Use the complementary event rule P(A') = 1 - P(A) to find probabilities, including problems where using the complement is the most efficient method.
Worked examples
Using the complementary event rule
Straightforward
Problem
The probability that it will rain tomorrow is 73. Find the probability that it will NOT rain tomorrow.
1
Identify the complementary event
The event is 'rain tomorrow'. Its complement is 'no rain tomorrow'. These two events are mutually exclusive and together cover all possible outcomes, so their probabilities must add to 1.
2
Apply the complementary event rule
P(not rain)=1−P(rain)=1−73=77−73=74
Answer
P(not rain)=74
Finding a complementary probability from a sample space
Moderate
Problem
A number is chosen at random from the integers 1 to 20 inclusive. Find the probability that the number chosen is NOT a multiple of 3.
1
Find P(multiple of 3)
Multiples of 3 from 1 to 20: 3,6,9,12,15,18 — that is 6 numbers.
P(multiple of 3)=206=103
2
Apply the complementary event rule
P(not a multiple of 3)=1−P(multiple of 3)=1−103=107
Answer
P(not a multiple of 3)=107
At least one outcome — using the complement
Challenging
Problem
Two fair six-sided dice are rolled. Find the probability that at least one die shows a 5.
1
Identify the complementary event
The complement of 'at least one 5' is 'no 5 on either die'. It is much easier to calculate the complement first.
2
Find P(no 5 on either die)
For each die, P(not a 5)=65. Since the dice are independent:
P(no 5 on either)=65×65=3625
3
Apply the complementary event rule
P(at least one 5)=1−P(no 5 on either)=1−3625=3611
Answer
P(at least one 5)=3611
Practise
Q1·Straightforward
The probability of an event A occurring is P(A)=83. What is P(A′), the probability that A does not occur?
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Explanation
P(A′)=1−P(A)=1−83=88−83=85
Q2·Straightforward
The probability of event B is P(B)=52. What is the probability that B does not occur?
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Explanation
P(B′)=1−52=55−52=53
Q3·Straightforward
A fair six-sided die is rolled. What is the probability of NOT rolling a 3?
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Explanation
There is 1 outcome that gives a 3 out of 6 possible outcomes, so P(3)=61.
P(not 3)=1−61=65
Q4·Straightforward
The probability that a player wins a game is 107. What is the probability that the player does not win?
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Explanation
P(not win)=1−P(win)=1−107=103
Q5·Moderate
A bag contains 3 red, 4 blue, and 2 yellow counters. One counter is chosen at random. Use the complementary event rule to find P(not yellow).
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Explanation
Total counters: 3+4+2=9. P(yellow)=92.
P(not yellow)=1−92=97
Q6·Moderate
The probability of event C is P(C)=207. What is P(C′)?
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Explanation
P(C′)=1−207=2020−207=2013
Q7·Moderate
A letter is chosen at random from the 26 letters of the English alphabet. The vowels are A, E, I, O and U. Using the complementary event rule, find P(not a vowel).
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Explanation
P(vowel)=265.
P(not a vowel)=1−265=2621
Q8·Moderate
A number is chosen at random from the integers 1 to 12 inclusive. Using the complementary event rule, find the probability that the number is NOT a multiple of 4.
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Explanation
Multiples of 4 from 1 to 12: 4,8,12 — that is 3 numbers. P(multiple of 4)=123=41.
P(not a multiple of 4)=1−41=43
Q9·Challenging
Three fair coins are tossed. Find the probability of getting at least one tail. (Hint: use the complementary event.)
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Explanation
The complementary event is 'all three coins show heads'. There is only 1 such outcome (HHH) out of 23=8 equally likely outcomes. P(all heads)=81.
P(at least one tail)=1−81=87
Q10·Challenging
A number is chosen at random from the integers 1 to 30 inclusive. Find the probability that the number is divisible by neither 2 nor 3. (Use the complementary event and inclusion–exclusion.)
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Explanation
Multiples of 2 up to 30: 15 numbers. Multiples of 3 up to 30: 10 numbers. Multiples of 6 (both): 5 numbers. By inclusion–exclusion: 15+10−5=20 numbers are divisible by 2 or 3. P(divisible by 2 or 3)=3020=32.
P(neither)=1−32=31
Q11·Challenging
Two fair six-sided dice are rolled. Find the probability that at least one die shows a 6. (Use the complementary event.)
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Explanation
P(no 6 on one die)=65. Since the dice are independent, P(no 6 on either die)=65×65=3625.
P(at least one 6)=1−3625=3611
Q12·Challenging
A spinner has 12 equal sections numbered 1 to 12. Find the probability that the number selected is NOT a perfect square.
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Explanation
Perfect squares from 1 to 12: 1,4,9 — that is 3 numbers. P(perfect square)=123=41.