Independence

Events can be considered independent if knowing one does not tell us anything about the other.

Two events $A$ and $B$ are independent, denoted $A \perp B$, if and only if:

Knowing that one has occurred has no bearing on the probability that the other occurs.

If $A \perp B$ and $P(B)>0$, we can easily see that:

thus, the probability of $A$ is unchanged by conditioning on $B$.

If $A \perp B$, it is also true that $\bar{A} \perp B, A \perp \bar{B}$, and $\bar{A} \perp \bar{B}$.

Example: Independence

Imagine we toss a coin twice. There are four outcomes in the sample space of this experiment: (1) both tosses land heads $(H H),(2)$ both tosses land tails $(T T),(3)$ first toss lands heads but the second one tails $(H T)$, and (4) the first toss lands tails but the second one heads $(T H)$. Each of these outcomes occur with equal probabilities.

If we define events $A$ and $B$ as the events where the first and second coin tosses result in heads respectively, each of these events would contain two outcomes resulting in probability of $\frac{2}{4}$ for both of them:

The joint event, $A \cap B$, has only one outcome, $H H$, and therefore probability $P(A, B)=\frac{1}{4}$, which is equal to $P(A) \times P(B)$ proving that events $P(A)$ and $P(B)$ are independent.

However, what if we define an event $C$ to include outcomes where at least one toss is heads; what can we infer about this event in relation to events $A$ and $B$ ?

Solution:

$C$ is not independent from either event $A$ or event $B$, as:

giving the probabilities: