Conditional Probability

Conditional probability measures the chance of an event given that another event has occurred.

P(A\mid B)=\frac{P(A\cap B)}{P(B)},\quad \text{provided } P(B)>0.

Key relationships:

Multiplication rule: $P(A\cap B)=P(A\mid B)\,P(B)=P(B\mid A)\,P(A)$ .
Independence: $A$ and $B$ are independent if $P(A\mid B)=P(A)$ which is equivalent to $P(A\cap B)=P(A)P(B)$ .
Law of Total Probability: If $B_1,\dots,B_k$ partition the sample space, then $P(A)=\sum_i P(A\mid B_i)P(B_i)$ .

Core Skills

Translate word statements into $P(\cdot)$ notation with the correct condition.
Use two way tables to compute conditional probabilities by dividing the appropriate cell by the given row or column total.
Use tree diagrams with branch probabilities to track multi step processes.
Test independence by checking $P(A\mid B)=P(A)$ or $P(A\cap B)=P(A)P(B)$ .

Example 1: From a Two Way Table

A survey records whether students prefer tea or coffee and whether they are juniors or seniors.

\begin{array}{c|ccc} & \text{Tea} & \text{Coffee} & \text{Total}\\ \hline \text{Junior} & 24 & 16 & 40\\ \text{Senior} & 21 & 39 & 60\\ \hline \text{Total} & 45 & 55 & 100 \end{array}

(a)

P(\text{Tea}\mid \text{Senior})=\dfrac{21}{60}=\dfrac{7}{20}

. (b)

P(\text{Senior}\mid \text{Coffee})=\dfrac{39}{55}

. (c)

P(\text{Tea and Junior})=\dfrac{24}{100}=0.24

Example 2: Independence Check

Using the table above,

P(\text{Tea})=45/100=0.45

P(\text{Tea}\mid \text{Senior})=21/60=0.35 \ne 0.45

. Not equal, so Tea preference and Senior status are not independent.

Example 3: Tree Diagram for a Multi Step Process

A two-level tree diagram: the first branch is rain vs. no rain with probabilities 0.3 and 0.7. From each, a branch for traffic jam with conditional probabilities 0.5 if rain and 0.2 if no rain. Compute

P(\text{Jam})

P(\text{Jam})=P(\text{Jam}\mid \text{Rain})P(\text{Rain})+P(\text{Jam}\mid \text{NoRain})P(\text{NoRain}) =0.5(0.3)+0.2(0.7)=0.15+0.14=0.29.

Example 4: Using the Multiplication Rule

A bag has 5 red and 7 blue marbles. Draw 2 without replacement.

P(\text{both red})=P(\text{R}_1)\,P(\text{R}_2\mid \text{R}_1)=\frac{5}{12}\cdot\frac{4}{11}=\frac{20}{132}=\frac{5}{33}.

Example 5: Bayes Type Update

A test is positive with probability 0.9 if a person has a condition and 0.1 if not. Prevalence is 5%.

P(\text{Has}\mid +)=\frac{0.9\cdot0.05}{0.9\cdot0.05+0.1\cdot0.95} =\frac{0.045}{0.045+0.095}=\frac{0.045}{0.14}\approx 0.321.

Interpretation: despite a high true positive rate, the low prevalence means a positive result is only about 32.1% likely to indicate the condition.

Example 6: SAT Style Card Context

From a standard deck, one card is drawn and observed to be a face card. What is the probability it is a heart? There are 12 face cards total and 3 are hearts.

P(\text{Heart}\mid \text{Face})=\frac{3}{12}=\frac{1}{4}.

Key Takeaways

Read $P(A\mid B)$ as “probability of $A$ within the world where $B$ is known to have occurred.”
For tables, divide the intersection cell by the given condition total.
For sequential processes, multiply along branches and add across paths.
Independence means conditioning does not change the probability.