In the previous lesson on probability you were introduced to a fundamental way to calculate probabilities. A probability, we learned, is a ratio of two count values, the number of outcomes in the event divided by the total outcomes possible. This method worked as long as all outcome are equally likely. But what if they aren't?
The answer to this question depends on the situation. You always need some assumption about the problem's scenario before you can determine a probability. In this lesson we will explore the case where you know the probabilities of an events through theory or a knowledgeable person. Once such information is given then you can use it to answer more complicated questions that involve more than one event. In this lesson we will learn some basic probability theory so that we can determine the probabilities of combined events. For example, in a typical summer weather report you often get information on the likelihood of thunderstorms. In this lesson you will learn the mathematics needed to determine the likelihood of a thunderstorm during the weekend.
After completing this lesson you should be able to:
Recall that a probability of an event is a measure ranging from 0 to 1. 0 implies the event is impossible and 1 implies that the event is certain to happen. Values larger than 1 and less than 0 are impossible. Using this information and some logic we can develop the mathematics of probabilities of combined events.
Visually we can make an analogy between probability and area. First we must define our space. Imagine a box that represents all possible outcomes, i.e. the sample space. It has an area of 1 or 100%. Next imagine a circle representing the event, A. It has an area of 30%. The box's area outside of the circle would thus be 70%.
Probability of the Union of two Events
If you combine two events with an "or" you will have to add probabilities and possibly make some adjustments. In the simple case the two events are disjoint, i.e. they cannot both occur at once. Here you just add two events. For example if the chance of winning Saturday's football game is 30% and the chance of a tie is 10% then the chance of either winning or tying is 40%.
However what if the two events are not disjoint? For example assume you are a nurse and a young patient comes to you with a stomach ache. Assume that from experience you know that 30% of such patients ate something that made them ill, 40% have a stomach virus, and 5% have both a stomach virus and ate something that didn't agree with them. What is the probability that the patient has a stomach ache for either of those two reasons?
Let A represents "having ate something bad" and B "having a stomach virus", then P(A ⋃ B) is represented pictorially by two overlapping circles.
In this example the probability of the patient either having eaten something "bad" or having a stomach virus is P(A ⋃ B) = 0.3 + 0.4 − 0.05 = 0.65 or 65%.
Two Parts make a Whole
If you observe the venn diagram and think about what the symbols mean you can derive another formula that can be useful.
The probability of A (i.e. area of circle A) can be partitioned into two pieces P(A ⋃ B) (i.e. region of A also in B) and P(A ⋃ B^{c} )
(i.e. region of A that is outside of B). Thus we get the formula:
Probability of the Intersection of two Events
Note from the previous two formulae, that the second one is more general. The first one can be considered a special case of the second one if you remember the following definition.
Events A and B are disjoint is equivalent to saying P(A⋂B) = 0.
A second, extremely important definition, is independence. Two events are independent if they are unrelated, i.e. knowing something about one outcome will not help you predict the other one. For example flipping two coins, a quarter and a dollar will involve two independent events since knowing heads came up on the quarter does not help you predict the outcome of the dollar flip. On the other hand, if you consider our first example of this lesson, knowing a game was a tie makes it certain that your team didn't win. In short the outcomes "Tie" and "Win" are dependent (as well as being disjoint.
If you know that two events A and B are independent then P(A⋂B) = P(A)×P(B). This relationship often does not hold in real world situations. If P(A⋂B) ≠ P(A)×P(B) then the two events are NOT independent, thus dependent. In those cases you need all three probabilities.
Note that if A and B are independent, then so are their complements. i.e. P(A^{c}⋂B^{c}) = P(A^{c})×P(B^{c}).
Problem solving hint
Before you solve a problem similar to those in this section, change
everything to symbols first. Rewrite the question using symbols,
and identify the appropriate formula. Then preceded to use the formula
to answer the question. Be sure never to forget to answer the question
in a clear statement.
For problems 1 & 2 use the formulae above. Problem 3 requires a little more logic.