Probabilities in terms of an unknown

Form an equation, solve, reject

In 7 of 20 recent exams
1% of the marks
Paper 1 or 2
3 to 6 marks

How it is set in the exam

Paper 1 (no calculator) mostly (4 exam versions on Paper 1 (no calculator), 3 on Paper 2 (calculator)), 3–6 marks, often split into "complete the tree" or "show that" parts before the solve. Common (7 of 20 exam versions). The equation is a quadratic in 5 of them, and a root is rejected in 3.

Spot it in the exam

Exam phrases to spot (5)

“given that $A$ and $B$ are independent, find $x$ ”
“the probability that … is $\frac{a}{b}$ ; show that $\ldots k^2 \ldots = 0$ ”
“find the value of $k$ ”
“find the possible values of $x$ ”
“find the draw on which the probability is …”

In disguise (4)

the unknown is a number of items, so the probabilities are fractions in $n$
" $A$ is four times as likely as $B$ ": one probability is a multiple of the other
two formulas share the same unknowns: the union and a conditional, each with $\mathrm{P}(B)$ in it
a cell of a two-way table is unknown, and a conditional probability fixes it

Variants (3)

the fact that gives the equation: independence, a path probability, a conditional, the union and a conditional combined, or a product of fractions that telescopes
the equation: linear or quadratic
rejected: a root above $1$ (or one that makes another branch, such as $2k$ , greater than $1$ ), a negative root or count, or none (both roots valid, and both are required)

Looks like it, but isn't (3)

$k$ is in a table of $\mathrm{P}(X = x)$ and nothing else is given: Probability distribution tables (they add to $1$ )
independence is to be tested, with nothing unknown: Are the events independent?
the least number of trials for $1 - (1 - p)^n$ (not every trial failing) to pass a target (an inequality in $n$ ): Find n; with two or more successes needed, a table: Binomial probabilities. A probability equal to a value, with $n$ inside the fractions (Your turn 4), stays here

Worked example

Venn regions in $x$ , independence given: both roots valid

Paper 1 (no calculator)

Question

Events $A$ and $B$ are independent. $\mathrm{P}(A \cap B') = 0.3$ , $\mathrm{P}(A' \cap B) = 0.2$ and $\mathrm{P}(A \cap B) = x$ .
(a) Find the possible values of $x$ . [5]
(b) For each value of $x$ , find $\mathrm{P}(A \mid B)$ . [2]

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Steps

Write every probability you need in the unknown:
Venn regions and the singles they add to; branches as complements ( $1 - k$ ); table cells and totals; fractions in $n$ .
2
Form the equation(s)
from the fact given:
- independence: $\mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B)$
- a path probability: the product(s) of branches $=$ the value
- a conditional: $\mathrm{P}(A \cap B) =$ the value $\times\,\mathrm{P}(B)$
- two facts that share unknowns: write each as an equation, then substitute one into the other. Markschemes mark each equation separately.
Solve:
a linear equation directly; a quadratic by factorising on , or on the GDC on (solve a polynomial equation). Find both roots before rejecting anything.
Check every probability, not just the unknown:
each region, branch and cell must lie in $[0, 1]$ , and a count must be a whole number that is not negative (the number of items in a bag must be at least the number of each kind in it). Reject with the reason (), or keep both roots when both are valid.

Common mistakes

Rejecting a root before finding both,
or rejecting with no reason: the is for the reason, after both roots are seen.
Leaving an invalid root in the final answer
(), or dropping a valid one: when both roots give valid probabilities, both are the answer (Example 5.1).
Checking only the unknown
in Example 5.4, $k = \frac{3}{4}$ is a valid probability on its own, but it makes the branch $2k = \frac{3}{2}$ .

More mistakes (3)

Using independence when it isn't given.
$\mathrm{P}(A \cap B) = \mathrm{P}(A)\,\mathrm{P}(B)$ needs the word "independent" (or its meaning) in the question.
A "show that" worked backwards
from the given quadratic: form the equation from the diagram and expand it forwards.
A count that is negative or not a whole number
$n$ counters can't be $-9$ or $7.5$ .

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