Given that X is in a range

Conditional probability with a distribution

In 11 of 20 recent exams
2% of the marks
Paper 1 or 2
3 to 5 marks

How it is set in the exam

Paper 2 (calculator), 3–5 marks, usually a later part whose denominator is an earlier part's answer. Very common (11 of 20 exam versions, 17 parts). Once on Paper 1 (no calculator), with a table.

Spot it in the exam

Exam phrases to spot (4)

“given that at least one …, find the probability that at most $k$ …”
“given that a bag weighs more than …, find the probability that it weighs less than …”
“given that the person is over a given age”
“one class is removed …”

In disguise (3)

no "given that": "of the bags that pass inspection, what proportion …", or only one class goes on to the next stage. The condition is the class that is kept (a reduced sample space).
"if it is known that …", "among those who …": a condition in other words
the condition is an earlier answer: "(b) find $\mathrm{P}(X \ge 1)$ ", then "(c) given that at least one …"

Variants (2)

source = a discrete table (), a grouped table in percentages, a binomial, a normal, a reduced sample space
ranges = nested (one inside the other), overlapping, or symmetric about the mean; the cut-offs strict or not

Looks like it, but isn't (2)

the events are Venn regions, table cells or tree paths, not ranges of one variable: Conditional probability
a conditional probability is given, and $\mu$ and $\sigma$ are unknown: Unknown mean or standard deviation (the conditional supplies the second fact)

Worked example

A table: both events are inequalities

Paper 1 (no calculator)

Question

The random variable $X$ has the probability distribution shown.

$x$	$\mathrm{P}(X = x)$
$1$	$\frac{3}{20}$
$2$	$\frac{1}{4}$
$3$	$\frac{3}{10}$
$4$	$\frac{1}{5}$
$5$	$\frac{1}{10}$

$x$	$1$	$2$	$3$	$4$	$5$
$\mathrm{P}(X = x)$	$\frac{3}{20}$	$\frac{1}{4}$	$\frac{3}{10}$	$\frac{1}{5}$	$\frac{1}{10}$

Find $\mathrm{P}(X > 2 \mid X \le 4)$ . [3]

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See it move: Given that: a smaller sample space

$P(X > 2 \mid X \le 4) = \frac{P(\{3, 4\})}{P(X \le 4)} = \frac{6 + 4}{3 + 5 + 6 + 4} = \frac{10}{18} = \frac{5}{9}$ (0.556).
The condition shrinks the sample space to the shaded bars, X ≤ 4: the denominator is their total, 18 of 20, not 20. The numerator is the overlap, the yellow bars: 5 is in the event but fails the condition, so its count doesn't count.
Strict words first: "more than 2" is $X \ge 3$ and "at most 4" is $X \le 4$ , so the overlap is $\{3, 4\}$ , not $\{3, 4, 5\}$ : $\frac{6 + 4}{3 + 5 + 6 + 4} = \frac{10}{18} = \frac{5}{9}$ . With every probability written over 20, the 20s cancel.
The other way round, $P(X \le 4 \mid X > 2) = \frac{10}{12} = \frac{5}{6}$ (0.833): the same overlap over a different condition.
Check: the answer is at least the overlap's own probability, $\frac{10}{20} = 0.500$ , and at most 1: dividing by a probability under 1 makes it bigger.

Steps

Write the conditional in context,
with the variable: $\mathrm{P}(X \le 5 \mid X \ge 3)$ , or $\mathrm{P}(\text{fewer than } 6 \mid \text{more than } 2)$ . A bare $\mathrm{P}(A \mid B)$ earns nothing.
Draw both ranges on a number line.
The intersection is the overlap: for nested ranges, the narrower one. For a discrete variable, convert strict words first: "more than $2$ " is $X \ge 3$ , "fewer than $6$ " is $X \le 5$ .
Numerator: the probability of the overlap. Denominator: the probability of the condition.
Take them from the table, the binomial pdf (a single value) or cdf (GDC), or the normal cdf (GDC). The denominator is often an earlier part: use its stored value (keep full-accuracy values).
Divide,
and simplify an exact fraction on .
Check:
the answer is at most $1$ and at least the numerator (dividing by a probability makes it bigger).

Common mistakes

The event alone as the numerator
in Example 5.2, $\mathrm{P}(X \le 5)$ instead of $\mathrm{P}(3 \le X \le 5)$ . An answer above $1$ gives it away there, but not always: in Example 5.1, $\frac{\mathrm{P}(X > 2)}{\mathrm{P}(X \le 4)} = \frac{2}{3}$ is below $1$ and still wrong. Always write the overlap.
An exact value given a range
in $\mathrm{P}(X = 2 \mid X < 5)$ the overlap is $X = 2$ itself: $\mathrm{P}(X = 2)$ from the pdf, not $\mathrm{P}(X \le 2)$ .
Strict words not converted
for a binomial, "more than $2$ " is $X \ge 3$ , before the cdf.

More mistakes (4)

A bare $\mathrm{P}(A \mid B)$
the recognition mark needs the actual events, in symbols or words.
The unconditional probability used later
after a class is removed, the new $p$ is the conditional one (Example 5.5).
Dividing the wrong way round
the condition goes in the denominator. Upside down, the result is always at least $1$ ; but a result above $1$ can also come from a numerator that is not the overlap.
A rounded earlier answer as the denominator
use the stored value.

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