3.3 Compound Events
Recall the different combinations of relationships between two events:
| Independent? | |||
| Yes | No | ||
| Disjoint? | Yes | 1* | 2 |
| No | 3 | 4 |
We must always go into a problem assuming two events are not mutually exclusive or independent. This “default” starting point is illustrated by the 4th position in the table above. Depending on the information your are given and assumptions you are able to make, you may move potions on this grid. Where you fall on the grid will dictate how we apply the rules we will discuss in this section to find probabilities of compound events.
There are two types of compound events we may be interested in, Unions and Intersections, each with their own set of rules and assumptions.
*Note: You will rarely, if ever, find yourself in this case
Finding Probabilities of Unions
To find a union we will typically use the addition rule. Intuitively the idea is that if we are looking for the outcomes in either event A or Event B, we should be able to simply add up the probabilities of each outcome. However, two events being mutually exclusive has big implications on how we apply the addition rule.
For Two Mutually Exclusive Events
The idea is simple when events are mutually exclusive. Picture a Venn diagram of two mutually exclusive events.
Not very exciting, but we need to note here that if A and B are mutually exclusive, then they have no shared outcomes. In other words no intersection exists between two disjoint events. In probability notation this means A ∩ B = Ø and P(A ∩ B) = 0.
In this case, the Union of A OR B is simply:
P(A ∪ B) = P(A) + P(B)
This is reflected in the previously mentioned Third Axiom of Probability (also called the disjoint addition rule).
Recall:
3. For each two events E1 and E2 with E1 ∩ E2 = Ø then P(E1 U E2) = P(E1) + P(E2)
Example
Klaus is trying to choose where to go on vacation. His two choices are: A = New Zealand and B = Alaska
- Klaus can only afford one vacation. The probability that he chooses A is P(A) = 0.6 and the probability that he chooses B is P(B) = 0.35.
- P(A AND B) = 0 because Klaus can only afford to take one vacation
- Therefore, the probability that he chooses either New Zealand or Alaska is P(A OR B) = P(A) + P(B) = 0.6 + 0.35 = 0.95.
- Note that the probability that he does not choose to go anywhere on vacation (the compliment) must then be 0.05.
For Two Non-Mutually Exclusive Events
When two events are not mutually exclusive it gets a bit trickier. Consider a Venn diagram of two non-mutually exclusive events.
Here we can see that when A and B are not mutually exclusive, then they do have shared outcomes, or an intersection. If we try to apply the addition rule, we need to be careful not to double count those shared outcomes
If A and B are defined on a sample space, then: P(A ∪ B) = P(A) + P(B) – P(A ∩ B).
Example
Carlos plays college soccer. He makes a goal 65% of the time he shoots. Carlos is going to attempt two goals in a row in the next game. A