For mutually exclusive outcomes, all probabilities sum to exactly 1 — forming a linear equation in the unknown.

Worked example: A spinner lands on Blue (0.15), Yellow ($2x$), Green ($x$), or Red (0.35). Find $x$.

Instead of calculating a probability from known quantities, you're given the final probability and an unknown total (e.g. $n$ counters in a bag) — your job is to form and solve an equation, usually a quadratic, for that unknown.

If a bag has $N$ total items, $n$ of one colour, and you pick without replacement:

Both numerator and denominator decrease by 1 — forgetting the numerator decrease (writing $\frac{n}{N-1}$) is the single most common conceptual error here.

This fraction, once cross-multiplied against a given probability, is what generates the quadratic equation.

A bag has $n$ sweets, 6 of them orange. Hannah eats one, then takes another. $P(\text{both orange})=\frac13$. (a) Show $n^2-n-90=0$. (b) Solve it.

(a) $P(\text{1st orange})=\frac6n$, $P(\text{2nd orange})=\frac{5}{n-1}$ (both the orange count and total drop by 1).

(b) Factorise: $(n-10)(n+9)=0 \implies n=10$ or $n=-9$. Since $n$ is a count of sweets, reject the negative solution: $n=10$.

A bag has $x$ red counters and 4 more blue than red. $P(\text{two blue})=\frac38$. Find $x$.

Define the total first — this is the step most often skipped: Total $=x+(x+4)=2x+4$.

Expand both brackets (numerator: $x^2+7x+12$; denominator: $4x^2+14x+12$), cross-multiply:

Quadratic formula: $x=\frac{7\pm\sqrt{49+240}}{4}=\frac{7\pm17}{4}$, giving $x=6$ or $x=-2.5$. Reject the negative: $x=6$.

"Same colour" means two mutually exclusive ways to succeed — red-red or blue-blue — so you add the two probabilities before setting the sum equal to the given value. This typically produces a messier quadratic, but the method is unchanged: define totals, build each probability, add them, cross-multiply, solve, and reject the invalid root.

1. The "calculator trap": typing the final equation into a calculator's solve function and writing only the answer. This scores zero — the question explicitly tests algebraic manipulation, and no working means no method marks, however correct the number is.
2. Treating dependent events as independent: writing $\frac{n}{N}\times\frac{n}{N}$ (with replacement logic) for a "without replacement" scenario. This is a conceptual failure, not an arithmetic slip, and also scores zero.

A bag has $n$ beads, 6 of them red, the rest blue. Ravi takes two beads. $P(\text{same colour})=\frac{9}{17}$. Find $n$.

Blue count $=n-6$.

Cross-multiply: $17(n^2-13n+72)=9(n^2-n) \implies 8n^2-212n+1224=0 \implies 2n^2-53n+306=0$

Formula: $n=\frac{53\pm\sqrt{2809-2448}}{4}=\frac{53\pm19}{4}$, giving $n=18$ or $n=8.5$.

Since $n$ must be a whole number of beads, reject $n=8.5$: the answer is $n=18$.