Probability · Study notes

MATConditional Probability & Independence

Conditional Probability & the Multiplication Theorem

Conditional probability $P(A\mid B)$ is the probability of event $A$ given that event $B$ has already occurred, found by restricting the sample space to $B$ . Rearranging its definition gives the multiplication theorem $P(A\cap B)=P(B)\,P(A\mid B)$ , the backbone of "draw without replacement" and "given that" problems that appear every year.

Definition of conditional probability

P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}, \quad P(B) \ne 0

P(A\mid B)

= probability of

A

given

B

has occurred;

P(A\cap B)

= probability both occur. Requires

P(B) > 0

Multiplication theorem of probability

P(A \cap B) = P(B)\,P(A \mid B) = P(A)\,P(B \mid A)

Valid for any two events with positive probability; choose the conditioning order that matches the order of the experiment.

Multiplication for three events (chain rule)

P(A \cap B \cap C) = P(A)\,P(B \mid A)\,P(C \mid A \cap B)

Used for successive draws without replacement; each factor conditions on all previous outcomes.

Successive draws without replacement

P(\text{both red}) = \dfrac{r}{n}\cdot\dfrac{r-1}{n-1}

r

= number of red balls,

n

= total balls; the numerator and denominator each drop by

1

on the second draw because no ball is replaced.

Reduced sample space view: $P(A\mid B)$ counts favourable outcomes for $A$ only among the outcomes already in $B$ .
Key properties: $0 \le P(A\mid B) \le 1$ , $P(B\mid B)=1$ , and $P(A'\mid B) = 1 - P(A\mid B)$ .
For drawing "one after another without replacement", multiply the probabilities of each stage, updating the counts after each draw.
In the two-children "given at least one boy" problem the reduced sample space is $\{BB, BG, GB\}$ (3 equally likely cases), giving $P(\text{two boys}\mid \text{at least one boy}) = \dfrac{1}{3}$ , not $\dfrac{1}{2}$ .
"Given that the card is a face card" restricts to the $12$ face cards; for example $P(\text{King}\mid\text{face card}) = \dfrac{4}{12} = \dfrac{1}{3}$ .
Identify the conditioning event ( $B$ ) carefully — it is the information stated after the words "given that".
$P(A\mid B)$ and $P(B\mid A)$ are generally different; do not interchange them.

Worked example · 2 marks

Evaluate

P(A\cup B)

given that

2P(A) = P(B) = \dfrac{5}{13}

and

P(A\mid B) = \dfrac{2}{5}

Read off the two marginals: $P(B) = \dfrac{5}{13}$ , and since $2P(A) = \dfrac{5}{13}$ , $P(A) = \dfrac{5}{26}$ .
Turn the conditional into an intersection with the multiplication theorem: $P(A\cap B) = P(A\mid B)\,P(B) = \dfrac{2}{5}\times\dfrac{5}{13} = \dfrac{2}{13}$ .
Addition theorem: $P(A\cup B) = P(A) + P(B) - P(A\cap B)$ .
Over a common denominator of $26$ : $\dfrac{5}{26} + \dfrac{10}{26} - \dfrac{4}{26}$ .

Answer

P(A\cup B) = \dfrac{11}{26}

Where the marks go

Confusing $P(A\cap B)$ (both happen) with $P(A\mid B)$ (one given the other) — the conditional divides by $P(B)$ .
Treating without-replacement draws as if the second draw had the same denominator as the first.
In "at least one shows a 3" type dice problems, forgetting to restrict the sample space to the conditioning event before counting favourable cases.
Reversing the multiplication order so that the conditioning does not match the actual sequence of the experiment.

How the board asks it

Applicationmultiplication theorem; drawing without replacement4 mkAsked 2025
A bag contains $5$ white and $7$ black balls. Two balls are drawn one after the other without replacement. Find the probability that both balls drawn are black.
Numericaldefinition of conditional probability2 mkAsked 2024
If $P(A) = 0.6$ , $P(B) = 0.5$ and $P(A \cap B) = 0.3$ , find $P(A \mid B)$ and $P(B \mid A)$ .
Multiple choicedefinition of conditional probability1 mkAsked 2026
If $A$ and $B$ are events such that $P(B) = 0.4$ and $P(A \cap B) = 0.16$ , then $P(A \mid B)$ equals: (a) $0.16$ (b) $0.4$ (c) $0.64$ (d) $0.24$ .
Numericalmultiplication theorem for three events1 mkAsked 2026
An urn contains $4$ red and $6$ green balls. Three balls are drawn successively without replacement. Find the probability that the first is red, the second is green and the third is red.
Give reasonsreduced sample space view
A die is thrown twice and the sum of the numbers appearing is observed to be $8$ . Find the conditional probability that the number $5$ has appeared at least once, clearly stating the reduced sample space.
Assertion–Reasonmutual exclusivity is not what makes $P(A\mid B) = P(B\mid A)$ 1 mkAsked 2025
Assertion: for events $A$ and $B$ with $n(A) = n(B)$ , $P(A\mid B) = P(B\mid A)$ . Reason: $A$ and $B$ are mutually exclusive.

Practise this topic

NextIndependent Events

Written for Sublevo. Question text quoted anywhere in these notes is the Council’s and carries its year and paper; the board’s own diagrams are not reproduced.