Conditional Probability Calculator

Conditional probability answers the question: how likely is A once you know that B has happened? It is written P(A | B) and read as the probability of A given B. Learning that B happened rules out every outcome outside B, so the sample space shrinks to B, and P(A | B) is the share of B in which A also occurs. Take a card from a standard deck: the probability of a king is 4 / 52, but the probability of a king given that the card is a face card is 4 / 12, because only the 12 face cards are still possible.

The condition can make an event more likely, less likely or leave it unchanged. When P(A | B) equals P(A), knowing B tells you nothing about A and the two events are independent.

The condition goes after the bar and supplies the denominator. P(A | B) and P(B | A) have the same numerator, P(A and B), but different denominators, so they differ unless P(A) = P(B) or P(A and B) = 0. The rule P(not A | B) = 1 − P(A | B) holds because both probabilities use the same condition. The condition cannot be swapped: P(A | not B) is not 1 − P(A | B). In the first example below P(A | B) = 0.6 and P(A | not B) = 0.2.

Suppose P(A and B) = 0.3, P(B) = 0.5 and P(A) = 0.4. Load example fills in these values.

1. P(A | B) = 0.3 / 0.5 = 0.6, and P(not A | B) = (0.5 − 0.3) / 0.5 = 0.4.
2. P(B | A) = 0.3 / 0.4 = 0.75. The two conditional probabilities differ because A is less common than B.
3. P(A | not B) = (0.4 − 0.3) / (1 − 0.5) = 0.1 / 0.5 = 0.2.
4. P(A or B) = 0.4 + 0.5 − 0.3 = 0.6.
5. Independence: P(A) × P(B) = 0.4 × 0.5 = 0.2, which is not the observed P(A and B) = 0.3, so the events are not independent.

Knowing that B happened raises the probability of A from 0.4 to 0.6. The joint table has 0.3 for A and B, 0.1 for A and not B, 0.2 for not A and B and 0.4 for neither, which adds up to 1.

Of 100 students, A means the student passed the exam and B means the student attended the review session. Forty students attended and passed, 30 passed without attending, 10 attended and failed and 20 did neither. Choose Two-way table of counts and enter 40, 30, 10, 20 (or use Load example).

1. Totals: 70 passed, 50 attended, so P(A) = 0.7, P(B) = 0.5 and P(A and B) = 40 / 100 = 0.4.
2. P(A | B) = 40 / 50 = 0.8: among the 50 who attended, 80% passed.
3. P(A | not B) = 30 / 50 = 0.6: among the 50 who did not attend, 60% passed.
4. P(B | A) = 40 / 70 = 0.5714: among the 70 who passed, 57.14% attended. This is a different question from P(A | B).
5. P(B | not A) = 10 / 30 = 0.3333: among the 30 who failed, a third attended.
6. Independence: 40 × 20 = 800 but 30 × 10 = 300. The cross-products differ, so passing and attending are not independent in this group.

Two events are independent when knowing one does not change the probability of the other: P(A | B) = P(A), or equivalently P(A and B) = P(A) · P(B). With P(A) = 0.5, P(B) = 0.4 and P(A and B) = 0.2 the calculator finds P(A | B) = 0.2 / 0.4 = 0.5 = P(A) and P(A) × P(B) = 0.2 = P(A and B), so the events are independent. Then P(A | not B) = 0.3 / 0.6 = 0.5 is also equal to P(A).

Independent is not the same as mutually exclusive. Events that cannot happen together have P(A and B) = 0, so if both have a positive probability they are dependent: once B has happened, A is impossible.

The test is exact. The probabilities are read as exact decimals, so 0.1 × 0.7 equals 0.07 here, although most software using floating-point numbers gets 0.06999999999999999 and would call the events dependent. Counts from a real sample are rarely exactly independent, though. A difference between the cross-products can arise by chance, and the chi-square test of independence decides whether it is larger than chance would explain.

• Swapping the condition. P(A | B) and P(B | A) are different questions. The share of people with a disease who test positive is not the share of positive tests that belong to people with the disease; the second depends on how rare the disease is. The Bayes' theorem calculator converts one into the other, and the article Bayes' theorem explained shows why.
• Dividing by the wrong total. P(A | B) is divided by P(B), the condition. Dividing P(A and B) by P(A) gives P(B | A).
• Treating not B as the complement of the whole conditional. P(not A | B) = 1 − P(A | B), but P(A | not B) is a separate quantity with a different condition.
• Entering percentages. Enter 40% as 0.4. Values above 1 are rejected.
• Conditioning on an impossible event. When P(B) = 0 there is no probability of A given B, and the calculator says so instead of returning a number.

With the condition in the rows, normalising over rows gives the probability of the column variable given the row value. Normalising over columns instead gives the reverse conditional probability P(B | A).