Independent and dependent events

What it is

Two events A and B are independent if knowing that A has happened does not change the probability of B. If it does change it, the events are dependent. The test is an equality:

P(A and B) = P(A) · P(B)

If it holds, the events are independent; otherwise they are dependent. When P(A) > 0 this is the same as saying that the conditional probability of B given A equals the probability of B: P(B|A) = P(B).

Independent events:

• Two coin tosses. Heads on the first does not change the second. P(heads both times) = 1/2 · 1/2 = 1/4.
• Two dice. A six on both has probability 1/6 · 1/6 = 1/36.
• Drawing with replacement. The ball goes back, so the urn looks the same for the next draw.
• Two events in the same roll. A is “even”, {2, 4, 6}, and B is “at most 4”, {1, 2, 3, 4}. P(A) = 1/2 and P(B) = 2/3. Both happen for {2, 4}, with probability 2/6 = 1/3, and 1/2 · 2/3 = 1/3. The events are independent although they belong to the same roll.

Dependent events:

• Drawing without replacement. The drawn ball is gone, so the urn looks different for the next draw; see the table below.
• The same roll, other events. A is “even” and B is “at least 4”, {4, 5, 6}. Both happen for {4, 6}, with probability 1/3, but 1/2 · 1/2 = 1/4. Someone told that the number is at least 4 knows that the chance of even has risen from 1/2 to 2/3.
• Mutually exclusive events. “One” and “two” in the same roll cannot both happen, so P(A and B) = 0, which is not 1/6 · 1/6. The events are dependent: if one happens, you know the other did not.

If A and B are independent, so are A and “not B”. In the “even” and “at most 4” example, “not B” = {5, 6}. Both happen for {6}, with probability 1/6, and 1/2 · 1/3 = 1/6. The probability of “not B” is 1 − P(B); see the complement rule.

Why it matters

Independence decides which rule applies.

• Multiplying. The multiplication rule P(A and B) = P(A) · P(B) holds for independent events. For dependent events, P(A and B) = P(A) · P(B|A), where the second factor is a conditional probability.
• Tree diagrams. The probability of a path through the tree is the product of its branch probabilities, and the branches at each step are given what happened before. With replacement they are the same after every outcome; without replacement they change. The probabilities of several paths are added.

An urn holds 4 red and 6 blue balls, and two balls are drawn:

Without replacement, P(two red) = 4/10 · 3/9 = 2/15, not 4/10 · 4/10. “At least one red” is easiest as 1 minus the probability of two blue: 1 − 6/10 · 5/9 = 2/3 without replacement and 1 − 0.6 · 0.6 = 0.64 with it.

Someone who does not know what the first ball was still gets 4/10 for red on the second draw: 4/10 · 3/9 + 6/10 · 4/9 = 36/90 = 4/10, the sum over the two paths by the law of total probability. The dependence only shows once you know what the first ball was.

• Large populations. Drawing one ball from an urn of millions barely changes the proportions, and drawing without replacement can be treated as drawing with it. That is why the binomial distribution approximates the hypergeometric distribution when the sample is small relative to the population.
• Safety. Two smoke alarms that each work with probability 0.9 both miss a fire with probability 0.1 · 0.1 = 0.01, provided they work independently of each other. If they share a power supply, both fail when the power goes, and the probability that both miss is greater than 0.01.

Common misconceptions

• “After five heads in a row, tails is due.” The tosses are independent, so the probability of heads on the sixth toss is 1/2 whatever happened before. Six heads in a row has probability (1/2)⁶ = 1/64 ≈ 0.016, but that holds before the first toss; after five tosses only one is left. The belief that outcomes even out is called the gambler’s fallacy. The relative frequency approaches the probability over many tosses because new tosses dilute a deviation, not because they correct it.
• “The probability is 4/10 on every draw, so two red is 4/10 · 4/10.” That holds with replacement. Without it, the second factor is 3/9 after a red, and two red has probability 2/15 ≈ 0.133.
• “Mutually exclusive events are independent.” These are two different ideas. Mutually exclusive events cannot happen together; independent events do not affect each other’s probability. Two mutually exclusive events that both have probability above 0 are dependent; see also the addition rule.
• “Independent means the events have nothing to do with each other.” The test is the probabilities, not whether the events concern the same object: “even” and “at most 4” in one roll are independent, “even” and “at least 4” are not. And dependence does not mean that one event causes the other, only that the probabilities are linked.

Common questions