Complementary Events
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Theory
Two events are complementary when every outcome belongs to exactly one of them, so their probabilities add to \(1\): \(P(\text{not A}) = 1 - P(\text{A})\). On a spinner numbered \(1\) to \(5\), \(P(\text{not } 4) = 1 - \dfrac{1}{5} = \dfrac{4}{5}\). This Year 8 maths page shows how.
The complement of an event A is “not A”: every outcome in the sample space that is not in A. It is also written \(\bar{\text{A}}\) or A′.
The complement rule. Nothing is in both A and not A, and together they cover everything, so \(P(\text{A}) + P(\text{not A}) = 1\) and \(P(\text{not A}) = 1 - P(\text{A})\). In every form: \(1 - \dfrac{2}{7} = \dfrac{5}{7}\), \(1 - 0.27 = 0.73\), \(100\% - 44\% = 56\%\).
Test a pair. Two events are complementary only when each outcome is in one of them and none is in both. “Win” and “lose” are not complementary if a draw is possible: the complement of “win” is “draw or lose”.
A shortcut. When “not A” has many more outcomes than A, count A and subtract from \(1\). The probabilities of all the outcomes add to \(1\), which also finds a missing one.
Neither. \(P(\text{neither A nor B}) = 1 - P(\text{A or B})\), with shared outcomes counted once.
From a probability to a number. Number in a group \(=\) probability \(\times\) total. Over many trials, the expected number \(=\) probability \(\times\) number of trials.
The complement rule in each form:
| P(A) | P(not A) |
|---|---|
| \(\dfrac{2}{7}\) | \(1 - \dfrac{2}{7} = \dfrac{5}{7}\) |
| \(0.27\) | \(1 - 0.27 = 0.73\) |
| \(44\%\) | \(100\% - 44\% = 56\%\) |
Using the complement rule
- Check that the two events cover every outcome with no overlap.
- Find the easier probability (usually the event with fewer outcomes).
- Subtract from \(1\) (or \(100\%\)).
- Multiply by a total if a number of outcomes is wanted.
(i) List the sample space.
(ii) Find \(P(4)\).
(iii) List the outcomes in the complement of spinning a \(4\).
(iv) Find \(P(\text{not } 4)\).
| \(P(\text{not } 4)\) | \(=\) | \(1 - \dfrac{1}{5} = \dfrac{4}{5}\) |
(i) \(\{1, 2, 3, 4, 5\}\)
(ii) \(\dfrac{1}{5}\)
(iii) \(\{1, 2, 3, 5\}\)
(iv) \(\dfrac{4}{5}\)
(i) \(P(\text{white})\)
(ii) \(P(\text{not white})\) as a fraction and a decimal.
| \(P(\text{white})\) | \(=\) | \(\dfrac{28}{80} = \dfrac{7}{20}\) |
| \(P(\text{not white})\) | \(=\) | \(1 - \dfrac{7}{20} = \dfrac{13}{20}\) |
| \(\dfrac{13}{20}\) | \(=\) | \(\dfrac{65}{100} = 0.65\) |
(i) \(\dfrac{7}{20}\)
(ii) \(\dfrac{13}{20} = 0.65\). Check: \(52\) cars are not white, and \(\dfrac{52}{80} = \dfrac{13}{20}\).
There are only \(5\) multiples of \(9\): \(9, 18, 27, 36, 45\).
| \(P(\text{multiple of } 9)\) | \(=\) | \(\dfrac{5}{45} = \dfrac{1}{9}\) |
| \(P(\text{not})\) | \(=\) | \(1 - \dfrac{1}{9} = \dfrac{8}{9}\) |
\(\dfrac{8}{9}\)
(i) What is the chance of no rain?
(ii) Over \(50\) days with this forecast, how many dry days would you expect?
| \(P(\text{no rain})\) | \(=\) | \(100\% - 18\% = 82\%\) |
| \(0.82 \times 50\) | \(=\) | \(41\) |
(i) \(82\%\)
(ii) \(41\) days. Check: \(9\) wet days and \(41 + 9 = 50\).
Even: \(12\) cards. Multiples of \(3\): \(8\) cards. Both (multiples of \(6\)): \(4\) cards.
| \(n\) | \(=\) | \(12 + 8 - 4 = 16\) |
| \(P(\text{neither})\) | \(=\) | \(1 - \dfrac{16}{24} = \dfrac{1}{3}\) |
\(\dfrac{1}{3}\)
(i) Find \(P(\text{blue})\).
(ii) How many pens are blue?
(iii) How many blue pens must be added to make \(P(\text{not blue}) = \dfrac{1}{2}\)?
| \(P(\text{blue})\) | \(=\) | \(1 - \dfrac{5}{9} = \dfrac{4}{9}\) |
| \(\dfrac{4}{9} \times 36\) | \(=\) | \(16\) |
| \(\dfrac{1}{2}\) | \(=\) | \(\dfrac{20}{40}\) |
(i) \(\dfrac{4}{9}\)
(ii) \(16\)
(iii) The \(20\) other pens stay, so the total must be \(40\): add \(4\) blue pens.
Common pitfalls
Frequently asked questions
What are complementary events?
Two events where every outcome belongs to exactly one of them, such as rolling a 6 and not rolling a 6.
What is the complement rule?
\(P(\text{not A}) = 1 - P(\text{A})\).
Do complementary probabilities always add to 1?
Yes, because together they cover every outcome exactly once.
Are winning and losing complementary?
Only if a draw is impossible. Otherwise the complement of winning is drawing or losing.
How do I find an expected number?
Multiply the probability by the number of trials: \(0.65 \times 300 = 195\).