Mutually Exclusive Events

Some events cannot happen at the same time. A single roll of a dice cannot be both a $2$2 and a $5$5; a card drawn from a pack cannot be both a heart and a club. Events that cannot occur together are called mutually exclusive, and for them there is a beautifully simple way to find the probability of "one or the other". This lesson introduces mutually exclusive and exhaustive events, the OR (addition) rule, and the complement rule $P(\text{not } A) = 1 - P(A)$P(not A)=1−P(A), which is really the addition rule in disguise.

This material sits in AO1 (applying the addition and complement rules) and AO2 (deciding whether events are mutually exclusive before adding). OCR command words here include Work out, Calculate, Write down and Show that. A frequent trap is adding probabilities of events that are not mutually exclusive, so the reasoning matters as much as the arithmetic.

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Key Vocabulary

Term	Definition
Mutually exclusive	Two events that cannot happen at the same time.
Exhaustive	A set of events that together cover all possible outcomes.
OR (addition) rule	For mutually exclusive events, $P(A \text{ or } B) = P(A) + P(B)$P(A or B)=P(A)+P(B).
Complement	The event "$A$A does not happen", written $A'$A′ or "not $A$A".
Complement rule	$P(\text{not } A) = 1 - P(A)$P(not A)=1−P(A).
Sum of probabilities	The probabilities of all outcomes of an experiment add to $1$1.

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Mutually Exclusive Events

Two events are mutually exclusive if they cannot both happen at once — there is no outcome common to both. For example, when rolling a dice, "rolling a $2$2" and "rolling a $5$5" are mutually exclusive, but "rolling an even number" and "rolling a number greater than $3$3" are not, because $4$4 and $6$6 satisfy both.

Worked Example 1

A card is drawn from a standard pack. State whether each pair of events is mutually exclusive. (a) "Drawing a heart" and "drawing a club". (b) "Drawing a king" and "drawing a heart".

Solution: (a) Mutually exclusive — a card cannot be both a heart and a club. (b) Not mutually exclusive — the king of hearts is both a king and a heart, so the events can happen together.

Common error: Assuming any two different-sounding events are mutually exclusive. Always check whether an outcome belongs to both.

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The OR (Addition) Rule

If events $A$A and $B$B are mutually exclusive, then

$P(A \text{ or } B) = P(A) + P(B)$P(A or B)=P(A)+P(B)

This works because no outcome is counted twice. (If the events can overlap, you must subtract the overlap — that is covered in the Venn diagram lesson.) The intuition is straightforward: if the favourable outcomes for $A$A and the favourable outcomes for $B$B are completely separate, then the number of outcomes satisfying "$A$A or $B$B" is simply the two counts added together, and dividing by the same total turns that into the sum of the two probabilities.

The single most important habit when using the addition rule is to check first that the events cannot overlap. The rule is only valid for mutually exclusive events. If even one outcome belongs to both events — like the king of hearts belonging to both "kings" and "hearts" — then adding the probabilities counts that outcome twice and gives an answer that is too big. So before you ever write down $P(A) + P(B)$P(A)+P(B), ask yourself: "Is there any single outcome that satisfies both events?" If the answer is no, add; if yes, you need the full rule with a subtraction, which you will meet with Venn diagrams.

Worked Example 2

A fair dice is rolled. Work out the probability of rolling a $2$2 or a $5$5.

Solution: These events are mutually exclusive.

$P(2 \text{ or } 5) = P(2) + P(5) = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}$P(2 or 5)=P(2)+P(5)=61​+61​=62​=31​

Worked Example 3

A bag contains $4$4 red, $3$3 blue and $5$5 green counters. One counter is taken at random. Work out the probability that it is red or green.

Solution: Total $= 12$=12. The colours are mutually exclusive.

$P(\text{red or green}) = \dfrac{4}{12} + \dfrac{5}{12} = \dfrac{9}{12} = \dfrac{3}{4}$P(red or green)=124​+125​=129​=43​

Worked Example 4

A spinner has sections coloured red, blue, green and yellow with probabilities $P(\text{red}) = 0.3$P(red)=0.3, $P(\text{blue}) = 0.25$P(blue)=0.25, $P(\text{green}) = 0.15$P(green)=0.15, $P(\text{yellow}) = 0.3$P(yellow)=0.3. Work out $P(\text{red or yellow})$P(red or yellow).

Solution: $P(\text{red or yellow}) = 0.3 + 0.3 = 0.6$P(red or yellow)=0.3+0.3=0.6

Worked Example 5

A fair spinner has $10$10 equal sections numbered $1$1 to $10$10. Work out the probability of spinning a multiple of $3$3 or a multiple of $5$5.

Solution: Multiples of $3$3: $\{3, 6, 9\}${3,6,9}. Multiples of $5$5: $\{5, 10\}${5,10}. There is no overlap, so the events are mutually exclusive.

$P = \dfrac{3}{10} + \dfrac{2}{10} = \dfrac{5}{10} = \dfrac{1}{2}$P=103​+102​=105​=21​

Common error: Adding when events do overlap. If the question had asked for multiples of $2$2 or multiples of $3$3, the number $6$6 is in both and you would have to avoid double-counting it.

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Exhaustive Events and the Sum to 1

A set of events is exhaustive if together they cover every possible outcome. If a set of events is both mutually exclusive and exhaustive, their probabilities add to exactly $1$1. This pairing is extremely common in exam questions: the colours of a spinner, the faces of a dice, or the categories in a survey are usually mutually exclusive (you land on only one) and exhaustive (you must land on one of them), so their probabilities must total $1$1 — which is exactly what lets you find a missing value by subtraction.

Worked Example 6

A spinner can land on red, blue or green only. $P(\text{red}) = 0.5$P(red)=0.5 and $P(\text{blue}) = 0.2$P(blue)=0.2. Work out $P(\text{green})$P(green).

Solution: The three colours are mutually exclusive and exhaustive, so they sum to $1$1.

$P(\text{green}) = 1 - (0.5 + 0.2) = 1 - 0.7 = 0.3$P(green)=1−(0.5+0.2)=1−0.7=0.3

Worked Example 7

A biased dice has these probabilities for some faces.

Score	1	2	3	4	5	6
Probability	0.1	0.15	0.2	0.2	0.15	?

Work out $P(6)$P(6).

Solution: All six outcomes are exhaustive, so they sum to $1$1.

$P(6) = 1 - (0.1 + 0.15 + 0.2 + 0.2 + 0.15) = 1 - 0.8 = 0.2$P(6)=1−(0.1+0.15+0.2+0.2+0.15)=1−0.8=0.2

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The Complement Rule: P(not A) = 1 − P(A)

The events "$A$A" and "not $A$A" are always mutually exclusive and exhaustive, so they sum to $1$1. Rearranging gives the complement rule:

$P(\text{not } A) = 1 - P(A)$P(not A)=1−P(A)

This is often the quickest route to an answer, especially when the event "$A$A" is awkward to count directly. The logic is airtight: every trial either makes $A$A happen or it does not, with no third possibility and no overlap, so the two probabilities must account for everything and therefore total $1$1. Subtracting $P(A)$P(A) from $1$1 then isolates the probability of the complement.

The complement rule earns its keep on questions where the event you want is fiddly but its opposite is simple. "At least one" is the classic case: rather than adding the probabilities of "exactly one", "exactly two", "exactly three" and so on, you compute $1 - P(\text{none})$1−P(none), which is usually a single quick multiplication. You will use this trick constantly once tree diagrams arrive, so it is well worth recognising the cue words — "at least", "not", "fewer than all" — that signal the complement is the easier route.

Worked Example 8

The probability that a train is late is $0.18$0.18. Work out the probability that it is not late.

Solution: $P(\text{not late}) = 1 - 0.18 = 0.82$P(not late)=1−0.18=0.82

Worked Example 9

A bag has counters with $P(\text{red}) = \dfrac{3}{7}$P(red)=73​. Work out the probability of drawing a counter that is not red.

Solution: $P(\text{not red}) = 1 - \dfrac{3}{7} = \dfrac{7}{7} - \dfrac{3}{7} = \dfrac{4}{7}$P(not red)=1−73​=77​−73​=74​

Common error: Writing $1 - \dfrac{3}{7} = \dfrac{1}{4}$1−73​=41​. You must use the common denominator $7$7, not subtract numerator and denominator separately.

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Extended Worked Examples

Worked Example 10

A bag contains red, blue, green and white counters. $P(\text{red}) = 0.2$P(red)=0.2, $P(\text{blue}) = 0.35$P(blue)=0.35 and $P(\text{green}) = 0.3$P(green)=0.3. (a) Work out $P(\text{white})$P(white). (b) Work out the probability that a counter is not blue. (c) Work out $P(\text{red or white})$P(red or white).