Conditional Probability

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

Definition

The probability of $A$ given $B$:

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

Read this as: “out of all the probability that belongs to $B$, what fraction also belongs to $A$?”

Breaking down the formula:

• Numerator $P(A \cap B)$: the probability that both $A$ and $B$ happen — the overlap
• Denominator $P(B)$: the total probability of $B$ — the whole region we’re conditioning on
• The ratio: what share of $B$’s probability is also in $A$?

The formula is only defined when $P(B) > 0$ — you can’t condition on an event that never happens.

Venn Diagram Intuition

Events can be visualized as regions inside a rectangle representing the sample space. Two circles $A$ and $B$ represent two events — their areas correspond to the probabilities of those events occurring. In the Venn diagram demo, these two circles sit inside the rectangle. The overlap between the circles is $A \cap B$.

The demo lets you adjust three values:

• P(A) — the probability of event $A$ (controls the size of circle A)
• P(B) — the probability of event $B$ (controls the size of circle B)
• P(A∩B) — the probability of both events occurring (controls how much the circles overlap)

The legend shows the probability of each region: A only, B only, A∩B, and neither.

The formulas at the bottom show two useful calculations:

P(A∪B) — the probability that $A$ or $B$ (or both) occurs:

$$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$

We subtract $P(A \cap B)$ because the overlap was counted twice — once in $P(A)$ and once in $P(B)$.

P(A|B) — the conditional probability of $A$ given $B$:

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

Try it: increase P(A∩B) while keeping P(B) the same. Watch how P(A|B) goes up — because a larger share of $B$ falls inside $A$.

The set $B$ becomes a reduced sample space — a smaller sample space that contains only the outcomes where $B$ happened. Conditional probability is just ordinary probability computed inside this reduced sample space.

Special case: set P(A∩B) = P(B). Then P(A|B) = 1 — given that $B$ occurred, $A$ is certain (because $A$ contains all of $B$).

Example: Card Drawing

A deck has 52 cards. What’s the probability of drawing a King given that the card is a face card?

$$P(\text{King} | \text{Face}) = \frac{4/52}{12/52} = \frac{4}{12} = \frac{1}{3}$$

We zoom into the 12 face cards — the reduced sample space — and ask: what fraction are Kings? 4 out of 12.

Uniform Shortcut

When all outcomes are equally likely, conditional probability reduces to counting:

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

No need to compute probabilities first — just count outcomes within the reduced sample space $B$.

Example: Rolling a Die

A fair six-sided die is rolled. What is the probability of getting an even number given that the roll is greater than 3?

• $B$: roll $> 3$ → outcomes {4, 5, 6} → $|B| = 3$
• $A$: even → outcomes {2, 4, 6}
• $A \cap B$: even and $> 3$ → {4, 6} → $|A \cap B| = 2$

$$P(\text{even} \mid > 3) = \frac{|A \cap B|}{|B|} = \frac{2}{3}$$

Notice: we never calculated $P(A)$ or $P(B)$. We just shrunk the sample space to $B$ = {4, 5, 6} and counted how many of those are even.

Rules Still Work Inside the “Given” World

Once you condition on $B$, you’re working in a smaller sample space — but all the probability rules you already know still apply:

Complement rule: $P(\text{not } A \mid B) = 1 - P(A \mid B)$

Addition rule: If $A_1$ and $A_2$ can’t both happen, then $P(A_1 \text{ or } A_2 \mid B) = P(A_1 \mid B) + P(A_2 \mid B)$

Range: $0 \leq P(A \mid B) \leq 1$

Example

A jar has 10 marbles: 4 red, 3 blue, 3 green. You draw one at random. Given that it’s not green, what’s the probability it’s red or blue?

• $B$: not green → 7 marbles (4 red + 3 blue)
• $P(\text{red} \mid \text{not green}) = \frac{4}{7}$
• $P(\text{blue} \mid \text{not green}) = \frac{3}{7}$
• Red and blue can’t both happen, so: $P(\text{red or blue} \mid \text{not green}) = \frac{4}{7} + \frac{3}{7} = 1$

This makes sense — if it’s not green, it must be red or blue!