF) • Mark Scheme (PDF)
Score: 0 / 10 (0 / 36 marks)
Probability Worksheets Next: Averages from tables →

Grade 4 Quizzes

24 Topics
4A - Compound interest4B - Indices (index laws)
4C - Prime factors HCF and LCM4D - Real life graphs
4E - Inequalities on a number line4F - Solving inequalities
4G - Forming and solving equations4H - nth term of a linear sequence
4J - Expanding and factorising4K - Pythagoras theorem
4L - Angles in parallel lines4M - Angles in polygons
4N - Surface area4P - Volume of a cuboid
4Q - Volume of a prism4R - Bearings
4S - Cylinders4T - Plans and elevations
4U - Scatter graphs4V - Probability
4W - Averages from tables4X - Reverse mean
4Y - Area of a trapezium4Z - Loci and construction