P(A) = 0.5, P(B) = 0.4, P(A and B) = 0.1. Find P(A or B). [2 points]

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When can we add probabilities directly, and what is the general addition rule?

Topic 4.4 Mutually Exclusive Events: identify mutually exclusive (disjoint) events and apply the addition rule, including the general addition rule that subtracts the overlap, to find the probability of a union.

A focused answer to AP Statistics Topic 4.4, defining mutually exclusive (disjoint) events, the addition rule for disjoint events, and the general addition rule that subtracts the intersection, with worked union calculations.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Two events are mutually exclusive (disjoint) if they cannot happen at the same time, so their intersection is empty: $P(A \text{ and } B) = 0$ . For disjoint events the addition rule is simply $P(A \text{ or } B) = P(A) + P(B)$ . When events can overlap, you must use the general addition rule,

P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B),

subtracting the overlap once because adding

P(A)

and

P(B)

counts the shared part twice. Disjoint events are just the special case where the overlap is zero. Mutually exclusive is about whether events can co-occur; do not confuse it with independence, which is about whether one affects the other's probability.

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What this topic is asking
Mutually exclusive events
The two addition rules
Why you subtract the overlap
Mutually exclusive is not independent
Try this

What this topic is asking

The College Board (Topic 4.4) wants you to identify mutually exclusive (disjoint) events and apply the addition rule, including the general addition rule that subtracts the overlap, to find the probability of a union (an "or" event).

Mutually exclusive events

The test is purely logical: could both events happen on one trial? If not, they are disjoint. "Rolling a $2$ " and "rolling a $5$ " on one die are disjoint; "rolling an even number" and "rolling a number greater than $3$ " are not, because a $4$ or $6$ satisfies both. Identifying overlap correctly is the whole battle, because it decides which version of the addition rule applies.

The two addition rules

This is why you only ever need to remember the general rule: it always works, and it collapses to plain addition when there is no overlap. The classic mistake is to add $P(A)$ and $P(B)$ for events that can overlap, inflating the answer by double-counting the shared region.

Why you subtract the overlap

A Venn-diagram picture makes the general rule obvious. Imagine two overlapping circles for $A$ and $B$ . The region " $A$ or $B$ " is everything inside either circle. If you add the area of circle $A$ to the area of circle $B$ , the lens-shaped overlap (where both happen) is inside both circles, so you have added it twice. Subtracting $P(A \text{ and } B)$ once removes the duplicate, leaving each region counted exactly once. So the subtraction is not a quirk to memorize but a direct consequence of not wanting to count the shared outcomes twice. This also explains the disjoint case visually: if the circles do not overlap, there is nothing to subtract, and you simply add. Carrying this picture into harder problems, three events, or "or" combined with conditions, keeps you from either omitting a needed subtraction or subtracting when there is no overlap.

Mutually exclusive is not independent

The exam relentlessly tests one confusion: mutually exclusive and independent are different, often opposite, ideas. Mutually exclusive means the events cannot co-occur ( $P(A \text{ and } B) = 0$ ). Independent (Topic 4.6) means one event's occurrence does not change the other's probability. In fact, two events with positive probabilities that are mutually exclusive are necessarily dependent: if $A$ happens, then $B$ cannot, so knowing $A$ occurred drives $B$ 's probability to $0$ , a clear influence. So you cannot be both disjoint and independent (for events with non-zero probability). Keeping the questions separate, "can they happen together?" for mutually exclusive versus "does one affect the other?" for independent, prevents the most common Unit 4 error, and it sets up Topic 4.6, where the addition rule for unions is combined with the multiplication idea.

Applying the addition rules

A single die is rolled. Let $A$ be "an even number" and $B$ be "a number greater than $4$ ". (a) Are $A$ and $B$ mutually exclusive? (b) Find $P(A \text{ or } B)$ . (c) Contrast with rolling and the events "a $1$ " and "a $6$ ".

step 1 Check for overlap (part a)

$A = \{2, 4, 6\}$ and $B = \{5, 6\}$ . They share the outcome $6$ , so $P(A \text{ and } B) = \frac{1}{6} \neq 0$ . The events are not mutually exclusive.

step 2 Apply the general addition rule (part b)

$P(A) = \frac{3}{6}$ , $P(B) = \frac{2}{6}$ , $P(A \text{ and } B) = \frac{1}{6}$ . So

P(A \text{ or } B) = \frac{3}{6} + \frac{2}{6} - \frac{1}{6} = \frac{4}{6} = \frac{2}{3}.

Adding

\frac{3}{6} + \frac{2}{6} = \frac{5}{6}

without subtracting would wrongly count the

6

twice.

step 3 Contrast with disjoint events (part c)

"Roll a $1$ " and "roll a $6$ " cannot both happen, so they are mutually exclusive with $P(\text{and}) = 0$ . Here $P(1 \text{ or } 6) = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$ , with nothing to subtract.

step 4 Interpret

When events can share an outcome (like the $6$ ), you must subtract the overlap, giving $\frac{2}{3}$ ; when they cannot (a $1$ versus a $6$ ), plain addition is correct. Checking whether the events can co-occur tells you which rule to use.

Try this

Q1. $P(A) = 0.5$ , $P(B) = 0.4$ , $P(A \text{ and } B) = 0.1$ . Find $P(A \text{ or } B)$ . [2 points]

Cue. General addition rule: $0.5 + 0.4 - 0.1 = 0.8$ .

Q2. Explain why two mutually exclusive events (with non-zero probabilities) cannot be independent. [1 point]

Cue. If $A$ occurs, $B$ cannot, so knowing $A$ happened changes $P(B)$ to $0$ ; that influence means they are dependent, not independent.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2018 (style)1 marksSection I (multiple choice). For a single card drawn from a standard deck, let

A

be 'the card is a king' and

B

be 'the card is a queen'. What is

P(A \text{ or } B)

? (A)

\frac{4}{52}

(B)

\frac{8}{52}

(C)

\frac{16}{52}

(D)

\frac{2}{52}

Show worked answer →

The correct answer is (B).

A card cannot be both a king and a queen, so $A$ and $B$ are mutually exclusive. The addition rule for disjoint events gives $P(A \text{ or } B) = P(A) + P(B) = \frac{4}{52} + \frac{4}{52} = \frac{8}{52}$ .

(A) is only the kings. (C) double counts. (D) is too small. Because the events cannot overlap, you add the probabilities with no subtraction.

AP 2021 (style)4 marksSection II (free response). In a class,

P(\text{plays sport}) = 0.6

P(\text{plays music}) = 0.3

, and

P(\text{plays both}) = 0.2

. (a) Find the probability a student plays sport or music. (b) Explain why you cannot simply add

0.6

and

0.3

. (c) Are 'plays sport' and 'plays music' mutually exclusive? Justify in context.

Show worked answer →

A 4-point question on the general addition rule.

(a) (2 points) $P(\text{sport or music}) = P(\text{sport}) + P(\text{music}) - P(\text{both}) = 0.6 + 0.3 - 0.2 = 0.7$ (1 point for the rule, 1 point for the value).
(b) (1 point) Adding $0.6 + 0.3 = 0.9$ double counts the $0.2$ who play both; the general addition rule subtracts the overlap once to correct this.
(c) (1 point) No; since $P(\text{both}) = 0.2 \neq 0$ , some students play both, so the events can occur together and are not mutually exclusive.

Markers reward the general addition rule with the overlap subtracted, the double-counting explanation, and recognizing that a non-zero intersection means the events are not disjoint.

Related dot points

Sources & how we know this

AP Statistics Course and Exam Description — College Board (2020)