Conditional Probability Explained: P(A | B) with Real-World Examples

What is conditional probability?

Conditional probability is the probability that one event happens given that another event has already happened. We write it $P(A \mid B)$, read "the probability of $A$ given $B$."

The formula is

where $P(A \cap B)$ is the probability that both $A$ and $B$ happen. Knowing that $B$ already occurred shrinks the sample space — instead of looking at all outcomes, we only look at outcomes inside $B$, and ask what fraction of those also lie inside $A$.

A first example. A standard deck has 52 cards. Let $A$ = the card is a Queen and $B$ = the card is a face card (Jack, Queen, or King — 12 cards total). Then

Among face cards, exactly one third are Queens — exactly what intuition says.

The multiplication rule

Rearranging the formula above gives the multiplication rule for the joint probability:

This is how you combine probabilities when events happen in sequence.

Example. A bag has 5 red and 3 blue marbles. You draw two marbles without replacement. What is the probability both are red?

• $P(\text{first red}) = \dfrac{5}{8}$.
• After taking a red, 4 reds remain out of 7 total, so $P(\text{second red} \mid \text{first red}) = \dfrac{4}{7}$.
• $P(\text{both red}) = \dfrac{5}{8} \cdot \dfrac{4}{7} = \dfrac{20}{56} = \dfrac{5}{14}$.

The conditional probability captures the fact that the first draw changes what is left in the bag.

Independent vs. dependent events

Two events are independent when knowing one happened tells you nothing about the other. Formally, $A$ and $B$ are independent when

Either condition implies the other, so you can use whichever is easier to check.

Example. Roll a fair six-sided die twice. Let $A$ = the first roll is a 4 and $B$ = the second roll is even. The two rolls do not affect each other, so

If two events are not independent, they are dependent. The marble example above is a classic dependent case: the second draw depends on what you already removed from the bag.

Bayes' theorem

Sometimes you know $P(B \mid A)$ but want $P(A \mid B)$. Bayes' theorem flips a conditional:

Here $P(A)$ is the prior (what you believed about $A$ before seeing $B$), $P(B \mid A)$ is the likelihood of the evidence under $A$, and $P(A \mid B)$ is the posterior (your updated belief).

Example: medical testing. A disease occurs in $2\%$ of a population. A test is $95\%$ sensitive (it gives a positive result $95\%$ of the time when the disease is present) and has a $5\%$ false-positive rate (it gives a positive result $5\%$ of the time when the disease is absent). If a random person tests positive, what is the probability they actually have the disease?

Let $D$ = "has disease" and $+$ = "tests positive." We want $P(D \mid +)$.

1. Total probability of a positive test: $P(+) = P(+ \mid D) P(D) + P(+ \mid \neg D) P(\neg D) = (0.95)(0.02) + (0.05)(0.98) = 0.019 + 0.049 = 0.068$.
2. Apply Bayes' theorem: $P(D \mid +) = \dfrac{P(+ \mid D) P(D)}{P(+)} = \dfrac{0.019}{0.068} \approx 0.279$.

So even after a positive result, there is only about a $28\%$ chance the person has the disease. The takeaway: when a condition is rare, even an accurate test produces many false positives compared to true positives. This is the kind of result Bayes' theorem makes precise.

A quick recipe

When a problem gives you conditional information, work through it in this order:

1. Identify the events. Name them $A$, $B$, $D$, $+$, etc., so the formulas don't get muddled.
2. Write down what you know in symbols — $P(A)$, $P(B \mid A)$, etc.
3. Decide which formula you need: definition of $P(A \mid B)$, multiplication rule, or Bayes' theorem.
4. Plug in and simplify. Always sanity-check that probabilities are between $0$ and $1$.

Common mistakes

• Confusing $P(A \mid B)$ with $P(B \mid A)$. They are different in general — Bayes' theorem is exactly the bridge between them.
• Assuming independence. "With replacement" or "the two trials don't influence each other" usually means independent. "Without replacement" or sequential physical processes usually do not.
• Forgetting the denominator $P(B) > 0$. You can only condition on events that actually have positive probability.
• Adding instead of multiplying. Use addition for "$A$ or $B$"; use multiplication (with a conditional) for "$A$ and $B$ in sequence."