Conditional Probability Formula with Examples

Key Points

• The conditional probability formula is $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$, valid whenever $P(A) > 0$.

• Conditioning on $A$ shrinks the sample space to $A$ alone, which is why the formula divides by $P(A)$.

• $P(B \mid A)$ and $P(A \mid B)$ are different quantities, and swapping them is the most consequential error in probability.

• Bayes' theorem links the two directions as $P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}$.

• When the events are independent, $P(B \mid A) = P(B)$, so conditioning changes nothing.

What Is the Conditional Probability Formula?

The conditional probability formula gives the chance of one event once another is known to have happened:

$$P(B \mid A) = \frac{P(A \cap B)}{P(A)}, \qquad P(A) > 0$$

It is the multiplication rule rearranged. MacTutor records Thomas Bayes, who lived from 1702 to 1761, as the English clergyman who set out his theory of probability in 1764, and notes his work as the first occurrence of a probability logic result involving conditional probability.

Why the Formula Divides by P(A)

Knowing that $A$ happened throws away every outcome outside $A$. The question is no longer about the whole sample space but about the part of it where $A$ is true.

Within that reduced space, the outcomes where $B$ also happens are exactly those in $A \cap B$. So the proportion wanted is the size of $A \cap B$ relative to the size of $A$, which is what the division computes.

That reading also explains the condition $P(A) > 0$. Conditioning on an impossible event leaves no sample space at all, so the formula has nothing to divide by.

Variable Key

Examples of the Conditional Probability Formula

Example 1

$P(A) = 0.4$ and $P(A \cap B) = 0.1$. Find $P(B \mid A)$.

$$P(B \mid A) = \frac{0.1}{0.4} = 0.25$$

Final answer: 0.25

Example 2

In a class, 60 per cent play football and 25 per cent play both football and tennis. A tennis player is chosen. What is the probability they play football, if 40 per cent play tennis?

The figures 0.25 and 0.60 are both to hand, so dividing the overlap by the football share looks right:

$$\frac{0.25}{0.60} \approx 0.417$$

But that answers a different question: the chance a footballer plays tennis. The person chosen is known to play tennis, so tennis is the condition and its probability must be the denominator.

$$P(\text{football} \mid \text{tennis}) = \frac{0.25}{0.40} = 0.625$$

Final answer: 0.625. The condition always goes in the denominator, and the two answers differ by half as much again.

Example 3

A die shows an even number. What is the probability it is a 6?

The even outcomes are 2, 4 and 6, so the reduced sample space has three members:

$$P(6 \mid \text{even}) = \frac{1/6}{1/2} = \frac{1}{3}$$

Final answer: $\frac{1}{3}$

Conditioning on an even result raised the chance of a six from one sixth to one third.

Example 4

$P(A) = 0.5$, $P(B) = 0.3$ and $P(A \cap B) = 0.15$. Find $P(B \mid A)$ and comment.

$$P(B \mid A) = \frac{0.15}{0.5} = 0.3$$

Final answer: 0.3, which equals $P(B)$, so the events are independent and conditioning on $A$ tells us nothing about $B$.

Example 5

A test detects a condition in 90 per cent of those who have it. The condition affects 1 per cent of people, and the test is positive for 10 per cent of everyone tested. Find the probability someone testing positive has the condition.

Use Bayes' theorem to reverse the given direction:

$$P(\text{condition} \mid +) = \frac{0.90 \times 0.01}{0.10} = \frac{0.009}{0.10} = 0.09$$

Final answer: 0.09, or 9 per cent.

A test that catches 90 per cent of cases still leaves most positives false, because the condition is rare. Ignoring that base rate is the classic error in medical-test reasoning.

Four Related Quantities

The first two rows share a numerator and differ only in the denominator. That is the whole of the reversal error: the same overlap divided by two different conditions.

Conditional Probability, Common Confusions Cleared Up

Mistake 1: Reversing the condition

Where it slips in: Both conditional probabilities use the same intersection, so the denominator is the only thing distinguishing them.

Don't do this: Compute $\frac{P(A \cap B)}{P(A)}$ when the question says $B$ is known.

The correct way: Identify what has already happened and put its probability underneath. Reading the question for the words "given", "if" or "known to be" locates the condition before any arithmetic.

Students who underline the condition in the question and write its probability as the denominator first, before filling in the numerator, almost never reverse it. Building the fraction from the bottom up forces the condition to be identified.

Mistake 2: Ignoring the base rate

Where it slips in: A high detection rate feels like it should mean a high probability of having the condition.

Don't do this: Treat a 90 per cent accurate test as giving a 90 per cent chance the condition is present.

The correct way: Use Bayes' theorem, which weights by how common the condition is. For a rare condition, most positive results are false, as Example 5 shows.

Mistake 3: Conditioning on an impossible event

Where it slips in: The formula is applied mechanically without checking the denominator.

Don't do this: Divide by $P(A)$ when $P(A) = 0$.

The correct way: The formula requires $P(A) > 0$. Conditioning on something that cannot happen leaves no sample space, so the quantity is undefined rather than zero.

The Short Version

• The conditional probability formula is $P(B \mid A) = \frac{P(A \cap B)}{P(A)}$.

• Conditioning shrinks the sample space to $A$, which is why $P(A)$ is the denominator.

• $P(B \mid A)$ and $P(A \mid B)$ differ only in that denominator, and confusing them is the costliest error here.

• Bayes' theorem reverses the direction and forces the base rate into the calculation.

• To work through probability with a teacher, explore a high school math tutor or math tuition.

A Practical Next Step

Given $P(A) = 0.6$, $P(B) = 0.5$ and $P(A \cap B) = 0.3$, find both conditional probabilities and check whether the events are independent. Then reverse one of them with Bayes' theorem.

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• What Is Probability

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