Using the addition rule, and union vs. intersection

 
 
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What is the addition rule, and when does it apply?

Sometimes we’ll need to find the probability that two events occur together within one experiment. Remember that an event is a specific collection of outcomes from the sample space.

For example, what’s the probability that we roll a pair of 66-sided dice and either get at least one 11, or an even sum when we add the dice together?

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When we roll two dice together, there are 3636 possible outcomes. There are 1111 rolls out of the 3636 where we get at least one 11:

 
sample space for dice rolls that include a 1
 

And there are 1818 possible outcomes where the sum of the dice is even.

 
sample space when the sum of the dice is even
 

So we might be tempted to say that the probability of getting at least one 11 or an even sum is P(1 or even)=(11+18)/36P(1\text{ or even})=(11+18)/36, or 29/3629/36. But we’ve neglected to consider that there’s some overlap between these two sets. We have the rolls 111-1, 131-3, 151-5, 313-1, and 515-1 in both sets, so we’re double-counting those in our probability calculation.

Which means we have to subtract out the values that are overlapping. Since there are 55 overlapping values, the probability calculation is actually

P(1 or even)=11+18536P(1\text{ or even})=\frac{11+18-5}{36}

P(1 or even)=2436P(1\text{ or even})=\frac{24}{36}

P(1 or even)=23P(1\text{ or even})=\frac23

A great way to illustrate this kind of overlapping probability is with a Venn diagram. We would build a Venn diagram to show that there are 1111 rolls where we get at least one 11, that there are 1818 rolls where the sum is even, and that there are 55 rolls where we get at least one 11 and the sum is also even.

 
the intersection of the sample spaces
 

Then, from the Venn diagram, we just add the 6+5=116+5=11 and the 5+13=185+13=18, and then subtract the overlapping 55, in order to get all of the outcomes that meet our criteria, but without double-counting any of the outcomes. And then our probability again is

P(1 or even)=11+18536=2436=23P(1\text{ or even})=\frac{11+18-5}{36}=\frac{24}{36}=\frac23


Addition rule

This idea of making sure that we don’t double-count the overlap is called the addition rule (or sum rule) for probability, and it’s given as:

P(A or B)=P(A)+P(B)P(A and B)P(A \text{ or } B)=P(A)+P(B)-P(A\text{ and }B)

Now what happens if there’s no overlap between AA and BB? In that case, AA and BB are called mutually exclusive (or disjoint), and P(A and B)P(A\text{ and }B) will be 00. Which means the addition rule will simplify this way:

P(A or B)=P(A)+P(B)P(A and B)P(A \text{ or } B)=P(A)+P(B)-P(A\text{ and }B)

P(A or B)=P(A)+P(B)0P(A \text{ or } B)=P(A)+P(B)-0

P(A or B)=P(A)+P(B)P(A \text{ or } B)=P(A)+P(B)

Which tells us that when events are mutually exclusive/disjoint, we can calculate the probability of either event AA happening or event BB happening simply by adding together the probability of each one happening individually.

For instance, the events in this Venn diagram are disjoint, since the circles don’t overlap:

 
venn diagram for disjoint events
 

Because there are 10+7=1710+7=17 total events, the probability of event AA is P(A)=10/17P(A)=10/17. And the probability of event BB is P(B)=7/17P(B)=7/17. So the probability that both events occur is

P(A or B)=P(A)+P(B)P(A\text{ or }B)=P(A)+P(B)

P(A or B)=1017+717P(A\text{ or }B)=\frac{10}{17}+\frac{7}{17}

P(A or B)=1717P(A\text{ or }B)=\frac{17}{17}

P(A or B)=1P(A\text{ or }B)=1

Union and intersection

In the first version of the addition rule formula, we use the words “or” and “and.” But we can also write the formula as:

P(AB)=P(A)+P(B)P(AB)P(A\cup B)=P(A)+P(B)-P(A\cap B)

This second formula is the same addition rule calculation, but we use the \cup and \cap symbols instead of the words “and” and “or.”

P(AB)P(A\cup B) is called the union of AA and BB, and it means the probability of either AA or BB or both occurring. P(AB)P(A\cap B) is called the intersection of AA and BB, and it means the probability of  AA and BB both occurring.

 
 

Using the addition rule to calculate probability


 
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Using the addition rule to calculate probabilities from a two-way data table

Example

We surveyed 100100 people about their favorite sport, and recorded their gender and favorite sport in a table.

gender vs. favorite sport
  1. What is the probability that a participant is male?

  2. What is the probability that a participant’s favorite sport is football?

  3. What is the probability that a participant is female or prefers a sport other than football or basketball?

We know from the table that 6060 of the 100100 participants are male, so the probability that a participant is male is

P(male)=60100=35P(\text{male})=\frac{60}{100}=\frac35

And from the table we can see that 3838 of the 100100 participants like football best, so the probability that a participant’s favorite sport is football is

P(football)=38100=1950P(\text{football})=\frac{38}{100}=\frac{19}{50}

The addition rule, and union vs. intersection for Probability and Statistics.jpg

This idea of making sure that we don’t double-count the overlap is called the addition rule (or sum rule) for probability.

These were both simple probability questions, but the third question requires us to use the addition rule. There are 4040 female participants, and 4141 participants who prefer a sport other than football or basketball.

But there are 1616 participants in the “overlap” group: the group of females who also prefer a sport other than football or basketball. Therefore, we’ll apply the addition rule and say that the probability that a participant is female or likes a sport other than football and basketball is

P(female or other)=40+4116100=65100=1320P(\text{female or other})=\frac{40+41-16}{100}=\frac{65}{100}=\frac{13}{20}

 
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