Two events and are said to be independent if the occurrence of makes it neither more nor less probable that occurs and, conversely, if the occurrence of makes it neither more nor less probable that occurs.
In other words, after receiving the information that will happen, we revise our assessment of the probability that will happen, computing the conditional probability of given ; if and are independent events, the probability of remains the same as it was before receiving the information:Conversely,
In standard probability theory, rather than characterizing independence by properties (1) and (2) above, we define it in a more compact way, as follows.
Definition Two events and are said to be independent events if and only if
It is easy to prove that this definition implies properties (1) and (2) above.
Suppose and are independent and (say) . Then,Note that we have assumed . When , things are more complicated (see the discussion about division by zero in the lecture on conditional probability and in the references therein). It is exactly because of the difficulties that arise in defining when that a general definition of independence is not given by using properties (1) and (2).
The following example shows how to check whether two events are independent in a simple probabilistic experiment.
Example An urn contains four balls , , and . We draw one of them at random. The sample space isEach of the four balls has the same probability of being drawn, equal to , i.e.,Define the events and as follows:Their respective probabilities areThe probability of the event isHence,As a consequence, and are independent events.
The definition of independence can be extended also to collections of more than two events.
Let , ..., be a collection of events. It is important to note that even if all the possible couples of events are independent (i.e., is independent of for any ), this does not imply that the events , ..., are jointly independent. This is proved with a simple counter-example.
Example Consider the experiment presented in the previous example (extracting a ball from an urn that contains four balls). Define the events , and as follows:It is immediate to show thatThus, all the possible couple of events in the collection , , are independent. However, the three events are not jointly independent. In fact,
On the contrary, it is obviously true that if , ..., are jointly independent, then is independent of for any .
If is a zero-probability event, then is independent of any other event .
Note thatAs a consequence, by the monotonicity of probability,But , so . Since probabilities cannot be negative, it must be . The latter fact implies independence:
Below you can find some exercises with explained solutions.
Suppose that we toss a die. Six numbers (from to can appear face up, but we do not yet know which one of them will appear. The sample space isEach of the six numbers is a sample point and is assigned probability . Define the events and as follows:Prove that and are independent events.
The probability of isThe probability of isThe probability of is and are independent events because
A firm undertakes two projects, and . The probabilities of having a successful outcome are for project and for project . The probability that both projects will have a successful outcome is . Are the two outcomes independent?
Denote by the event "project is successful", by the event "project is successful" and by the event "both projects are successful". The event can be expressed asIf and are independent, it must be thatTherefore, the two outcomes are not independent.
A firm undertakes two projects, and . The probabilities of having a successful outcome are for project and for project . What is the probability that neither of the two projects will have a successful outcome if their outcomes are independent?
Denote by the event "project is successful", by the event "project is successful" and by the event "neither of the two projects is successful". The event can be expressed as:where and are the complements of and . Using De Morgan's law () and the formula for the probability of a complement, we obtainBy using the formula for the probability of a union, we obtainFinally, since and are independent, we have that
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