Independent Events: Definition, Formula & Example

Q: What's the quickest way to test independence on an exam?

Compute P(A) P(B) and compare it to P(A B). If they're equal, the events are independent. If they're different — even slightly — the events are dependent, and you need conditional probability.

Q: Can knowing one event happened make another event MORE likely?

Yes. If learning A makes B more likely, the events are positively associated (a kind of dependence). If learning A makes B less likely, they're negatively associated. Either way, they're not independent.

Section 1

Quick Answer

Two events are independent when one outcome has zero effect on the chance of the other. Flipping a coin and rolling a die is the classic example: whether the coin lands heads tells you nothing about the die. For independent events, the chance that both happen equals the chance of one multiplied by the chance of the other.

Section 2

Why This Matters

Independence is the dividing line between two completely different probability tools. If events are independent, you can multiply their probabilities directly — that's how every coin/die/spinner/dart problem becomes simple. If they are NOT independent, you must use conditional probability and the multiplication rule with $P(A \mid B)$ , which is much harder. Getting this distinction wrong is the single most common source of error on probability tests.

Section 3

Intuitive Explanation

Imagine you flip a coin in your right hand and roll a die in your left. The coin can't see the die. The die can't see the coin. So whatever happens on one side, the other side's probabilities don't change at all — that's independence.

Now imagine drawing two cards from a deck without replacing the first one. The first card is gone, so the second draw has different odds. Knowing what came out first updates what's possible second — that's NOT independent.

The test is simple: after you learn one event happened, does your estimate of the other event's probability change? If yes → dependent. If no → independent.

Core idea

Independent does not mean “separate topics.” It means one event gives no probabilistic information about the other.

Section 4

When to Use

Use independence when the problem describes two separate physical or random processes that genuinely don't influence each other — different coins, different dice, different days, different people picked at random with replacement. Watch for the words “separately,” “independently,” “with replacement,” or two distinct devices/processes. Do NOT assume independence just because two events sound different — always check whether one truly affects the other.

✨ Pro tip

Ask: after I learn one event happened, does the probability of the other event stay the same or change?

Section 5

How to Recognize It

Before you reach for $P(A)P(B)$ , run through this checklist. If any answer is no, the events are dependent and you need conditional probability instead.

Does the first event change the sample space for the second event?

If the sample space (the set of possible outcomes) shrinks or shifts after the first event, the events are dependent. Drawing a card without replacement is the canonical example — the deck has changed.
Is there replacement — i.e. does each trial start from the same starting state?

Drawing with replacement, repeating an experiment, or rolling dice multiple times typically preserves the sample space. Without replacement is a strong signal of dependence.
Are the two events produced by separate random processes?

Different physical devices (a coin AND a die), different days, different people sampled independently — these usually produce independent events. One device that produces two correlated outcomes does not.
Does knowing one result change the probability of the other?

This is the formal definition, and it is written with a bar: $P(A \mid B)$ reads “the probability of A, given B.” Everything to the right of the bar is what you already know happened, and it becomes the new total you count out of. Independent means the bar makes no difference — $P(A \mid B) = P(A)$ . Test it on a class of 20 students where 10 wear glasses, 10 play piano, and 5 do both. Conditioning on glasses, the 10 glasses-wearers are the whole group and 5 of them play piano, so $P(\text{piano} \mid \text{glasses}) = \frac{5}{10} = 0.5$ . Ignoring glasses, $P(\text{piano}) = \frac{10}{20} = 0.5$ . Same number, so knowing about the glasses told you nothing and the events are independent. If those two numbers come out different, they are dependent.

Section 6

Independent vs Dependent vs Mutually Exclusive

These three terms get mixed up constantly. They are NOT synonyms — they describe different relationships between two events.

Independent vs Dependent vs Mutually Exclusive
Type	Meaning	Can both happen?	Formula	Example
Independent	One event has no effect on the probability of the other.	Yes	$P(A \cap B) = P(A)P(B)$	Flip a coin AND roll a die
Dependent	One event changes the probability of the other.	Yes	$P(A \cap B) = P(A)\,P(B \mid A)$	Draw two cards without replacement
Mutually exclusive	The two events cannot both happen at the same time.	No	$P(A \cap B) = 0$	Roll one die and get both a 2 and a 5

Independent

Meaning: One event has no effect on the probability of the other.
Can both happen?: Yes
Formula: $P(A \cap B) = P(A)P(B)$
Example: Flip a coin AND roll a die

Dependent

Meaning: One event changes the probability of the other.
Can both happen?: Yes
Formula: $P(A \cap B) = P(A)\,P(B \mid A)$
Example: Draw two cards without replacement

Mutually exclusive

Meaning: The two events cannot both happen at the same time.
Can both happen?: No
Formula: $P(A \cap B) = 0$
Example: Roll one die and get both a 2 and a 5

Section 7

Formula & Notation

P(A \cap B) = P(A)P(B) \quad \text{and} \quad P(A \mid B) = P(A)

What each part means

$P(A)$: chance event A happens
$P(B)$: chance event B happens
$P(A \cap B)$: chance both happen together
$P(A \mid B)$: chance of A given B already happened

Start from the definition of conditional probability:

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

, valid whenever

P(B) > 0

. Independence says the left side is just

P(A)

— knowing

B

changes nothing. Substitute that in and multiply both sides by

P(B)

:

P(A \cap B) = P(A)\,P(B)

. That is where the product form comes from; it is not a separate rule. All three statements say the same thing, so proving any one proves the others:

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

,

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

, and

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

.

How to read it: $P(A \mid B)$ reads “the probability of A, given B.” The bar $\mid$ means “given” or “assuming”: whatever sits to the RIGHT of the bar is what you already know happened, and it becomes the new total you count out of. So $P(A \mid B)$ asks: out of the times B happens, how often does A happen too? $P(B \mid A)$ flips the roles — it counts out of the times A happens — and usually gives a different number. Independence is the special case where conditioning changes nothing: $P(A \mid B) = P(A)$ . Multiply that statement through by $P(B)$ and you get the familiar $P(A \cap B) = P(A)P(B)$ — the same fact written a different way.

Section 8

Worked Examples

Example 1 — Coin + die: classic independent events

Easy

Problem

A fair coin is flipped and a fair six-sided die is rolled. Find the probability that the coin lands heads AND the die shows a 4.

Solution

Identify the two events.

$A$ = 'coin lands heads', $B$ = 'die shows 4'.
Decide whether they are independent.

The coin and the die are physically separate. The coin's outcome can't change the die's outcome. So $A$ and $B$ are independent.
Find each individual probability.

$P(A) = 1/2$ for a fair coin. $P(B) = 1/6$ for a fair die.
Multiply the probabilities.

Because the events are independent, $P(A \cap B) = P(A) \cdot P(B) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12}$ .

Answer

$P(\text{heads and 4}) = \dfrac{1}{12}$

Takeaway: When two events come from separate random processes, multiply their probabilities directly.

Example 2 — Reading the bar: what P(A | B) and P(B | A) actually mean

Easy

Problem

In a group of 40 students, 20 ride a bike to school, 10 ride a skateboard, and 5 do both. Let $A$ = “rides a bike” and $B$ = “rides a skateboard.” (a) Find $P(A \mid B)$ and $P(B \mid A)$ . (b) Decide whether $A$ and $B$ are independent.

Solution

Name the events and say the bar out loud. $P(A \mid B)$ is read “the probability of A, given B.”

The event to the right of the bar is the one you already know happened. It stops being something you are predicting and becomes the group you count inside.
Compute $P(A \mid B)$ . Given B, only the 10 skateboarders are still in play, and 5 of them also ride a bike: $P(A \mid B) = \frac{5}{10} = 0.5$ .

The bar replaced the total. The denominator is 10 — the skateboarders — not 40, the whole group.
Flip the bar and compute $P(B \mid A)$ . Given A, the 20 bike riders are the group, and 5 of them also skateboard: $P(B \mid A) = \frac{5}{20} = 0.25$ .

Same 5 students on top, different total underneath. That is the whole reason $P(A \mid B)$ and $P(B \mid A)$ are usually different numbers — swapping which event is “given” swaps the denominator.
Compare each conditional with its unconditional twin. $P(A) = \frac{20}{40} = 0.5$ , and $P(A \mid B) = 0.5$ . $P(B) = \frac{10}{40} = 0.25$ , and $P(B \mid A) = 0.25$ .

Independence is exactly this: conditioning changed nothing. Learning that a student skateboards did not move the chance that the student rides a bike.
Confirm with the product form. $P(A \cap B) = \frac{5}{40} = 0.125$ , and $P(A)P(B) = 0.5 \times 0.25 = 0.125$ .

The two tests always agree, because $P(A \cap B) = P(A)P(B)$ is just $P(A \mid B) = P(A)$ multiplied through by $P(B)$ .

Answer

$P(A \mid B) = \dfrac{5}{10} = 0.5$ and $P(B \mid A) = \dfrac{5}{20} = 0.25$ . Each matches its unconditional twin ( $P(A) = 0.5$ , $P(B) = 0.25$ ), and $P(A \cap B) = 0.125 = P(A)P(B)$ , so $A$ and $B$ are independent.

Takeaway: Read the bar as “given,” and read whatever follows it as the new total. $P(A \mid B)$ and $P(B \mid A)$ share a numerator but not a denominator, so they are usually different — and independence is the case where each one equals its plain, unconditioned version.

Example 3 — Cards without replacement: a dependent contrast

Standard

Problem

Two cards are drawn one after the other from a standard 52-card deck WITHOUT replacement. Are the events 'first card is a heart' and 'second card is a heart' independent?

Solution

Compute the first probability.

There are 13 hearts in 52 cards, so $P(\text{1st heart}) = 13/52 = 1/4$ .
Compute the conditional probability after a heart is drawn.

If the first card was a heart, only 12 hearts remain in 51 cards: $P(\text{2nd heart} \mid \text{1st heart}) = 12/51 \approx 0.235$ .
Compare the unconditional and conditional probabilities.

Unconditional: $P(\text{2nd heart}) = 13/52 = 0.25$ . Conditional: $\approx 0.235$ . These are different — knowing the first draw changed the probability of the second.
Apply the definition of independence.

Independence requires $P(B \mid A) = P(B)$ . Here $0.235 \neq 0.25$ , so the events are NOT independent.

Answer

The events are dependent. You must use $P(A \cap B) = P(A) \cdot P(B \mid A) = \dfrac{1}{4} \cdot \dfrac{12}{51} = \dfrac{1}{17}$ .

Takeaway: Without replacement is the strongest signal of dependence in card / ball / lottery problems. Always check the sample space after the first draw.

Example 4 — “At least one” means 1 minus none

Standard

Problem

A safety system has two sensors that fail independently, each with probability $0.02$ . Find the probability that at least one sensor fails.

Solution

Notice how many cases “at least one” covers: the first fails, the second fails, or both fail.

Counting three branches separately is slow, and it is easy to double-count the “both” case. The opposite event has only one branch.
Name the opposite event. If $A$ = “at least one sensor fails,” then $A^c$ = “no sensor fails.” The superscript $c$ means complement — $A^c$ is “A does not happen,” and $P(A^c) = 1 - P(A)$ .

Every outcome either lands in $A$ or in $A^c$ , never both and never neither, so their probabilities add to 1.
Compute $P(A^c)$ . Each sensor survives with probability $1 - 0.02 = 0.98$ , and the sensors are independent, so $P(A^c) = 0.98 \times 0.98 = 0.9604$ .

Independence is what licenses the multiplication here — one sensor failing tells you nothing about the other.
Subtract from 1: $P(A) = 1 - 0.9604 = 0.0396$ .

Sanity check: a single sensor fails only 2% of the time, so “at least one of two” should be a little under 4%. It is.

Answer

$P(\text{at least one fails}) = 1 - 0.98^2 = 1 - 0.9604 = 0.0396$

Takeaway: “At least one” is almost always a complement problem: find the probability of none, then subtract from 1. In symbols, $P(\text{at least one}) = 1 - P(\text{none})$ .

Example 5 — “A or B” when A and B are independent

Standard

Problem

$A$ and $B$ are independent. (a) If $P(A) = 0.6$ and $P(B) = 0.5$ , find $P(A \cup B)$ . (b) If instead $P(A \cup B) = 0.7$ and $P(B) = 0.5$ , find $P(A)$ .

Solution

Start from the addition rule, which holds for any two events: $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ .

Adding $P(A)$ and $P(B)$ counts the overlap twice, so you subtract it once to undo the double count.
Use independence to price the overlap. Because $P(A \cap B) = P(A)P(B)$ , the rule becomes $P(A \cup B) = P(A) + P(B) - P(A)P(B)$ .

Independence does not change the addition rule — it only tells you what the overlap term is worth, so “or” problems never turn into a plain product.
(a) Substitute: $P(A \cup B) = 0.6 + 0.5 - (0.6)(0.5) = 1.1 - 0.3 = 0.8$ .

Check the size: a union must be at least as big as either event and no bigger than 1. Here $0.8 \ge 0.6$ and $0.8 \le 1$ .
(b) Run the same identity backwards, with $P(A)$ as the unknown: $0.7 = P(A) + 0.5 - 0.5\,P(A)$ .

Every quantity except $P(A)$ is given, so the identity becomes a one-variable equation instead of a formula to plug into.
Collect like terms and solve: $0.7 = 0.5\,P(A) + 0.5$ , so $0.5\,P(A) = 0.2$ and $P(A) = 0.4$ .

Substitute back to check: $0.4 + 0.5 - (0.4)(0.5) = 0.7$ , which matches the given union.

Answer

(a) $P(A \cup B) = 0.8$ . (b) $P(A) = 0.4$ .

Takeaway: For “or,” add and then subtract the overlap; independence only tells you the overlap equals $P(A)P(B)$ . The same identity works forwards for a union and backwards for a missing single probability.

Example 6 — Exactly one red in three spins: where the factor of 3 comes from

Hard

Problem

A spinner lands on red with probability $0.25$ on each spin, and the spins are independent. Find the probability of getting exactly one red in three spins.

Solution

Price one specific arrangement first — red, not-red, not-red: $0.25 \times 0.75 \times 0.75 = \frac{1}{4} \cdot \frac{9}{16} = \frac{9}{64}$ .

Independence lets you multiply straight across a sequence of spins, so a single fully-specified arrangement is easy to evaluate.
Ask where else the one red could land. It could be the first, second, or third spin: (red, no, no), (no, red, no), (no, no, red). That is 3 arrangements.

“Exactly one” never says which spin was red. Each position is a different outcome of the experiment, and no two of them can happen at once, so their probabilities add.
Notice that all three arrangements have the same probability. Each multiplies one factor of $0.25$ and two factors of $0.75$ ; only the order differs, and multiplication does not care about order.

That is why you get one multiplier instead of three separate calculations — this is exactly where the factor of 3 comes from.
Add the three equal pieces, which is the same as multiplying by 3: $P(\text{exactly one red}) = 3 \times 0.25 \times 0.75^2 = 3 \times \frac{9}{64} = \frac{27}{64}$ .

$\frac{27}{64} \approx 0.422$ . The 3 is a count of positions, not a probability, which is why the answer can be bigger than any single arrangement.

Answer

$P(\text{exactly one red}) = 3 \times 0.25 \times 0.75^2 = \dfrac{27}{64} \approx 0.422$

Takeaway: For exactly $k$ successes in $n$ independent trials, price one arrangement by multiplying, then multiply by how many arrangements there are. Dropping that count is the most common way this problem goes wrong.

Section 9

Common Mistakes

Common slip-up

Assuming independence just because two events sound unrelated

The right idea

always check whether the first event's outcome actually changes the second event's probability before applying $P(A)P(B)$ .

Common slip-up

Using the multiplication rule $P(A)P(B)$ when events are actually dependent

The right idea

for dependent events you must use $P(A) \cdot P(B \mid A)$ instead.

Common slip-up

Confusing mutually exclusive with independent

The right idea

mutually exclusive events CAN'T both happen so $P(A \cap B) = 0$ , while independent events have $P(A \cap B) = P(A)P(B)$ , which is nonzero.

Section 10

Mini Practice

Try these on your own. Tap Reveal when you want to check.

A spinner is spun and a coin is flipped. Are the two outcomes independent or dependent?

Hint: Ask: does the spinner's result change anything about the coin?
Two cards are drawn from a deck without replacement. Are the two draws independent?

Hint: Without replacement: the sample space shrinks after the first draw.
A fair die is rolled twice. What is the probability of rolling a 6 on both rolls?

Hint: Two rolls of the same die are independent. Multiply.
Events $A$ and $B$ have $P(A) = 0.4$ , $P(B) = 0.5$ , and $P(A \cap B) = 0.2$ . Are they independent?

Hint: Compute $P(A) \cdot P(B)$ and compare to $P(A \cap B)$ .
A fair coin is flipped twice. What is the probability of getting at least one head? Take the complement route: first find the probability of no heads at all.

Hint: “At least one” has three branches (HT, TH, HH); “none” has just one. Price the one, then subtract from 1.
$A$ and $B$ are independent with $P(A) = 0.5$ and $P(B) = 0.2$ . What is $P(A \cup B)$ — the probability that at least one of them happens?

Hint: Add the two probabilities, then subtract the overlap once. For independent events the overlap is $P(A)P(B)$ .
Two independent free throws each go in with probability $0.8$ . What is the probability of making exactly one of the two? Write down the two arrangements before you calculate.

Hint: “Exactly one” does not say which throw went in. There are two orders, and each is priced by multiplying.

Want the full set?

50 practice questions for this concept — free to try, every one with a complete worked solution showing the why, not just the answer.

Practice this concept → Take the 3-question mastery check

Section 11

Frequently Asked Questions

How can two events be independent if they happen at the same time?

Independence isn't about timing — it's about whether one event's outcome changes the probability of the other. A coin flip and a die roll happen at roughly the same time, but neither outcome shifts the other's odds, so they're independent.

Are mutually exclusive events also independent?

No — and this is a famous trap. If two events are mutually exclusive (they can't both happen), then knowing $A$ happened forces $P(B) = 0$ , which is a huge change. So they are dependent. The only edge case is when one of the events is impossible to begin with.

Is drawing cards independent if I shuffle between draws?

If you shuffle a full deck back together (i.e. draw with replacement), each draw is independent because the sample space is identical every time. If you draw without replacement, the events are dependent — even shuffling the remaining cards doesn't restore the missing card.

What's the quickest way to test independence on an exam?

Compute $P(A) \cdot P(B)$ and compare it to $P(A \cap B)$ . If they're equal, the events are independent. If they're different — even slightly — the events are dependent, and you need conditional probability.

Does independence work the same for more than two events?

Pairwise independence (every pair satisfies the rule) is necessary but not sufficient for full (mutual) independence of three or more events. For three independent events $A, B, C$ , you need $P(A \cap B \cap C) = P(A) \cdot P(B) \cdot P(C)$ AND every pair to be independent.

Can knowing one event happened make another event MORE likely?

Yes. If learning $A$ makes $B$ more likely, the events are positively associated (a kind of dependence). If learning $A$ makes $B$ less likely, they're negatively associated. Either way, they're not independent.

Section 12

Learning Path

← Before

Conditional Probability

Independent Events

You are here

Next →

Multiplication Rule Expected Value

Before this, students should be comfortable with basic probability and conditional probability — independence is defined in terms of conditional probability not changing. After this, the multiplication rule and expected value problems become much safer to apply because you can correctly choose between multiplying directly versus using a conditional adjustment.

Section 13