Conditional Probability and Independence

By the end of this lesson, you will be able to compute conditional probabilities using the conditional probability formula, determine independence of events, apply Bayes' Theorem to reverse conditional probabilities, and interpret conditional probability in context.

The Big Idea

Conditional probability asks: "Given that event $B$ has already occurred, how likely is $A$?" The known occurrence of $B$ restricts our sample space and changes the probabilities. This is fundamental to medical testing, legal reasoning, and machine learning.

The Conditional Probability Formula

$$P(A|B) = \frac{P(A \cap B)}{P(B)} \quad \text{(when } P(B) > 0\text{)}$$

Read: "probability of $A$ given $B$."

Rearranged: $P(A \cap B) = P(B) \cdot P(A|B)$ — this is the multiplication rule for dependent events.

Independence

Events $A$ and $B$ are independent if and only if:

$$P(A|B) = P(A)$$

Knowing $B$ occurred gives no information about $A$.

Equivalent condition: $P(A \cap B) = P(A) \cdot P(B)$

Testing for independence: compute both $P(A) \cdot P(B)$ and $P(A \cap B)$ — if equal, independent.

Example 1 — Computing $P(A|B)$

From a survey: $P(\text{exercise}) = 0.6$, $P(\text{healthy}) = 0.7$, $P(\text{exercise and healthy}) = 0.5$.

$P(\text{healthy | exercise}) = \dfrac{0.5}{0.6} = \dfrac{5}{6} \approx 0.833$

$P(\text{exercise | healthy}) = \dfrac{0.5}{0.7} \approx 0.714$

Example 2 — Independence Test

Two dice rolled. $A$ = first die shows even, $B$ = sum is 7. Are $A$ and $B$ independent?

$P(A) = 1/2$; $P(B) = 6/36 = 1/6$

$P(A \cap B)$: even first die AND sum 7: $(2,5), (4,3), (6,1)$ → $3/36 = 1/12$

$P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$ = $P(A \cap B)$ → independent.

Example 3 — Two-Way Table

	Passed	Failed	Total
Studied	42	8	50
Didn't study	18	32