Definition 4.7.1. Independent Events.
Events A and B are independent provided
\begin{equation*}
P(A \cap B) = P(A) P(B)
\end{equation*}
Section 4.7 Independence
You have seen when repeatedly sampling without replacement leads to a change the the likelihood of some event in successive trials. Indeed, this is what conditional probabilities above illustrate. However, when sampling with replacement you may find a different situation arises. Indeed, you easily notice that when flipping a coin, P(Heads) = 1/2 regardless of the outcome of any previous flip. In situations such as this where the probability of an event is not affected by the occurrence (or lack of occurrence) of some other event determining the probability of compound events can be greatly simplified.
Corollary 4.7.2. Independence and Conditional Probability.
Given independent events 4.7.1 A and B,
\begin{equation*}
P(B | A) = P(B)
\end{equation*}
and
\begin{equation*}
P(A | B) = P(A).
\end{equation*}
Proof.
By the multiplication rule and the definition of independence, for any events A and B
\begin{equation*}
P(A) \cdot P(B) = P(A \cap B) = P(A) \cdot P(B | A) .
\end{equation*}
Therefore, if P(A) is non-zero, canceling yields the first result. Switching around notation provides the second.
Checkpoint 4.7.3. WebWork - Independence.
For two events \(A\) and \(B\text{,}\) \(P(A) = 0.7\) and \(P(B) = 0.2\text{.}\)
(a)\(\) If \(A\) and \(B\) are independent, then
\(P(A \cap B)\) \(=\)
\(P(A \cup B)\) \(=\)
\(P(A|B)\) \(=\)
(b)\(\) If \(A\) and \(B\) are dependent and \(P(A|B) = 0.2\text{,}\) then
\(P(B|A)\) \(=\)
\(P(A \cap B)\) \(=\)
Basically you just multiply individual probabilities together. Independence is often assumed since it makes computations easier. That said, you should remember to consider each time whether independence should or should not be assumed.
Corollary 4.7.4. Independence and Mutual Exclusivity.
If events A and B are both independent 4.7.1 and mutually exclusive 4.3.1, then at least one of them has zero probability.
Proof.
By independence, \(P(A \cap B) = P(A) \cdot P(B)\text{.}\) However, by mutually exclusivity, \(A \cap B = \emptyset \Rightarrow P(A \cap B) = 0\) gives
\begin{equation*}
P(A) \cdot P(B) = 0.
\end{equation*}
Hence, one or the other (or both) must be zero.
Corollary 4.7.5. Successive Independent Events.
Given a sequence of mutually independent events \(A_1, A_2, A_3, ...\text{,}\)
\begin{equation*}
P(\cap_{k \in R} A_k) = \prod_{k \in R} P(A_k)
\end{equation*}
