Probability

Probability underpins every Statistics topic in H2 Maths. You need to move fluently between set notation, tree diagrams, and two-way tables, and to keep a clear head about the difference between independent events and mutually exclusive events — two ideas students frequently confuse.

What you need to know

• Addition rule: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ for any two events $A$ and $B$.
• Mutually exclusive events: $P(A \cap B) = 0$, so $P(A \cup B) = P(A) + P(B)$.
• Conditional probability: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, provided $P(B) > 0$.
• Independent events: $P(A \cap B) = P(A)\,P(B)$, equivalently $P(A \mid B) = P(A)$.
• Independence $\neq$ mutual exclusivity: if $A$ and $B$ are mutually exclusive and both have positive probability, they cannot be independent (knowing one occurred tells you the other did not).
• Tree and Venn diagrams are complementary tools; a tree is clearest when events occur in sequence, a Venn diagram when you need to read off union/intersection probabilities at a glance.

Worked example 1 — Conditional probability from a two-way table

A group of 200 students sat two papers. 120 passed Paper 1, 90 passed Paper 2, and 60 passed both.

Find (a) $P(\text{passed Paper 2} \mid \text{passed Paper 1})$, and (b) whether passing Paper 1 and passing Paper 2 are independent.

Solution.

Let $A$ = passed Paper 1, $B$ = passed Paper 2.

(a)

(b) Check whether $P(A \cap B) = P(A)\,P(B)$:

Since $0.27 \neq 0.30$, the events are not independent.

Worked example 2 — Tree diagram and addition rule

A bag contains 4 red and 6 blue counters. Two counters are drawn without replacement. Find the probability that both counters are the same colour.

Solution.

Draw a tree. Branch probabilities (first draw / second draw):

• Red then Red: $\dfrac{4}{10} \times \dfrac{3}{9} = \dfrac{12}{90}$
• Blue then Blue: $\dfrac{6}{10} \times \dfrac{5}{9} = \dfrac{30}{90}$

These outcomes are mutually exclusive, so:

Check: the four branch-pair probabilities are $\tfrac{12}{90} + \tfrac{24}{90} + \tfrac{24}{90} + \tfrac{30}{90} = \tfrac{90}{90} = 1$. ✓

Worked example 3 — Independence and the addition rule

Events $C$ and $D$ are independent with $P(C) = 0.4$ and $P(D) = 0.5$. Find $P(C \cup D)$.

Solution.

Because $C$ and $D$ are independent, $P(C \cap D) = P(C)\,P(D) = 0.4 \times 0.5 = 0.2$.

Common mistakes

• Confusing independent with mutually exclusive. Mutually exclusive events with positive probability are never independent — ruling one out gives certainty about the other.
• Forgetting to subtract $P(A \cap B)$ in the addition rule. It is only safe to add probabilities directly when events are mutually exclusive.
• Using replacement probabilities for without-replacement draws. Always adjust the denominator after each draw when items are not replaced.
• Reversing conditional probability. $P(A \mid B) \neq P(B \mid A)$; read the conditioning event (after "$\mid$") carefully before substituting.