Probability of the Complement

3. Probability: Probability

Probability of the Complement

Besides calculating the probability that event $A$ will happen, you can also calculate the probability that $A$ will not happen. In other words: the probability that the complement of A happens.

Rules

The probability that an event $A$ happens plus the probability that its complement $A^c$ happens is always equal to $1$ .

$\mathbb{P}(A) + \mathbb{P}(A^c) = 1$
Complement Rule:

$\mathbb{P}(A^c) = 1 - \mathbb{P}(A)$
Subtraction Rule:

$\mathbb{P}(A) = 1 - \mathbb{P}(A^c)$

Sometimes when calculating the probability of event $A$ , it is easier to calculate the probability of its complement $A^c$ first.

Consider the random experiment of rolling $2$ normal, 6-sided dice.

What is the probability of rolling two different scores?

$\mathbb{P}(A)=\cfrac{5}{6}$

Two different scores are for example getting a $2$ and a $3$ , or a $1$ and a $6$ . This makes for quite a long list of possible outcomes:

$A= \{(1,2)(1,3),(1,4),(1,5),(1,6),(2,1),\ldots, (6,5)\}$ .

But the complement, $A^c$ , (the event that two outcomes are the same) only has $6$ outcomes:

$A^c = \{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6)\}$ .

The sample space of rolling two dice contains $6^2=36$ possible outcomes. The probability of $A^c$ occurring is:

$\mathbb{P}(A^c)=\cfrac{\text{number of outcomes where two scores are the same}}{\text{total number of possible outcomes}}=\cfrac{6}{36}=\cfrac{1}{6}$

Now, by applying rule 3, it is possible to calculate the probability of $A$ :

$\mathbb{P}(A)=1-\mathbb{P}(A^c)=1-\cfrac{1}{6}=\cfrac{5}{6}$

New example