Probability equation questions Quiz (Grade 8+)

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Score: 0 / 10

Probability equation questions — Notes & Worked Examples Show / Hide

Explanation:

Algebraic probability questions give the number of counters in terms of n or x (or a ratio) and state the numerical probability of a two-step event without replacement. Write the branch probabilities as algebraic fractions, remembering to subtract 1 from the total denominator on the second pick.

Multiply (and add if there are multiple valid paths), equate your algebraic fraction to the given probability, cross-multiply to clear fractions, and rearrange into a quadratic equation an² + bn + c = 0. Always reject negative or non-integer roots.

Example 1: A bag has n red counters and 3 blue counters. Two are taken without replacement. Given P(both red) = 514, find n.

Solution:

nn + 3 × n − 1n + 2 = 514 ⇒ 14n(n − 1) = 5(n + 3)(n + 2)

3n² − 13n − 10 = 0 ⇒ (3n + 2)(n − 5) = 0 ⇒ Answer: n = 5

Example 2: A bag has x green and (10 − x) yellow balls. Two are picked without replacement. Given P(both green) = 13, find x.

Solution:

Total balls = 10: P(GG) = x10 × x − 19 = 13

x(x − 1)90 = 13 ⇒ x² − x = 30

x² − x − 30 = 0 ⇒ (x − 6)(x + 5) = 0

Answer: x = 6

One Page Guide (PDF) Probability equation questions Worksheets

Question 1

[1 mark]

Work out
A bag contains 9 counters, of which x are red. Two counters are taken at random without replacement.
The probability that both counters are red is 16. Form and solve a quadratic equation to find x.

x =

Worked Solution & Answer

x(x - 1)9(8) = 16 ⇒ x(x - 1) = 12 ⇒ x = 4 B1

Answer: x = x = 4

Question 2

[1 mark]

Work out
A bag contains 10 counters, of which x are red. Two counters are taken at random without replacement.
The probability that both counters are red is 13. Form and solve a quadratic equation to find x.

x =

Worked Solution & Answer

x(x - 1)10(9) = 13 ⇒ x(x - 1) = 30 ⇒ x = 6 B1

Answer: x = x = 6

Question 3 IGCSE Jan 2020 2HR Q16

[2 marks]

Cody has two bags of counters, bag **A** and bag **B**.
Each of the counters has either an odd number or an even number written on it.
There are 10 counters in bag **A** and 7 of these counters have an **odd** number written on them.
There are 12 counters in bag **B** and 7 of these counters have an **odd** number written on them.
Cody is going to take at random a counter from bag **A** and a counter from bag **B**.

(a) Complete the probability tree diagram. [2 marks]

Worked Solution & Answer

(a)

Bag A Even = 3/10 M1
Bag B branches = 7/12 Odd, 5/12 Even A1

Answer (a): Bag A: 3/10; Bag B: 7/12, 5/12 and 7/12, 5/12

Question Paper (PDF) • Mark Scheme (PDF)

Question 4

[3 marks]

There are only red sweets and yellow sweets in a bag.
There are n red sweets in the bag.
There are 8 yellow sweets in the bag.
Sajid is going to take at random a sweet from the bag and eat it.
He says that the probability that the sweet will be red is 710

(a) Show why the probability cannot be 710 [3 marks]

Worked Solution & Answer

(a)

n / (n + 8) = 7/10 M1
⇒ 10n = 7n + 56 M1
⇒ 3n = 56
⇒ n = 18.67 (not an integer) A1

Answer (a): nn + 8 = 710 ⇒ 10n = 7n + 56 ⇒ 3n = 56 ⇒ n = 563 = 18.6̇, which is not a whole number.

Question Paper (PDF) • Mark Scheme (PDF)

Question 5 IGCSE Nov 2021 1F Q15

[4 marks]

There are 54 fish in a tank.
Some of the fish are white and the rest of the fish are red.
Jeevan takes at random a fish from the tank.
The probability that he takes a white fish is 49

(a) Work out the number of white fish originally in the tank. [2 marks]

(b) Work out the number of white fish Jeevan put into the tank. [2 marks]

Worked Solution & Answer

(a)

(4/9) × 54 = 24 white fish (and 30 red fish) M1 A1

Answer (a): 24

(b)

Since red fish = 30 and P(white) = 1/2, new number of white fish = 30 M1
⇒ 30 – 24 = 6 A1

Answer (b): 6

Question Paper (PDF) • Mark Scheme (PDF)

Question 6

[4 marks]

There are 12 counters in a bag.
There is an equal number of red counters, blue counters and yellow counters in the bag.
There are no other counters in the bag.
3 counters are taken at random from the bag.

(a) Work out the probability of taking 3 red counters. [2 marks]

(b) Is it now less likely or equally likely or more likely that the 3 counters will be red?
You must show how you get your answer. [2 marks]

Worked Solution & Answer

(a)

(4/12) × (3/11) × (2/10) = 24/1320 = 1/55 M1 A1

Answer (a): 155 (or 241320)

(b)

With 5 of each (15 total): (5/15) × (4/14) × (3/13) = 60/2730 = 2/91 = 0.0219... > 1/55 (0.0181...), so more likely M1 A1

Answer (b): More likely (e.g. with 15 counters: 515 × 414 × 313 = 291 ≈ 0.022 > 155 ≈ 0.018)

Question Paper (PDF) • Mark Scheme (PDF)

Question 7 IGCSE Jun 2023 1FR Q28

[5 marks]

Moeen has a biased 6-sided dice.
The table gives information about the probability that, when the dice is thrown, it will land on each number.

(a) Show that x + y = 0.19 [2 marks]

(b) work out the value of x and the value of y
Show clear algebraic working. [3 marks]

Worked Solution & Answer

(a)

1 – (0.15 + 0.5 + 0.13 + 0.03) = 1 – 0.81 = 0.19 M1 C1

Answer (a): x + 0.15 + 0.5 + y + 0.13 + 0.03 = 1 ⇒ x + y + 0.81 = 1 ⇒ x + y = 0.19

(b)

Adding equations: 4x = 0.28 ⇒ x = 0.07 M2
y = 0.19 – 0.07 = 0.12 A1

Answer (b): x = 0.07, y = 0.12

Question Paper (PDF) • Mark Scheme (PDF)

Question 8

[5 marks]

There are some counters in a bag.
The counters are red or white or blue or yellow.
Bob is going to take at random a counter from the bag.
The table shows each of the probabilities that the counter will be blue or will be yellow.

There are 18 blue counters in the bag.
The probability that the counter Bob takes will be red is twice the probability that the counter will be white.

(a) Work out the number of red counters in the bag. [4 marks]

(b) Explain why. [1 mark]

Worked Solution & Answer

(a)

P(red) + P(white) = 1 – (0.45 + 0.25) = 0.30 ⇒ P(red) = 0.20, P(white) = 0.10 M2
Total counters = 18 ÷ 0.45 = 40 M1
Red counters = 0.20 × 40 = 8 A1

Answer (a): 8

(b)

Because half (0.5) of the total number of marbles must be a whole number

Answer (b): Because half (0.5) of the total number of marbles must be a whole number

Question Paper (PDF) • Mark Scheme (PDF)

Question 9 Eduqas Nov 2024 2F Q10

[5 marks]

Zara makes necklaces by threading beads onto lengths of string.

(a) She uses a bag of beads that contains pink, yellow, green and blue beads.
In the bag:
• half of the beads are pink
• there are half as many yellow beads as pink beads
• there are an equal number of green beads and blue beads.
Zara selects a bead at random from the bag.
Complete the table below to show the probability of selecting each colour:
@@TABULAR₀@@ [2 marks]

(b) Zara makes necklaces of two different lengths.
The length of string for each long necklace is y cm.
The string for each short necklace is 11 cm shorter than the string for a long necklace.
Write an expression, in terms of y, for the total length of string needed to make one long necklace and two short necklaces.
Simplify your answer. [3 marks]

Worked Solution & Answer

(a)

Pink = 1/2 M1
Yellow = 1/2 × 1/2 = 1/4
Remaining = 1 - (1/2 + 1/4) = 1/4
Green = Blue = 1/4 ÷ 2 = 1/8. A1

Answer (a): Pink: 1/2 (0.5); Yellow: 1/4 (0.25); Green: 1/8 (0.125); Blue: 1/8 (0.125)

(b)

Short necklace = y - 11 M2
Total length = y + 2(y - 11) = y + 2y - 22 = 3y - 22. A1

Answer (b): 3y - 22

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Question 10 OCR Jun 2024 4H Q11

[7 marks]

A student attempts two tasks.
The result of each task is either “Pass” or “No pass”.
The probability of the student passing the first task is 0.6.
The probability of the student passing the second task is x.

(a) Complete the tree diagram. [2 marks]

(b) Write down the mathematical assumption that has been made about the two tasks. [1 mark]

x =

Worked Solution & Answer

(a)

1 − 0.6 = 0.4 on first task branch M1
1 − x on both second task No pass branches. A1

Answer (a): First task No pass = 0.4; Second task No pass = 1 − x on both branches.

(b)

The probability of passing the second task is independent of the result of the first task. C1

Answer (b): The two tasks are independent.

(c)

P(pass just one) = 0.6(1 − x) + 0.4x = 0.528. M3
0.6 − 0.6x + 0.4x = 0.528 ⇒ 0.6 − 0.2x = 0.528 ⇒ 0.2x = 0.072 ⇒ x = 0.36. A1

Answer (c): x = 0.36

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Score: 0 / 10 (0 / 37 marks)

Probability equation questions Worksheets Next: Invariant points →

Grade 8+ Quizzes

14 Topics

8A - Quad. sim. equations	8B - Transforming graphs y=f(x)
8C - Algebraic proof	8D - Completing the square
8E - Quadratic inequalities	8F - Quadratic sequences
8G - Velocity time graphs	8H - Circles and tangents
8J - Circle theorem proof	8K - Vector proof
8L - Probability equation questions	8M - Invariant points
8N - Spheres, cones and cylinders	8P - Volume of a frustum