Probability and Algebra

This lesson combines probability with algebra — setting up and solving equations from probability statements, working with unknowns in probability problems, and tackling "show that" or proof questions. These are primarily Higher tier topics on the Edexcel GCSE Mathematics (1MA1) specification and often appear as the final parts of 5–6 mark questions.

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Setting Up Equations from Probability Statements

Many Edexcel questions present probability information in words and require you to form and solve an equation.

General Approach

1. Let the unknown be a variable (usually x or n).
2. Write expressions for the number of favourable outcomes and the total number of outcomes.
3. Set up an equation using the given probability.
4. Solve for the unknown.

Worked Example 1

A bag contains x red counters and 8 blue counters. A counter is chosen at random. The probability that it is red is 3/7. Find the value of x.

Solution:

P(red) = x / (x + 8)

Set this equal to 3/7: x / (x + 8) = 3/7

Cross-multiply: 7x = 3(x + 8) 7x = 3x + 24 4x = 24 x = 6

Check: 6 red, 8 blue, total 14. P(red) = 6/14 = 3/7 ✓

Worked Example 2

A bag contains n balls. 5 are black and the rest are white. The probability of choosing a white ball is 3/4. Find the value of n.

Solution:

Number of white balls = n − 5

P(white) = (n − 5) / n = 3/4

Cross-multiply: 4(n − 5) = 3n 4n − 20 = 3n n = 20

Check: 20 balls total, 5 black, 15 white. P(white) = 15/20 = 3/4 ✓

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Probability with Quadratic Equations

When problems involve "without replacement" and an unknown total, you often get a quadratic equation.

Worked Example 3

A bag contains n balls, of which 6 are red. Two balls are drawn at random without replacement. The probability that both balls are red is 1/3. Show that n² − n − 90 = 0 and find the value of n.

Solution:

P(both red) = (6/n) × (5/(n − 1)) = 30 / (n(n − 1))

Set equal to 1/3: 30 / (n(n − 1)) = 1/3

Cross-multiply: 90 = n(n − 1) 90 = n² − n n² − n − 90 = 0 ✓ (shown)

Factorise: (n − 10)(n + 9) = 0 n = 10 or n = −9

Since n must be positive: n = 10

Check: P(both red) = $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$106​×95​=9030​ = 1/3 ✓

Edexcel Exam Tip: "Show that" questions require you to present every step of working — you must arrive at the given equation. Do not skip steps or work backwards from the answer. Show the cross-multiplication clearly.

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Probability with Algebraic Fractions

Worked Example 4

A bag contains (2x + 1) red counters and (x + 2) blue counters. The probability of picking a red counter is 3/5. Find x.

Solution:

Total counters = (2x + 1) + (x + 2) = 3x + 3

P(red) = (2x + 1) / (3x + 3) = 3/5

Cross-multiply: 5(2x + 1) = 3(3x + 3) 10x + 5 = 9x + 9 x = 4

Check: Red = 2(4) + 1 = 9, Blue = 4 + 2 = 6, Total = 15. P(red) = 9/15 = 3/5 ✓

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Probability Involving Ratios

Worked Example 5

A bag contains red and blue counters in the ratio 2 : 3. Two more red counters are added. The probability of picking a red counter is now 1/2. How many counters were in the bag originally?

Solution:

Let the original numbers be 2k red and 3k blue (total = 5k).

After adding 2 red: (2k + 2) red and 3k blue, total = 5k + 2.

P(red) = (2k + 2) / (5k + 2) = 1/2

Cross-multiply: 2(2k + 2) = 5k + 2 4k + 4 = 5k + 2 2 = k

Original bag: 2(2) = 4 red, 3(2) = 6 blue, total = 10 counters

Check: After adding 2 red: 6 red, 6 blue, total 12. P(red) = 6/12 = 1/2 ✓

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Show That / Proof Questions

Edexcel often asks you to "show that" a particular equation or result follows from a probability scenario. The key is to:

1. Define variables clearly.
2. Write down probability expressions step by step.
3. Show all algebraic manipulation.
4. Arrive at the given expression.

Worked Example 6

A bag has n counters. 4 are white and the rest are black. Two counters are drawn without replacement. Show that the probability of getting two black counters is:

(n − 4)(n − 5) / n(n − 1)

Solution:

Number of black counters = n − 4.

P(1st black) = (n − 4) / n

After removing one black counter: (n − 5) black remain, total (n − 1).

P(2nd black | 1st black) = (n − 5) / (n − 1)

P(both black) = P(1st black) × P(2nd black | 1st black) = ((n − 4) / n) × ((n − 5) / (n − 1)) = (n − 4)(n − 5) / n(n − 1) ✓

Worked Example 7

A bag contains n counters, 6 of which are red. Two counters are drawn without replacement. The probability that both are red is 1/3. Show that n² − n − 90 = 0 and find n.

Solution:

P(both red) = (6/n) × (5/(n − 1)) = 30/(n(n − 1))

Set equal to 1/3: 30/(n(n − 1)) = 1/3

Cross-multiply: 90 = n(n − 1) 90 = n² − n n² − n − 90 = 0 ✓ (shown)

Factorise: (n − 10)(n + 9) = 0 n = 10 or n = −9

Since n must be positive: n = 10

Check: P(both red) = $\frac{6}{10} \times \frac{5}{9} = \frac{30}{90}$106​×95​=9030​ = 1/3 ✓

Worked Example 8 — Algebraic Ratio Problem

A bag has (x + 3) green counters and (2x − 1) red counters. The probability of picking a green counter is 2/5. Find x and the total number of counters.

Solution:

Total counters = (x + 3) + (2x − 1) = 3x + 2.

P(green) = (x + 3) / (3x + 2) = 2/5.

Cross-multiply: 5(x + 3) = 2(3x + 2) 5x + 15 = 6x + 4 11 = x x = 11

Substituting back: green = 11 + 3 = 14, red = 2(11) − 1 = 21, total = 14 + 21 = 35 counters.

Check: P(green) = 14/35 = 2/5 ✓

Worked Example 9 — Quadratic from "Exactly One"

A bag has n counters, 3 of which are white. Two counters are drawn without replacement. The probability that exactly one is white is 1/2. Show that n² − 13n + 36 = 0 and find n.

Solution:

Let W' mean "not white" (so n − 3 of them).

P(exactly one white) = P(WW') + P(W'W) = (3/n)((n − 3)/(n − 1)) + ((n − 3)/n)(3/(n − 1)) = 2 × 3(n − 3) / (n(n − 1)) = 6(n − 3) / (n(n − 1))

Set equal to 1/2: 6(n − 3) / (n(n − 1)) = 1/2

Cross-multiply: 12(n − 3) = n(n − 1) 12n − 36 = n² − n n² − n − 12n + 36 = 0 n² − 13n + 36 = 0 ✓ (shown)

Factorise: (n − 4)(n − 9) = 0, so n = 4 or n = 9.