Conditional probability and independence

Content updated: October 22, 2026

Learning that an event has occurred changes the space under consideration. Conditional probability measures the intersection as a proportion of the conditioning event, rather than of the original whole space.

Learning goals: calculate conditional probabilities; test independence by a probability identity; describe two different drawing procedures precisely.

Prerequisites: intersections of events, classical probability, and combinations.

On this page: Definition · Method · Examples · Practice · Further learning

1. Definition when the conditioning event has positive probability

For $P(B)>0$, the probability of $A$ given that $B$ occurs is defined by

This notation corresponds to $P(A/B)$ in some older lessons. The condition follows the vertical bar. In a finite uniform model, $P(A\mid B)=|A\cap B|/|B|$ because the equal probability weights cancel. Do not automatically use uniform counting if the model outcomes are not equally likely.

If $P(B)=0$, the ratio is undefined; do not assign it the value 0. Conditional probability is generally not symmetric, since the denominator of $P(A\mid B)$ differs from that of $P(B\mid A)$.

2. Independence and the new space

Two events are independent if $P(A\cap B)=P(A)P(B)$. This symmetric definition also applies when a probability is zero. When $P(B)>0$, it is equivalent to $P(A\mid B)=P(A)$; if also $P(A)>0$, it is equivalent to $P(B\mid A)=P(B)$ as well.

To calculate a conditional probability, specify the information event, check that its probability is positive, and identify the possible outcomes or box contents under that condition. Mutually exclusive events are not independent if both have positive probability: their intersection then has probability zero but their product is positive.

For draws from two separate boxes, the model must specify that the selections are independent. If balls are transferred between boxes before a later draw, the later box contents depend on the transferred colors; use the contents appropriate to the condition.

3. Four worked examples

Example 1. A die given an even result

Roll a fair die. Let $M=\{2,4,6\}$; let $A=\{1,2,3\}$ mean fewer than 4 dots, $B=\{3,4,5,6\}$ more than 2 dots, and $C=\{1\}$ fewer than 2 dots. Given $M$, the new space has 3 equally likely outcomes. The intersections are $\{2\}$, $\{4,6\}$, and the empty set, respectively, so

Example 2. Which event is independent of an even result?

Since $P(B)=4/6=2/3=P(B\mid M)$, the events $B,M$ are independent. By contrast, $P(A)=1/2\ne1/3$ and $P(C)=1/6\ne0$, so $A,M$ are not independent, nor are $C,M$. The fact that $B$ contains both even and odd faces does not prove independence; the probability identity is the test.

Example 3. Independent selections from two boxes

Box I has 6 red and 4 white balls; box II has 7 red and 3 white balls. Choose 3 uniformly without replacement from each box, with the two selections independent. Let $A$ mean 2 red and 1 white from box I, and $B$ mean 1 red and 2 white from box II. The events concern independent selections, so

Both marginal probabilities are positive, so both conditional probabilities are defined. Their different values show that independence does not require the reverse conditional probabilities to agree.

Example 4. Transferring balls from box I to box II

Keep the same initial box contents. This time choose 3 uniformly from box I and transfer them to box II, then choose 3 uniformly from box II after the transfer, sampling without replacement within each selection. Let $A$ mean transferring 2 red and 1 white, and $B$ mean obtaining 1 red and 2 white in the final selection. Find $P(B\mid A)$.

When $A$ occurs, box II contains 13 balls: 9 red and 4 white. Its groups of 3 are equally likely, so

The denominator 13 concerns the box after transfer, not the initial 10. To check that this differs from the probability without knowing $A$, partition by the number of transferred red balls, $K=0,1,2,3$. Their probabilities are $1/30,3/10,1/2,1/6$, respectively, and the corresponding final-selection probabilities are $105/286,80/286,54/286,30/286$. The weighted sum is $P(B)=119/572$, different from $27/143=108/572$; thus $A,B$ are not independent in this experiment.

4. Practice

1. For a fair die, calculate $P(\text{even}\mid\text{number of dots}>3)$.
2. Two independent events satisfy $P(A)=1/2$ and $P(B)=1/3$. Calculate $P(A\cap B)$ and $P(A\mid B)$.
3. May the ratio $P(A\cap B)/P(B)$ be used when $B=\varnothing$?

5. Common mistakes

Do not reverse the conditioning event and the event whose probability is sought. Do not identify mutual exclusivity with independence. Do not retain the old box contents after a transfer. A conditional-probability ratio with denominator zero is not a valid number under this definition.

6. Further learning

Random experiments and events (English lesson) · Probability of an event (English lesson) · The addition rule of probability (English lesson) · The multiplication rule of probability (English lesson).