**Independent Events**

In probability, two events (\(A\) and \(B\)) are said to be __independent__ if the fact that one event (\(A\)) occurred does not affect the probability that the other event (\(B\)) will occur

Consider the experiment rolling a die twice. The outcome of the first roll does not affect the outcome of the second roll.

Example: In both rolls there is a \({1}/{6}\) probability of rolling any particular number regardless of the outcome of the other roll. Therefore, these events are considered independent.

In contrast, two events are considered dependent if the outcome of one event impacts the probability of the second event occurring.

Consider the experiment drawing 2 cards without replacement from a standard 52 card deck. The probability of drawing a specific 2^{nd} card depends on the outcome of the 1^{st} draw.

Example: Before any card is drawn the probability of drawing a heart is \({13}/{52}\). But if the king of hearts is the first card drawn then the probability of drawing a heart on the second draw becomes \({12}/{51}\).

If \(A\) and \(B\) are independent events, then the following are also independent events:

- A and B’
- A’ and B
- A’ and B’

And therefore,

$$ P\left( A\cap { B }^{ ’ } \right) =P\left( A \right) \ast P\left( { B }^{ ’ } \right) $$

$$ P \left(A^{ ’ }\cap { B } \right)=P\left( A’ \right) \ast P\left( { B } \right) $$

$$ P \left(A^{ ’ }\cap { B }^{ ’ }\right)=P\left( A’ \right) \ast P\left( { B }^{ ’ } \right) $$

**Probability of Independent Events**

If events \(A\) and \(B\) are independent, then

The probability of \(A\) and \(B\) both occurring is the product of the probability of \(A\) occurring and the probability of \(B\) occurring, as shown in the formula below:

$$ \begin{align*} P \left(A \cap B \right) & = P \left(A \right) \ast P \left(B \right) \\ \end{align*} $$

**Example**

What is the probability of rolling a 1 and then another 1 on two successive rolls of a die?

Since these events are independent,

$$ \begin{align*} P \left(A \cap B \right) & = P \left(A \right) \ast P \left(B \right) \\ & = {1}/{6} \ast {1}/{6} = {1}/{36} \\ \end{align*} $$

Events \(A\), \(B\) and \(C\) are __mutually independent__ if and only if the flowing two conditions are true:

- \(A\), \(B\) and \(C\) are pairwise independent, meaning

\(P \left(A \cap B \right) = P \left(A \right) \ast P \left(B \right) \)

\(P \left(A \cap C \right) = P \left(A \right) \ast P \left(C \right),\) and

\(P \left(B \cap C \right) = P \left(B \right) \ast P \left(C \right) \) - \( P \left(A \cap B \cap C \right) = P \left(A \right) \ast P \left(B \right) \ast P \left(C \right) \)

**Example**

What is the probability of flipping tails on 3 successive flips of a coin?

Since these are all independent events,

$$ \begin{align*} P \left(A \cap B \cap C \right) & = P \left(A \right) \ast P \left(B \right) \ast P \left(C \right) \\ & = { \frac {1}{2}} \ast { \frac {1}{2}} \ast { \frac {1}{2}} = { \frac {1}{8}} \\ \end{align*} $$

If \(A\), \(B\) and \(C\) are mutually independent, then the following are also independent events:

- \(A\) and \(\left(B \cap C \right)\)
- \(A\) and \(\left(B \cup C\right)\)
- A’ and \(\left(B \cap C \right)\)
- A’, B’ and C’

Also,

$$ P{ \left( A\cap B\cap C \right) }^{ \prime }=1-P\left( A\cap B\cap C \right) $$

which says that the probability that a set of mutually independent events “fails” is 1 minus the probability that all three mutually independent events do occur.

**Example**

Consider a machine that fails if 3 mutually independent systems fail each with probability .2. Then the probability that the machine does not fail is

$$ P{ \left( A\cap B\cap C \right) }^{ \prime }=1-P\left( A\cap B\cap C \right) =1-P\left( A \right) \ast P\left( B \right) \ast P\left( C \right) =1-.2\ast .2\ast .2=.992 $$

**Learning Outcome**

**1.c Topic: General Probability – Define independence and calculate probability of independent events.**