Complementary event

In probability theory, the complement of any event A is the event [not A], i.e. the event that A does not occur. The event A and its complement [not A] are mutually exclusive and exhaustive. Generally, there is only one event B such that A and B are both mutually exclusive and exhaustive; that event is the complement of A. The complement of an event A is sometimes denoted A′.

Simple examples

• A coin is flipped and one assumes it cannot land on its edge. It can either land on "heads" or on "tails" Because these two events are complementary, we have

$\Pr(\mathrm {heads} )+\Pr(\mathrm {tails} )=1.$

• Three plastic balls are in a bag. One is blue and two are red. Assuming that each has an equal chance of being pulled out of the bag,

$\Pr(\mathrm {blue} )=1/3\ {\mbox{and}}\ \Pr(\mathrm {red} )=2/3.$

Example of the utility of this concept

Suppose one throws an ordinary six-sided die eight times. What is the probability that one sees a "1" at least once?

It may be tempting to say that

Pr(["1" on 1st trial] or ["1" on second trial] or ... or ["1" on 8th trial])

= Pr("1" on 1st trial) + Pr("1" on second trial) + ... + P("1" on 8th trial)

= 1/6 + 1/6 + ... + 1/6.

= 8/6 = 1.3333... (...and this is clearly wrong.)

That cannot be right because a probability cannot be more than 1. The technique is wrong because the eight events whose probabilities got added are not mutually exclusive.

Instead one may find the probability of the complementary event and subtract:) it from 1, thus:

Pr(at least one "1") = 1 − Pr(no "1"s)

= 1 − Pr([no "1" on 1st trial] and [no "1" on 2nd trial] and ... and [no "1" on 8th trial])

= 1 − Pr(no "1" on 1st trail) × Pr(no "1" on 2nd trial) × ... × Pr(no "1" on 8th trial)

= 1 −(5/6) × (5/6) × ... × (5/6)

= 1 − (5/6)⁸

= 0.7674...

See also

• Exclusive disjunction
• Binomial probability