Probability of Combined Events

Objective

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:


Key Concepts

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.

Notation
P( Ac ) = the probability that event A does NOT occur.  (Alternative notation: P(~A) or P(A). )
P(  AB ) = P( A or B ) = the probability that either event A or B (or both) occur.
P(  AB ) = P( A and B ) = the probability that both events A and B occur.
Probability of a Negation
The first and simplest formula to remember is the negation (or complement) of an event. If the probability of an event is known, for example the chance of an afternoon storm, then you also know the chance of no thunderstorm. Since you will either have the storm or won't have a storm, and the two probabilities add up to 100%. If there is a 30% chance of storm, then there is a 70% of no storm. Mathematically we write:
P( Ac ) = 1 − P( A ).

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%.

Simple Venn Diagram with one event.
This type of diagram is a useful tool when solving probability problems, and can be expanded to help you with more difficult problems.

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%.

Venn Diagram of two disjoint events
Visual representation of two disjoint events.

Formula for the union of two disjoint events is P(AB) = P(A) + P(B).

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(AB) is represented pictorially by two overlapping circles.

Visual representation of two non-disjoint events.
Visual representation of two non-disjoint events.

The overlap can be represented by "AB" thus the general formula for the probability of the union or two events is:
P(AB) = P(A) + P(B) − P(AB).

In this example the probability of the patient either having eaten something "bad" or having a stomach virus is P(AB) = 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(AB) (i.e. region of A also in B) and P(ABc ) (i.e. region of A that is outside of B). Thus we get the formula:

P(A) = P(AB) + P(ABc).

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(AB) = 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(AB) = P(AP(B). This relationship often does not hold in real world situations. If P(AB) ≠ P(AP(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(AcBc) = P(AcP(Bc).

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.


Practice Problems

For problems 1 & 2 use the formulae above.  Problem 3 requires a little more logic. 

  1. Let "A" be the event that it will rain, let "B" be the event that I find a parking space near the building. Assume you know that P( A ) = 0.4 and P( B ) = 0.7, and that A and B are independent. Find:
    1. Probability that it won't rain = P( Ac ) = ANSWER
    2. Probability that it won't rain and I find a parking space = P( Ac ⋂ B )  = ANSWER
    3. P( Ac ⋃ B ) = ANSWER
    4. P( A ⋂ Bc )  = ANSWER
    5. Interpret the symbols P( A ⋂ BcANSWER


  2. Use the following probabilities:  the chance that it will snow on Christmas Day in Western, NC is 20%, the chance that it will snow on New Year's Eve in that same area is 25%. and that the chance that it will snow on both days is 10%, to do the following.
    1. State the two events of interest in this problem. ANSWER
    2. Find the probability that it will NOT snow on Christmas. ANSWER
    3. Find the probability that it will NOT snow on Christmas, but will snow on New Year's Eve. ( Hint )  ANSWER
    4. Find the probability that it will NOT snow on either day.  ANSWER 
    5. Are the two events independent? ANSWER


  3. Given the following situation at a car lot.  At "Super deal Auto" 400 out of 1000 of the customers are women, (600 are men), 50 customers ended up buying a car, and according to the manager  of these 23 were women. Using the above information answer the following questions.
    1. What percentage of customers end up not buying a car at "Super deal Auto?"
    2. What percentage of customers are both women and eventual car buyers.
    3. What percentage of the male customers buy a car at this dealership.
    4. If a car is sold what is the chance that the buyer was a women.
    5. Are events "customer is a woman" and "customer will buy car" independent?


  4. Given a pair of fair dice.  Let A be the event that when rolling the dice you get a pair.  Let  B be the event that the largest number is smaller than 4.  Determine each of the listed probabilities.   Hint: write down all 36 possible outcomes, and count how many outcomes are in each given event. Then use the basic rule of probability to find:
    1. Probability of rolling a pair, P( A ) = ANSWER 
    2. Probability that largest number is less than 4, P( B )  = ANSWER
    3. Probability of a pair and largest value less than 4, P( A ⋂ B )  = ANSWER
    4. P( Ac ⋃ Bc  ) = ANSWER
    5. Are A and B independent? ANSWER