2.1.3.2.4 - Complements

2.1.3.2.4 - Complements
Complement

The probability that the event does not occur. The complement of $P(A)$ is $P(A^C)$. This may also be written as $P(A')$.

In the diagram below we can see that $A^{C}$ is everything in the sample space that is not A.

Complement of A
$P(A^{C})=1−P(A)$

Example: Coin Flip

When flipping a coin, one can flip heads or tails. Thus, $P(Tails^{C})=P(Heads)$ and $P(Heads^{C})=P(Tails)$

Example: Hearts

If you randomly select a card from a standard 52-card deck, you could pull a heart, diamond, spade, or club. The complement of pulling a heart is the probability of pulling a diamond, spade, or club. In other words: $P(Heart^{C})=P(Diamond,\; Spade,\;\;Club)$

The complement of any outcome is equal to one minus the outcome. In other words: $P(A^{C})=1-P(A)$

It is also true then that: $P(A)=1-P(A^{C})$

Example: Rain

According to the weather report, there is a 30% chance of rain today: $P(Rain) = .30$

Raining and not raining are complements.

$P(Not \:rain)=P(Rain^{C})=1-P(Rain)=1-.30=.70$

There is a 70% chance that it will not rain today.

Example: Winning

The probability that your team will win their next game is calculated to be .45, in other words:

$P(Winning)=.45$

Winning and losing are complements of one another. Therefore the probability that they will lose is:

$P(Losing)=P(Winning^{C})=1-.45=.55$

The sum of all of the probabilities for possible events is equal to 1.

Example: Cards

In a standard 52-card deck there are 26 black cards and 26 red cards. All cards are either black or red.

$P(red)+P(black)=\frac{26}{52}+\frac{26}{52}=1$

Example: Dominant Hand

Of individuals with two hands, it is possible to be right-handed, left-handed, or ambidextrous. Assuming that these are the only three possibilities and that there is no overlap between any of these possibilities:

$P(right\;handed)+P(left\;handed)+P(ambidextrous) = 1$

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