Venn Diagrams and Probability Notation

New Videos: Videos #1 and #2 on Probability for Actuarial Exam 1/P

Visualizing probabilities as areas in a Venn diagram.

Sometimes people think that fancy notation and formulas must always be relied upon to solve probability problems. This is definitely not the case, however. Sometimes visuals, such a Venn diagrams, are good enough to help you get the right answers. These diagrams can also provide intuitive insight into the nature of the problems.

In videos #1 and #2 of my new series of problem solving videos on Probability for Actuarial Exam 1 (Exam P), I emphasize this with two problems whose solutions are best worked out by thinking about Venn diagrams. In fact, in video #2, the problem can be solved very quickly by using just a bit of logic and mental arithmetic.

However, near the end of video #2, I also emphasize that “event notation” and “probability notation” will be used in future videos and are helpful for doing more complicated kinds of problems. One more thing that I emphasize is that it can be helpful to think of probabilities in terms of relative areas of the events in the Venn diagram. This will be especially important when doing problems about conditional probability, which can be a tricky topic when first encountered.

Videos with Venn Diagrams

These two videos are embedded below. In the first video, problem #1 from the online list of Society of Actuaries (SOA) Sample Exam P Questions is done. Its solution involves using a Venn diagram with three circles (events). In the second video, problem #2 from the same list is done. It is actually a bit easier to solve because it only involves two circles (events).

Actuarial Exam 1/P Prep: Use a Venn Diagram with Three Circles to Solve a Percentage Problem. Probability for Actuarial Exam 1/P, Video #1. Online SOA Sample Exam P Questions, Problem #1.
Actuarial Exam 1/P Prep: Venn Diagram Problem with Two Events, Event & Probability Notation. Probability for Actuarial Exam 1/P, Video #2. Online SOA Sample Exam P Questions, Problem #2.

Events and Probability Notation

For the videos above, the letters that label the circles are shorthands for “events”. In a random experiment, an event can informally be thought of as just something that can happen.

In the first video, the random experiment consists of picking a person at random from a group of people and asking what kinds of sporting events they watched in the past year from a list of three possibilities: gymnastics, baseball, and soccer (i.e., football in every country but the United States). The letters G, B, and S represent the events that the person watched gymnastics, baseball, and soccer in the last year, respectively.

In the second video, the random experiment consists of picking a person who visited a primary care physician’s office and finding out if they were referred to a specialist and/or if they had lab work done (or neither). The letters S and L were used to refer to these events, respectively.

In the second video, it is then emphasized that mathematicians use either the notation P[L] or P(L) to refer to the probability of the event L occurring. This is essentially function notation: the “P” represents the function name and the “L” represents the input. The entire expression P[L] (or P(L)) then represents the output of the function, which is a real number. As a probability, we can say that 0=0\%\leq P[L]\leq 1=100\% for any event L.

Set Intersection as Shorthand for “And”

The informal idea of an event as just “something that can happen” is not very helpful for mathematical modeling. It will ultimately be best for us to think of events as sets: more specifically, as subsets of an entity know as a sample space. Because of this, we will find that set operations will be useful.

The set intersection operator \cap, which takes two sets and forms a new set of their common elements, is a specific example of this. However, in the informal setting above, we can just think of this as shorthand for the word “and“. The easiest way to remember this is the fact that the symbol \cap looks like a rounded “A” (as shorthand for “And“), but without the horizontal bar.

Therefore, the probability that both events S and L occur will be denoted by P[S\cap L]. This is the quantity that must be solved for in Video #2 and it is found that the answer is P[S\cap L]=0.05.

Set Union as Shorthand for “Or”

Again, in video #2, I found the answer very quickly with some intuitive logic, based on the diagram. We can, however, represent the method used in the video in a symbolic way based on some important formulas.

The set union operator \cup, which takes two sets and forms a new set of their combined elements, can be used to represent one important piece of information: that P[S\cup L]=0.65. The reason is because the probability of neither S nor L occurring is given to be 0.35, implying that the probability of either S or L (or both) occurring is 1-0.35=0.65. In other words, the set union operator \cup can be thought of as shorthand for the word (inclusive) “or”.

Finally, since we are given P[S]=0.30 and P[L]=0.40, we can say that the final answer is P[S\cap L]=0.30+0.40-0.65=0.70-0.65=0.05.

This last line can be derived from the so-called general addition rule: P[S\cup L]=P[S]+P[L]-P[S\cap L].

Set Complement as Shorthand for “Not”

The equation P[S\cup L]=1-P[\mbox{neither }S\mbox{ nor } L]=1-0.35=0.65 from the previous section is also sometimes written as P[S\cup L]=1-P[(S\cup L)']. In general, for an event A the symbol A' represents the event that A has not occurred.

In set theory, the symbol A' represents the set complement of A. We take complements always with respect to some larger set (in this case, the sample space), which is visually represented as the bounding box in the Venn diagram. The general relationship represented in the equations of the previous paragraph is sometimes written P[A]+P[A']=1. Also note that (A')'=A.

We end this post by noting the existence of one of De Morgan’s Laws: (S\cup L)'=S'\cap L' (be careful to take note of the fact that there are various notations for set complements that you will find online and in textbooks). Therefore, we can also write, for example, P[S\cup L]=1-P[S' \cap L'].

You might be interested in a challenge problem: prove this De Morgan Law on your own.