Conditional Probability on Positive-Probability Events

Conditioning as a Change of Reference Event

A conditional probability answers a narrower question than an unconditional probability. Instead of asking how likely an event \(A\) is across all possible outcomes, we ask how likely \(A\) is among outcomes known to lie in another event \(B\). The calculation keeps only the part of \(A\) inside \(B\), then measures that part relative to the probability of \(B\).

This interpretation requires \(B\) to have positive probability. If \(P(B)=0\), the normalization below would divide by zero, so this elementary definition does not assign a conditional probability \(P(A\mid B)\). Conditioning on values or events of probability zero requires other ideas, treated separately.

Definition: Let \(A\) and \(B\) be events in a probability space, with \(P(B)>0\). The conditional probability of \(A\) given \(B\) is

P(A\mid B)=\frac{P(A\cap B)}{P(B)}.

The numerator is the probability of outcomes satisfying both events; the denominator rescales that probability relative to the event \(B\).

The definition has a direct counting interpretation when all outcomes are equally likely: restrict attention to the outcomes in \(B\), and count what fraction of them also belong to \(A\). The probability-space definition works more generally, whether outcomes are equally likely or not.

Why the Normalization Gives a Probability Measure

Fix an event \(B\) with \(P(B)>0\). For each event \(A\), define \(Q(A)=P(A\mid B)\). This new function \(Q\) is itself a probability measure on the original event space. In particular, it assigns total probability one to the whole sample space, and it is countably additive. This explains precisely why conditioning can be understood as changing the reference event to \(B\).

Theorem: If \(P(B)>0\), then \(Q(A)=P(A\cap B)/P(B)\), defined for every event \(A\), satisfies the probability axioms.

Proof. Since \(A\cap B\) is an event, \(Q(A)\geq 0\). Also,

Q(\Omega)=\frac{P(\Omega\cap B)}{P(B)} =\frac{P(B)}{P(B)}=1.

Now let \(A_1,A_2,\ldots\) be pairwise disjoint events. The events \(A_i\cap B\) are also pairwise disjoint: if an outcome belonged to both \(A_i\cap B\) and \(A_j\cap B\), it would belong to both \(A_i\) and \(A_j\), contradicting disjointness when \(i\ne j\). Further,

\left(\bigcup_{i=1}^{\infty} A_i\right)\cap B =\bigcup_{i=1}^{\infty}(A_i\cap B).

Countable additivity of \(P\), followed by division by the fixed positive number \(P(B)\), gives

Q\left(\bigcup_{i=1}^{\infty} A_i\right) =\frac{P\left(\bigcup_{i=1}^{\infty}(A_i\cap B)\right)}{P(B)} =\sum_{i=1}^{\infty}\frac{P(A_i\cap B)}{P(B)} =\sum_{i=1}^{\infty}Q(A_i).

Thus \(Q\) is nonnegative, has total mass one, and is countably additive, as required. \(\square\)

Worked Example: Conditioning Two Dice on the First Die

Roll two fair six-sided dice. Let \(A\) be the event that their sum is even, and let \(B\) be the event that the first die shows more than 3. Find \(P(A\mid B)\).

There are 36 equally likely ordered outcomes in all. The event \(B\) consists of the 18 outcomes whose first coordinate is 4, 5, or 6, so \(P(B)=18/36=1/2\).

For first die 4, the sum is even when the second die is 2, 4, or 6: three outcomes. For first die 5, it is even when the second die is 1, 3, or 5: three outcomes. For first die 6, it is even for second die 2, 4, or 6: three outcomes. Thus \(A\cap B\) has 9 outcomes, and \(P(A\cap B)=9/36=1/4\). Therefore

P(A\mid B)=\frac{P(A\cap B)}{P(B)} =\frac{1/4}{1/2}=\frac12.

Equivalently, among the 18 outcomes compatible with \(B\), 9 give an even sum. Both calculations use the same restriction to \(B\).

The Multiplication Rule

Rearranging the definition gives a useful way to calculate the probability that two events occur together. The condition \(P(B)>0\) is essential for defining \(P(A\mid B)\); once it is defined, the identity follows by multiplying both sides of the definition by \(P(B)\).

Theorem: For events \(A\) and \(B\) with \(P(B)>0\),

P(A\cap B)=P(A\mid B)P(B).

If \(P(B)=0\), then \(P(A\cap B)=0\), but \(P(A\mid B)\) is not defined by the positive-probability definition.

The rule can also be applied in the reverse order when \(P(A)>0\), yielding \(P(A\cap B)=P(B\mid A)P(A)\). These are two expressions for the same intersection; neither says that the conditional probability equals the unconditional probability. That equality holds when the relevant events are independent, not in general.

Worked Example: Drawing Two Colors Without Replacement

An urn contains 4 red balls and 3 blue balls. Two balls are drawn one after the other without replacement. Let \(A\) be the event that the first ball is red, and \(B\) the event that the second ball is blue. Find \(P(A\cap B)\) using the multiplication rule.

The first draw is red with probability \(P(A)=4/7\). Given that the first draw was red, 6 balls remain, including all 3 blue balls. Hence \(P(B\mid A)=3/6=1/2\). The multiplication rule, applied with \(A\) as the conditioning event, gives

P(A\cap B)=P(B\mid A)P(A) =\frac12\cdot\frac47=\frac27.

A direct check gives the same result: there are \(4\cdot3=12\) ordered red-then-blue pairs among \(7\cdot6=42\) possible ordered draws, so \(12/42=2/7\). The second-draw probability is conditional on the first draw; using \(3/7\) for it after observing a red first draw would ignore the change in the urn.

Using a Partition of the Sample Space

A partition separates all outcomes into mutually exclusive cases. If an event can occur through several disjoint cases, its probability is the sum of its probabilities within those cases. Conditional probabilities make the contribution of each case explicit.

Theorem: Let \(E_1,\ldots,E_k\) be a finite partition of the sample space: the events are pairwise disjoint and their union is \(\Omega\). For any event \(A\),

P(A)=\sum_{\{i:\,P(E_i)>0\}}P(A\mid E_i)P(E_i) +\sum_{\{i:\,P(E_i)=0\}}P(A\cap E_i).

In particular, when every partition event has positive probability,

P(A)=\sum_{i=1}^{k}P(A\mid E_i)P(E_i).

Proof. Since the \(E_i\) partition \(\Omega\), intersecting \(A\) with their union gives

A=A\cap\Omega=A\cap\left(\bigcup_{i=1}^{k}E_i\right) =\bigcup_{i=1}^{k}(A\cap E_i).

The events \(A\cap E_i\) are pairwise disjoint because the \(E_i\) are pairwise disjoint. Finite additivity therefore gives \(P(A)=\sum_{i=1}^{k}P(A\cap E_i)\). If \(P(E_i)>0\), the definition of conditional probability gives \(P(A\cap E_i)=P(A\mid E_i)P(E_i)\). If \(P(E_i)=0\), then \(0\leq P(A\cap E_i)\leq P(E_i)=0\), so that intersection has probability zero. Separating the positive- and zero-probability terms proves the stated formula. \(\square\)

Worked Example: Defect Probability Across Two Machines

A factory's output comes from two machines. Machine 1 produces 60% of the items, and 2% of its items are defective. Machine 2 produces the remaining 40%, and 5% of its items are defective. Find the probability that a randomly selected item is defective.

Let \(E_1\) and \(E_2\) denote production by machines 1 and 2, and let \(D\) denote a defective item. The machine events form a partition, and both have positive probability. The given values are \(P(E_1)=0.60\), \(P(E_2)=0.40\), \(P(D\mid E_1)=0.02\), and \(P(D\mid E_2)=0.05\). Applying the partition formula,

\begin{aligned} P(D) &=P(D\mid E_1)P(E_1)+P(D\mid E_2)P(E_2)\\ &=(0.02)(0.60)+(0.05)(0.40)\\ &=0.012+0.020\\ &=0.032. \end{aligned}

Thus 3.2% of all items are defective. The answer is a weighted sum: the two machine-specific defect rates are weighted by their shares of production, not simply averaged as \((0.02+0.05)/2=0.035\). An unweighted average would incorrectly give the machines equal production shares.

Complements Under a Fixed Condition

Once \(B\) is fixed as the conditioning event, \(A\) and its complement divide the outcomes in \(B\) into disjoint pieces. This gives a complement rule for conditional probabilities. It is often simpler to calculate the probability of an event by first finding the probability of its complement.

Theorem: If \(P(B)>0\), then

P(A^c\mid B)=1-P(A\mid B).

Proof. Within \(B\), the events \(A\cap B\) and \(A^c\cap B\) are disjoint and their union is \(B\). Therefore \(P(B)=P(A\cap B)+P(A^c\cap B)\). Divide by \(P(B)>0\) and use the definition on each term:

1=\frac{P(A\cap B)}{P(B)}+\frac{P(A^c\cap B)}{P(B)} =P(A\mid B)+P(A^c\mid B).

Rearranging proves the identity. \(\square\)

Worked Example: An Even Result Given That a Die Is Not 1

Roll a fair six-sided die. Let \(B\) be the event that the result is not 1, and \(A\) the event that the result is even. Find \(P(A\mid B)\), then find the conditional probability of an odd result.

The event \(B=\{2,3,4,5,6\}\) has probability \(5/6\), and \(A\cap B=\{2,4,6\}\) has probability \(3/6\). Therefore

P(A\mid B)=\frac{3/6}{5/6}=\frac35.

An odd result is the complement of \(A\), so the complement rule gives \(P(A^c\mid B)=1-3/5=2/5\). Directly, the odd outcomes in \(B\) are \(\{3,5\}\), two of the five outcomes. This verifies that \(3/5+2/5=1\).

Common Mistakes and Practical Checks

Conditional probability is straightforward once the reference event is kept visible throughout a calculation. Several errors arise from changing that reference event or treating an assumption as a consequence of the definition.

Using the wrong denominator. In \(P(A\mid B)\), the denominator is \(P(B)\), not \(P(A)\). The numerator is the overlap \(P(A\cap B)\).
Dropping the conditioning information. In draws without replacement, the composition of the remaining items changes after a draw. Use the probability appropriate to what is already known.
Assuming independence. In general, \(P(A\mid B)\ne P(A)\). Equality follows when \(A\) and \(B\) are independent and \(P(B)>0\); it is not part of the definition.
Conditioning on a zero-probability event. The formula requires \(P(B)>0\). If \(P(B)=0\), the ratio is undefined, even though \(P(A\cap B)=0\).
Averaging partition-specific rates without weights. The contribution from case \(E_i\) is \(P(A\mid E_i)P(E_i)\), so cases with different probabilities do not receive equal weights.

A reliable procedure is to name the target event and the conditioning event, identify their intersection, and check that the conditioning event has positive probability. For a partition calculation, verify that the cases are disjoint, cover the sample space, and have probabilities summing to one. These checks catch many errors before arithmetic begins.

Check Your Understanding

Use the definition and results above to answer these questions.

If \(P(A\cap B)=0.12\) and \(P(B)=0.30\), calculate \(P(A\mid B)\). What is the corresponding intersection probability according to the multiplication rule?
Why does the positive-probability definition not give a value for \(P(A\mid B)\) when \(P(B)=0\)?
A partition has two cases with probabilities 0.7 and 0.3. The probabilities of \(A\) within the cases are 0.1 and 0.4. Calculate \(P(A)\).
If \(P(A\mid B)=0.65\) and \(P(B)>0\), what is \(P(A^c\mid B)\)? State the result used.
In the two-machine example, explain why the two conditional defect rates should not be averaged with equal weights.