4.6 Combined, conditional and independent events (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Organising outcomes: Venn diagrams, trees and tables

Several tools help you count outcomes and find probabilities. On this platform they are described in words, but you should picture them.

Venn diagram: overlapping sets $A$ and $B$ inside the sample space $U$ ; fill in the intersection first, then 'only $A$ ', 'only $B$ ', and 'neither'.
Tree diagram: one set of branches per stage; multiply along a branch, add the ends of different branches.
Sample space diagram / table of outcomes: for two dice, a $6\times6$ grid of 36 outcomes.

Key termsintersectionuniontree diagram

Exam tip

With '120 members, 20 use neither', start from $n(A \cup B) = 120 - 20 = 100$ .

Section 2

Combined events

The addition rule (in the formula booklet) is $P(A \cup B) = P(A) + P(B) - P(A \cap B).$ The overlap is subtracted because it has been counted twice. In probability, 'or' is inclusive: ' $A$ or $B$ ' includes the outcomes where both happen.

$A$ and $B$ are mutually exclusive if they cannot happen together: $P(A \cap B) = 0$ , so $P(A \cup B) = P(A) + P(B)$ .

Key termsmutually exclusiveaddition rule

Common mistake

Adding $P(A) + P(B)$ for events that can happen together. Subtract the overlap unless they are mutually exclusive.

Section 3

Conditional probability

$P(A \mid B)$ is the probability of $A$ given that $B$ has happened: $P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$ The condition shrinks the sample space to $B$ . With counts, $P(W \mid S) = \frac{n(S \cap W)}{n(S)} = \frac{25}{70}$ .

From a tree you can also reverse the condition: if $P(\text{bus} \cap L) = 0.1$ and $P(L) = 0.16$ , then $P(\text{bus} \mid L) = \frac{0.1}{0.16} = 0.625$ .

Key termsconditional probability

Common mistake

Dividing by the wrong event. In $P(A \mid B)$ you divide by $P(B)$ , the event that is given.

Exam tip

'Given that the socks are the same colour' means the denominator is $P(\text{same colour})$ .

Section 4

Independent events

$A$ and $B$ are independent if one happening does not change the probability of the other: $P(A \cap B) = P(A)P(B), \quad \text{equivalently } P(A \mid B) = P(A).$ To test for independence, calculate both sides and compare. Do not confuse this with mutually exclusive: mutually exclusive events with non-zero probabilities are never independent, because one happening makes the other impossible.

Key termsindependent events

Common mistake

Assuming independence without checking. Show the calculation and state the conclusion.

Section 5

With and without replacement

With replacement, the second draw is independent of the first and the probabilities stay the same: $\left(\frac{5}{12}\right)^2$ .

Without replacement, the second draw depends on the first: both the number of that colour and the total go down by one: $\frac{5}{12}\times\frac{4}{11}$ .

For 'one of each colour', remember both orders: $P(WB) + P(BW)$ . For 'at least one', use the complement: $1 - P(\text{none})$ .

Key termswith replacementwithout replacement

Must know

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$ ; 'or' includes both.
Mutually exclusive: $P(A \cap B) = 0$ .
$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$ .
Independent: $P(A \cap B) = P(A)P(B)$ .
Without replacement: reduce the counts on the second draw.
'At least one' $= 1 - P(\text{none})$ .

That's the notes covered.

Carry on to the next subtopic.