Combined and Conditional Probability

Two events are independent if the outcome of one does not affect the probability of the other.

Multiplication rule for independent events:

$P(A\text{ and }B)=P(A)\times P(B)$

Worked example — two fair dice are rolled. Find $P(\text{both show a six})$.

$P(\text{both six})=\frac{1}{6}\times \frac{1}{6}=\frac{1}{36}$ ✓

Tree diagram with replacement — drawing maintains the same probabilities at every stage because the original composition is restored.

Worked example — a bag contains 4 red and 6 blue balls. Two draws with replacement. Find $P(\text{at least one red})$.

$P(\text{at least one red})=1-P(\text{no red})=1-P(BB)=1-\frac{6}{10}\times \frac{6}{10}=1-\frac{36}{100}=\frac{64}{100}=\frac{16}{25}$ ✓

Underlying assumption: independence requires that drawing one ball does not change the composition of the bag — only valid when sampling with replacement.

When an item is not replaced, the probabilities on the second draw depend on what happened first.

Worked example — a bag has 5 red and 3 blue balls. Two balls are drawn without replacement. Find $P(\text{both red})$.

First draw: $P(R)=5/8$

Second draw (given red drawn first): $P(R\mid \text{first red})=4/7$ (4 red remain out of 7 total)

$P(RR)=\frac{5}{8}\times \frac{4}{7}=\frac{20}{56}=\frac{5}{14}$ ✓

Worked example — find $P(\text{one of each colour})$:

$P(RB)=\frac{5}{8}\times \frac{3}{7}=\frac{15}{56}$; $P(BR)=\frac{3}{8}\times \frac{5}{7}=\frac{15}{56}$

$P(\text{one of each})=\frac{15}{56}+\frac{15}{56}=\frac{30}{56}=\frac{15}{28}$ ✓

The conditional probability $P(A\mid B)$ is the probability of $A$ given that $B$ has occurred:

$P(A\mid B)=\frac{P(A\cap B)}{P(B)}$

Worked example — from a two-way table:

	Coffee	Tea	Total
Male	15	10	25
Female	8	17	25
Total	23	27	50

$P(\text{coffee given male})=\frac{15}{25}=\frac{3}{5}$

$P(\text{male given coffee})=\frac{15}{23}$

Note: $P(\text{coffee}\mid \text{male})\ne P(\text{male}\mid \text{coffee})$ — the conditioning event changes the denominator.

Using a Venn diagram with known frequencies, conditional probabilities are found by restricting to the given set.

Worked example — $\xi =50$ students: $A=$ studies Art ($n(A)=22$), $B=$ studies Biology ($n(B)=18$), $n(A\cap B)=7$.

$P(A\mid B)=\frac{n(A\cap B)}{n(B)}=\frac{7}{18}$ — given the student studies Biology, look only within the Biology set.

$P(B\mid A)=\frac{7}{22}$ ✓

Expected frequencies approach (P9 Higher): instead of fractions, set up a 100-person (or 1000-person) table with expected counts matching the given probabilities, then read off the conditional probability directly.

Worked example — $P(A)=0.4$, $P(B\mid A)=0.3$, $P(B\mid A^{\prime })=0.1$. Use a frequency tree of 1000 people to find $P(A\mid B)$.

From 1000: $A$: 400 people; $B\mid A$: $400\times 0.3=120$; $A^{\prime }$: 600 people; $B\mid A^{\prime }$: $600\times 0.1=60$.

Total $B$: $120+60=180$. $P(A\mid B)=120/180=2/3$ ✓

For mutually exclusive events (cannot both happen): $P(A\cup B)=P(A)+P(B)$

For non-mutually exclusive events: $P(A\cup B)=P(A)+P(B)-P(A\cap B)$

Worked example — from a pack of 52 cards, find $P(\text{heart or king})$.

$P(\text{heart})=13/52$; $P(\text{king})=4/52$; $P(\text{heart and king})=1/52$ (king of hearts)

$P(\text{heart or king})=13/52+4/52-1/52=16/52=4/13$ ✓

Checking independence: $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)\times P(B)$.

1. Without replacement — not adjusting the second probability

After drawing a red ball from a bag of 5R and 3B without replacing it, the second draw has 7 balls remaining, not 8. Failing to reduce both the favourable count and the total is a very common error.

2. Conditional probability — using the wrong denominator

$P(A\mid B)$ has denominator $P(B)$ (or $n(B)$ in frequency form), not the total sample size. Read "given $B$" as "restrict to the $B$ circle/row/column."

3. Adding independent probabilities instead of multiplying

$P(A\text{ and }B)$ for independent events uses multiplication. Adding gives $P(A\text{ or }B)$ for mutually exclusive events — a different quantity.

4. Assuming all combined events are independent

Cards, balls without replacement, and real-world paired events are often dependent. Confirm whether items are replaced or whether the events can influence each other before choosing a method.

Mistake	Correction
"P(red twice without replacement) = $\frac{5}{8}\times \frac{5}{8}$"	Second draw: $\frac{4}{7}$, not $\frac{5}{8}$ — one red is gone
"$P(A\mid B)=P(B\mid A)$"	These are equal only in special cases; use $P(A\mid B)=P(A\cap B)/P(B)$
"P(heart or ace) = P(heart) + P(ace) = $\frac{13}{52}+\frac{4}{52}=\frac{17}{52}$"	Subtract the overlap: $P(\text{heart and ace})=\frac{1}{52}$; answer $=\frac{16}{52}$