# Probability of Compound Events

## Key Questions

• Compound events are combinations of elementary events. For example, when rolling a dice, an elementary event is any concrete result of a single rolling - a number on top from 1 to 6. An example of a compound event is an event of having rolled two dice getting 6 on each.

A compound event can consist of independent elementary events. This is the case when the result of any one elementary event does not have any influence on the result of another. An example with two dice above is such a compounded event.

Alternatively, we might have a situation of elementary events that depend on each other. For example, a compounded event of having a sum of two numbers rolled on two dice being less than 6.

In case of independent elementary events that are composed into a compounded event, the probability of a compounded event equals to a product of probabilities of its elementary parts. Thus, the probability of having rolled two 6 with two dice equals to a product of probability of having 6 on the first dice (that is, $\frac{1}{6}$) by probability of having 6 on the second one (also, $\frac{1}{6}$), that is the answer is $\frac{1}{36}$.

More complex case of probability of a compounded event that consists of dependent elementary events requires the knowledge of conditional probabilities and equals to a product of probability of one event by a conditional probability of another under condition that the first elementary event has occurred.

More detailed analysis of events, probabilities, conditional probabilities and other concept can be found in the chapter Probability at Unizor - free on-line course of advanced mathematics for teenagers.

• It matters whether the events are independent or not.

Example
If you roll a die twice, the second roll is independent of the first, because a die doesn't have memory.

Probabilty of rolling two sixes
First: $P \left(1 s t = 6\right) = 1 / 6$
Second: $P \left(2 n d = 6\right) = 1 / 6$

Probability of two sixes means:
First is 6 AND second is 6:
$P \left(1 s t = 6\right) \cdot P \left(2 n d = 6\right) = 1 / 6 \cdot 1 / 6 = 1 / 36$

Probability of rolling more than 4 (i.e. 5 OR 6) in one try:
$P \left(5\right) = 1 / 6$
$P \left(6\right) = 1 / 6$
$P \left(> 4\right) = P \left(5\right) + P \left(6\right) = 1 / 6 + 1 / 6 = 1 / 3$

General rule
If it's an AND situation multiply the chances
If it's an EITHER.. OR situation, you add