# Probability of A Given B: Conditional Probability Formula and Examples

The probability of A given B answers a narrower question than the plain probability of A. It asks how likely A is once you already know that B happened, so the sample space shrinks from everything that could occur to only the outcomes inside B. The formula is $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, and it is valid whenever $P(B) > 0$ [1].

## Quick Answer

- The probability of A given B is written $P(A \mid B)$ and read "the probability of A given B" [2].
- Formula: $P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}$, the probability that both events occur divided by the probability of the condition [1].
- The condition B becomes the new sample space. You count only outcomes inside B [2].
- If A and B are independent, knowing B changes nothing, so $P(A \mid B) = P(A)$ [3].
- Rearranged, the same rule gives the multiplication rule $P(A \cap B) = P(A) \cdot P(B \mid A)$ [4].

## The Formula

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

Each symbol means something specific:

| Symbol | Meaning |
|---|---|
| $A$ | The event you want the probability of |
| $B$ | The event you already know happened, called the condition |
| $A \cap B$ | The intersection, meaning both A and B occur together |
| $P(A \cap B)$ | The joint probability of A and B |
| $P(B)$ | The unconditional probability of B |
| $P(A \mid B)$ | The conditional probability of A given B |

The denominator is the key. Dividing by $P(B)$ rescales the joint probability so that the total probability inside B becomes 1. That is why the condition is described as a restricted or reduced sample space [1].

The formula only works when $P(B) > 0$. If B cannot happen, conditioning on it is undefined [1].

## How to Calculate It Step by Step

1. Define A and B in words before touching any numbers. Ambiguity here causes most errors.
2. Find $P(B)$, the probability of the condition. This is the denominator.
3. Find $P(A \cap B)$, the probability that both events happen together. This is the numerator.
4. Divide the joint probability by the probability of B.
5. Check that the answer lies between 0 and 1. A result outside that range means an arithmetic or setup error.
6. Sanity-check the direction. $P(A \mid B)$ and $P(B \mid A)$ are usually different numbers.

If you have raw counts instead of probabilities, you can skip the probability step entirely. Count the cases where both A and B occur, then divide by the count of cases where B occurs. The probabilities cancel out, so the ratio is identical.

## Worked Example

A clinic tests 200 patients for a disease. Each patient has a test result and a confirmed disease status, giving four groups.

| Test Result | Disease Status | Count |
|---|---|---|
| Positive | Disease | 90 |
| Positive | No Disease | 20 |
| Negative | Disease | 10 |
| Negative | No Disease | 80 |

The four counts are 90 true positives, 20 false positives, 10 false negatives, and 80 true negatives, for 200 patients total.

Let A be "has the disease" and B be "tests positive." We want the probability of A given B, which is the chance a patient actually has the disease given a positive test.

**Step 1. Total patients.** $N = 200$.

**Step 2. Probability of disease given a positive test.** Restrict to the 110 patients who tested positive. Of those, 90 have the disease.

$$P(\text{Disease} \mid \text{Positive}) = \frac{90}{90 + 20} = \frac{90}{110} = 0.8182$$

**Step 3. Probability of a positive test given disease.** Now restrict to the 100 patients who have the disease. Of those, 90 tested positive.

$$P(\text{Positive} \mid \text{Disease}) = \frac{90}{90 + 10} = \frac{90}{100} = 0.9000$$

**Step 4. Unconditional probabilities.** $P(\text{Disease}) = \dfrac{90 + 10}{200} = \dfrac{100}{200} = 0.5000$ and $P(\text{Positive}) = \dfrac{90 + 20}{200} = \dfrac{110}{200} = 0.5500$.

**Step 5. Joint probability.** $P(\text{Disease and Positive}) = \dfrac{90}{200} = 0.4500$.

**Step 6. Apply the formula.** $\dfrac{0.4500}{0.5500} = 0.8182$, matching Step 2.

**Step 7. Bayes check.** The same answer comes from the other direction: $\dfrac{P(\text{Positive} \mid \text{Disease}) \cdot P(\text{Disease})}{P(\text{Positive})} = \dfrac{0.9000 \times 0.5000}{0.5500} = 0.8182$.

Notice that $P(\text{Disease} \mid \text{Positive}) = 0.8182$ and $P(\text{Positive} \mid \text{Disease}) = 0.9000$ are different numbers. The test is more sensitive than the result is conclusive, because 20 healthy patients still tested positive.

## How to Interpret the Result

A value of 0.8182 means that among patients who test positive, about 81.8% actually have the disease. It does not mean 81.8% of all patients are sick, and it does not mean the test is 81.8% accurate. Those are different quantities.

The condition defines the group you are talking about. Every conditional probability is a statement about a subgroup, so always name the subgroup when you report it. "82% of positive-test patients have the disease" is a clear claim. "82% have the disease" is a different and false claim about this dataset.

Conditional probability is also the foundation of Bayes' theorem, which flips the condition to go from $P(B \mid A)$ to $P(A \mid B)$ [3]. That reversal is exactly what Step 7 did.

## Doing It in Software

With counts, the calculation is a single division. This Python snippet uses the four counts from the table.

```python
TP, FP, FN, TN = 90, 20, 10, 80
p_disease_given_pos = TP / (TP + FP)
p_pos_given_disease = TP / (TP + FN)
print(f"P(Disease | Positive) = {p_disease_given_pos:.4f}")
print(f"P(Positive | Disease) = {p_pos_given_disease:.4f}")
```

Output:

```text
P(Disease | Positive) = 0.8182
P(Positive | Disease) = 0.9000
```

In Excel, if the true positive count sits in cell A1 and the false positive count in A2, the formula `=A1/(A1+A2)` returns 0.8182. In R, the same division works on numeric vectors, and `prop.test` handles the interval estimation when you need a confidence interval around the proportion.

For a quick check on any conditional probability problem, the [Probability Calculator](/tools/probability-calculator) lets you enter the joint and marginal probabilities and returns the conditional value directly.

## Common Mistakes

- **Flipping the condition.** $P(A \mid B)$ and $P(B \mid A)$ are different quantities. In the example they are 0.8182 and 0.9000. Fix: write the condition in words before substituting numbers.
- **Dividing by the wrong total.** Using the full sample size instead of the count inside B gives 90/200 = 0.45, which is the joint probability, not the conditional one. Fix: the denominator is always the size of the condition group.
- **Assuming independence without checking.** If A and B are independent, $P(A \mid B) = P(A)$ [3]. That is a special case, not a default. Fix: verify independence or use the full formula.
- **Confusing the joint with the conditional.** $P(A \cap B)$ is "both happen." $P(A \mid B)$ is "A happens inside the world where B happened." Fix: check whether your denominator is 1 or $P(B)$.
- **Conditioning on a zero-probability event.** If $P(B) = 0$, the formula is undefined [1]. Fix: confirm the condition can actually occur before computing.
- **Ignoring base rates.** A high $P(\text{Positive} \mid \text{Disease})$ does not guarantee a high $P(\text{Disease} \mid \text{Positive})$ when the disease is rare. Fix: always bring in the unconditional probability of the condition.

## Limitations

Conditional probability describes association inside a defined group. It does not establish causation, and it cannot correct for a biased sample. If the 200 patients were not representative of the population you care about, the 0.8182 figure applies only to a group like this one.

The formula also depends on knowing $P(B)$ accurately. When the condition is rare, small errors in that estimate swing the conditional probability a lot. And when events are not independent, you cannot multiply their plain probabilities to get the joint probability. You need the conditional form of the multiplication rule instead [4].

## Frequently Asked Questions

### What is the difference between P(A|B) and P(B|A)?

They condition on different events and usually give different answers. $P(A \mid B)$ restricts the sample space to B, while $P(B \mid A)$ restricts it to A. In the worked example, $P(\text{Disease} \mid \text{Positive}) = 0.8182$ but $P(\text{Positive} \mid \text{Disease}) = 0.9000$. Bayes' theorem is the tool that converts one into the other [3].

### What does "probability of A given B" mean in plain words?

It means the chance that A happens, calculated only among the cases where B already happened. The condition removes every outcome outside B from consideration. If you roll a die and learn the result is even, the probability of a 6 becomes 1/3 instead of 1/6, because only three outcomes remain [2].

### Can conditional probability be greater than the unconditional probability?

Yes. Conditioning can raise or lower a probability. If B is positively associated with A, then $P(A \mid B)$ exceeds $P(A)$. If the two are independent, they are equal [3]. The only hard rule is that the result stays between 0 and 1.

### What happens if P(B) equals zero?

The conditional probability is undefined, because the formula divides by zero [1]. Conditioning on an impossible event has no meaning. In practice, check that the condition has a nonzero probability before you compute anything.

### How is conditional probability used in real analysis?

It appears wherever you filter data by a known characteristic. Medical testing, credit scoring, spam filtering, and survey subgroup analysis all rely on it. The multiplication rule $P(A \cap B) = P(A) \cdot P(B \mid A)$ extends the idea to chains of dependent events, such as drawing cards without replacement [4]. Related distributions such as the [binomial distribution](/blog/data-analysis/binomial-distribution-formula-examples) build directly on repeated independent trials, and [Bayes' theorem](/blog/data-analysis/bayes-theorem-definition-formula-examples) handles the reversal of the condition.

## References

1. [Conditional probability - Wikipedia](https://en.wikipedia.org/wiki/Conditional_probability)
2. [Conditional Probability](https://www.usu.edu/math/schneit/StatsStuff/Probability/probability5)
3. [Conditional Probability](http://www.stat.yale.edu/Courses/1997-98/101/condprob.htm)
4. [3.6: Conditional Probability - Statistics LibreTexts](https://stats.libretexts.org/Workbench/Statistics_for_Behavioral_Science_Majors/03%3A_Probability/3.06%3A_Conditional_Probability)

## Further Reading

- [NIST/SEMATECH e-Handbook of Statistical Methods](https://www.itl.nist.gov/div898/handbook/index.htm)
- [OpenStax. Introductory Statistics 2e](https://openstax.org/details/books/introductory-statistics-2e)

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