Two events

In Figure 3 Venn diagrams depict the union and intersection of events $E_1$ and $E_2$.

Figure 3: Intersection and Union of 2 sets

• The union of $E_1$ and $E_2$ contains the sample points from both events and is written $E_1 \cup E_2$. We would refer to $E_1 \cup E_2$ as the set of sample points that belong to $E_1$ or $E_2$ or both.

• The intersection of $E_1$ and $E_2$, denoted by $E_1 \cap E_2$, is the set of sample points which are common to $E_1$ and $E_2$.

Figure 4 assigns probabilities to the events; $P(E_1)=\frac{1}{4}$ and $P(E_2)=\frac{1}{3}$.

Figure 4: Probabilities of 2 events

(a)	(b)

The probability that $E_1$ and $E_2$ occur jointly is $P( E_1 \cap E_2)$. In Figure 4(a) there are no sample points in common so that the probability of $E_1$ and $E_2$ occurring together is zero. Figure 4(b) indicates that $P(E_1 \cap E_2) = \frac{1}{9}$.

The probability that

(i)

$E_1$ occurs or

(ii)

$E_2$ occurs or

(iii)

both occur

is written $P(E_1 \cup E_2)$ and

$$P( E_1 \cup E_2) = P(E_1) + P(E_2) - P( E_1 \cap E_2) .$$ (3)

The adjustment from the last term, $- P( E_1 \cap E_2)$, removes the one unit of the shared contribution since it is in both $E_1$ and $E_2$. If the two events are mutually exclusive and share no common information, their intersection is the empty set and with $P(E_1 \cap E_2) =0$,

$$P( E_1 \cup E_2) = P(E_1) + P(E_2) .$$

In Figure 4(a) the events are mutually exclusive,

$$P( E_1 \cup E_2) = \frac{1}{4} + \frac{1}{3}$$

and in Figure 4(b),

$$P( E_1 \cup E_2) = \frac{1}{4} + \frac{1}{3} - \frac{1}{9} .$$

When $E_1$ and $E_2$ are independent, information about one does not affect the information about the other. For instance, the event of rain on Jan 1st does not allow estimation of whether it will rain on July 1st as these events are far enough apart in time to be regarded as independent. The probability that the 2 independent events both occur is,

$$P( E_1 \cap E_2) = P(E_1) \times P(E_2) .$$ (4)

This relationship would not hold for dependent events because some of the information contained in $E_1$ is also contained in $E_2$ and the product would ``double dip''. The maximum temperature on Jan 1st is a reliable predictor of the maximum temperature on Jan 2nd and the probability that say the temperature exceeds $35^\circ$C on both days is really determined by the Jan 1st event, Jan 2nd is not an entirely new event.

Independence and mutually exclusive are different concepts,

Independence $\Rightarrow$	$P(E_1 \cap E_2) = P(E_1) \times P(E_2)$.
Mutually exclusive $\Rightarrow$	$P(E_1 \cup E_2) = P(E_1) + P(E_2)$.

The probability of $E_1$ when $E_2$ is known is termed probability of $E_1$, conditional on $E_2$ and written as $P(E_1 \vert E_2)$. Then the probability that both events occur is,

$$P( E_1 \cap E_2) = P(E_1 \vert E_2) \times P(E_2) .$$ (5)

In Figure 4(b), $P(E_1 \vert E_2) = \frac{1}{9} \div \frac{1}{3} = \frac{1}{3}$. That illustrates that if we gain extra information by knowing $E_2$ and that $P(E_1 \cap E_2) \neq 0$, the risk associated with $E_1$ is reduced.

The relationships of the probability of events are collated in Table 1. This provides us with enough basic principles to start probability analysis. Whilst the introductory examples may appear artificial, they are metaphors for ``real'' problems.

Table 1: Probability relationships

	1	$0 \leq P(E) \leq 1$
	2	$P(\bar E) =1 - P(E)$
	3	$P( E_1 \cup E_2) = P(E_1) + P(E_2) - P( E_1 \cap E_2)$
Mutually exclusive		$P(E_1 \cup E_2) = P(E_1) + P(E_2)$
Independence	4	$P(E_1 \cap E_2) = P(E_1) \times P(E_2)$
	5	$P( E_1 \cap E_2) = P(E_1 \vert E_2) \times P(E_2)$