Take two problems:

  1. Andrew is 35, and the probability he will be alive in 10 years is .72. Ellen is 35, for her, .92. Assuming these are independent, what is the probability they both will be alive in 10 years?

    Answer: .66, Method: Use the multiplication rule .72*.92 = .66

  2. Suppose your street has two traffic lights. The chance that the first light is red is .40, and the second light .30. The chance of them being red at the same time is .10. What is the the probability that neither light is red?

    Answer: .40, Method: Subtract .1 from .4 and .3, and then subtract all 3 from 1.0 = .4

Why doesn't the multiplication rule work for #2? In other words, why can't I multiply .40 and .30 to get .12? And then furthermore, if .12 is the chance they will both be red at the same time shouldn't (1-.12)=.88 be the chance that neither is on?

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    $\begingroup$ Because the two events (first light being red, second light being red) aren't independent. $\endgroup$ – mark999 Jan 14 '13 at 6:31
  • $\begingroup$ the second problem could have involved independence if they didn't supply the $.1$. The additive rule says $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. It's giving you the last term, but you could've found it by multiplying $.4$ and $.3$ if they told you the events were independent. $\endgroup$ – Taylor Jan 14 '13 at 6:36
  • $\begingroup$ math.stackexchange.com/questions/1898596 $\endgroup$ – BCLC May 26 '18 at 17:13

Let's draw pictures in which regions depict events (such as "the first light is red") and their areas are proportional to the probabilities of those events. Taking care to show areas accurately extends the Venn diagram metaphor in a useful quantitative way.

For the traffic light problem, I will divide a unit square (representing the total probability) into four parts. The left-right division will reflect the possibilities for the first light (set to red at the left, non-red at the right) and the top-bottom division will reflect the possibilities for the second light (red at the bottom, non-red at the top).


In the left figure, the divisions have been made in a 40-60 ratio and a 30-70 ratio, respectively. Where the red rectangle (of width 40%) and blue rectangle (of height 30%) intersect they form a purple rectangle of area 30% * 40% = 0.3 * 0.4 = 12% of the total area. Independent events can always be drawn in this way as separate overlapping rectangles. (When you think about what this means--overlapping rectangles are a geometric way to multiply quantities--it becomes clear that this is the very definition of independence.)

The right figure shows the actual information in the problem, which tells us the purple rectangle has an area of only 10% and asks us to find the area not covered by either rectangle: the white portion to the upper right, depicting the event "Light 1 is not red and light 2 is not red." This indicates lack of independence: now it takes more than just two rectangles to carve up the square correctly. (There's more than one way to do this. For instance, I could have left the blue rectangle alone and adjusted the two halves of the red rectangle, making the bottom skinnier and--to keep its total area at 40%--the top fatter. Either way works.)


Starting with the 10% purple rectangle, notice that the rest of the blue rectangle (at the right) has to include the remaining 20% = 30% - 10% of the time the second light is red. Similarly, the rest of the vertical red rectangle has to include the remaining 30% = 40% - 10% of the time the first light is red. This gives three rectangles of known area: 10%, 20%, and 30%. They sum to 60%. Consequently, because the sum of all areas must be 100%, the white area is 100% - 60% = 40%. This represents the probability to be found.


  • The white region in the first (left) figure is that of a rectangle with base 60% = 100% - 40% and height 70% = 100% - 30%, whence its area is 0.6 * 0.7 = 42% (and not 40%).

  • This area = probability method extends to more than two criteria: at What is the probability that this person is female? it is used to analyze a problem with three separate criteria.

  • Any two-by-two contingency table can (and usually should) be visualized this way, after translating its counts into frequencies relative to the total.

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    $\begingroup$ These figures were produced with the help of Simon Woods' code (slightly modified) posted at mathematica.stackexchange.com/a/11355. $\endgroup$ – whuber Jan 15 '13 at 19:05
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    $\begingroup$ FYI, besides purely personal aesthetic preferences, there is some current work being done in information visualization that would suggest the XKCD style theme has other benifits. See for example Sketchy rendering for information visualization (Wood et al. 2012). And some other related research via Robert Kosara. $\endgroup$ – Andy W Jan 15 '13 at 20:39

In problems like this it is important that you read specifically which information is given to you and which is the problem. So let's start with assignment 1.

The probability that Andrew is still alive - let's call that $P(A)$ and the probability that Ellen is still alive is $P(B)$.
What we are looking for is the event that both are still alive, so both events occur simultaneously, which we call $A \cap B$.
As noted, they tell you these events are independent. That means:
By definition, the probability of this event is
$P(A \cap B) = P(A) * P(B) $

So one of your question was, how is this related to the Venn diagramme?
Well, independence is not readily apparent in the venn diagramm. A Venn diagramme needs to look a certain way if there is independence, but even if it looks that way you only know events are independent if the problem set tells you! This is a bit tricky but stay with me.
If you'd draw a Venn diagramme for this first exercise, you would see that: $$P(A \cap B) / P(A) = P(B) / 1$$ Which is the same as $$P(A \cap B) / P(A) = P(B) / P(\Omega)$$ In essence that means that the chance of "hitting" $B$ if you are already in $A$ is the same as the chance of hitting $B$ in the first place - or said differently - no matter if you hit $A$ - the chance of hitting $B$ is always the same.
If $P(A)$ was 0,5, then the probability of $A \cap B$ would have to be 50% of the size of $A$. Only then can the events be independent. Try to draw this up, it is a nice exercise.

Also - are you aware that if the size of $A \cap B$ is zero in the Venn diagramme, then the events can not be independent? Think about why.

From this point on maybe you want to draw Venn diagramms as you follow along.

Okay so far so good. Now let's see what about problem two?
Well firstly, note that these events are not independent. You are, in fact, given given the probability that both lights are red at the same time and this probability does not fit what you know about independence. For you to see the difference to the first example, let us see what can happen at that intersection. We want to know when both lights are green. What is the complement of this event? It's NOT "both lights are red". In fact the two complementary events are:

  1. Both lights are green
  2. At least one light is red

It is either option 1 or 2. Option 1 is what we are after. If we can figure out the probability of option 2, then we know that option 1 is the complement/opposite of it, right? So let's figure out event 2.

Okay so why is there another probability given in the assignment? Well think about it: Event 2 can be described differently.

  • The first light can be red, let us call this $A$
  • The second light can be red, let's call this one $B$

Okay so this is probably where you are right now. The probabilities for these two events, let's call them $P(A)$ and $P(B)$ are given: $.4$ and $.3$ We also know:

  • The first and the second light are red at the same this - this is $A \cap B$ : $.1$

How does this fit together?
Well $A$ and $B$ are not enough to calculate when both lights are green (event 1), because we will need the second event - which we called: "Either light is red" and then take its complement (you already figured this out).

Now - why can we not multiply the probabilities? First: because they are not independent. The multiplication rule can not be used because the ratio of $A \cap B$ to either $A$ or $B$ is not as it was in the first example.
Second: Because even if they were independent, then multiplication gives us the event $A \cap B$ - the event that both occur at the same time. But we want the event that either light is red.

Let's think about how this event can also be described. The name you are looking for is $A \cup B$.
What does it include? $A \cup B = $

-Light $A$ can be red AND light $B$ can be red. We will call this $A \cap B$ (we have this!)
-Light $A$ can be red and light $B$ can be green. This is called $A \cap \neg B$
-Light $A$ can be green and light $B$ can be red. This is called $\neg A \cap B$

(the $\neg$ means negation or complement) So the problem is we have $A$ and $B$ and $A \cap B$, but not the probabilities for either light beeing red, while the other is green, $A \cap \neg B$ and $\neg A \cap B$
To make this short, here is where you apply the additive rule.

You can easily see why if you look at your Venn diagramms. The difference between $A$ and $A \cap \neg B$ is what?
Well it's the part of the intersection, $A \cap B$ which is in $A$.
The same goes for $B$. So if we add $A$ and $B$ instead of $(A \cap \neg B)$ and $(\neg A \cap B)$, we have added two things too much. First: the part of the intersection that lies in $A$ and second: the part of the intersection that lies in $B$.
Go ahead and draw this in your circles.

What you want to see is: What is the difference between $$A+B$$ and $$(A \cup B) = (A \cap B) + (A \cap \neg B) + (\neg A \cap B)$$ ?

Well the difference is exactly one time the intersection. So to get the probability of our event "At least one light is red", all we need to do is to add $P(A)$ and $P(B)$ and substract the intersect $P(A \cap B)$ and we have $$P(A) + P(B) - P(A \cap B) = P(A \cup B)$$

Alright and this is the same as "At least one light is red". And what's the opposite of this? It's "Both lights are green" - the solution to the problem.


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