Probability - By Complement | Brilliant Math & Science Wiki

In a probability experiment, the probability of all possible events (the sample space) must total to 1— that is, some outcome must occur on every trial. For two events to be complements, they must be mutually exclusive and exhaustive, meaning that one or the other must occur. Therefore, the probabilities of an event and its complement must always total to 1.

For any given event \(A\):

\[P(A) + P(A^c) = 1\]

Equivalently, \(P(A^c) = 1- P(A)\).

Note: this does not, however, mean that any two events whose probabilities total to 1 are each other's complements; complementary events must also fulfil the condition of mutual exclusivity.

Reveal the answer

If the probability that it rains tomorrow is 0.6, what is the probability that it does not rain tomorrow?

The correct answer is: 0.4

Suppose a fair six-sided die is rolled 10 times. What is the probability that a 1 is rolled at least once?

Recall that "at least one" events can often be expressed as the complement of "none" events.

Let \(A\) be the event that a 1 is rolled no times in 10 rolls.

As the 10 die rolls are mutually independent, we can calculate \(P(A)\) using the rule of product. Each die roll has a \(\dfrac{5}{6}\) probability of being a number other than 1. There are 10 rolls, and so

\(P(A)=\left(\dfrac{5}{6}\right)^{10} \approx 0.161506\)

\(A^c\) will be the event that a 1 is rolled at least once in 10 rolls.

\(P(A^c)=1-P(A)\approx 0.838494\)

There is approximately a \(\boxed{0.838494}\) probability that there will be at least one roll of 1 among the 10 rolls.

\[P(A)=\frac{1}{5},\frac{5}{6}, P(B)= \frac{4}{5},\frac{1}{6}\] \[P(A)=\frac{1}{5},\frac{7}{11}, P(B)= \frac{4}{5},\frac{4}{11}\] \[P(A)=\frac{3}{8},\frac{2}{7}, P(B)= \frac{5}{8},\frac{4}{7}\] None of the above \[P(A)=\frac{2}{3},\frac{5}{6}, P(B)= \frac{1}{3},\frac{1}{6}\] Reveal the answer

IF \(A\) and \(B\) are two independent events such that \(P (A^c \cap B ) =\dfrac2{15} \), and \( P (A \cap B^c) = \dfrac16 \), find \(P(A) \) and \(P(B) \).

The correct answer is: \[P(A)=\frac{1}{5},\frac{5}{6}, P(B)= \frac{4}{5},\frac{1}{6}\]