#### Motivation

In the previous lesson we learned about probability of one event. However, in real life, we often encounter situations with mixed events. For example, lets say we have a bag full of fruits (green and red apples) and vegetables (tomato and carrot). The questions we could ask are:

1. Probability of picking a fruit
2. Probability of picking a red fruit or vegetable
3. Probability of getting a red fruit
4. Probability of  getting something red or a vegetable…

For situations like this, we’ll use one of the basics in probability theory: addition rule. Lets explain it in the following example.

#### Example

In a group of $126$ students $57$ are seniors, $62$ are female and $32$ are female seniors. Find the probability that a student picked from this group at random is either a senior or female.

Solution

Firstly, lets see the probability of picking a senior. There are $126$ student in total, and $57$ of them are seniors. Consequently,  $P\{$senior$\displaystyle{\}=\frac{57}{126}}$. Similarly, the probability of picking a female is $P\{$female$\displaystyle{\}=\frac{62}{126}}$. However, we can see that we also have one more information and that is a number of seniors who are female. Similarly as before, the probability of picking a female senior is $P\{$senior and female$\displaystyle{\}=\frac{32}{126}}$. Lets use Venn diagram to show given data. Notice that event $\{$senior and female$\}$ is where events $\{$senior$\}$ and $\{$female$\}$ overlap. Which we know intuitively, some female seniors belong to both groups, group senior and female.

Finally, lets find the probability of picking a female or a senior. We know that denominator is still $126$. Lets find numerator. There are $62$ females and $57$ seniors. However, we can’t just add those two numbers. If we did, we would have counted female seniors twice, since they are in both groups. Consequently, we need to subtract their intersection.

$P\{$senior or female$\displaystyle{\}=\frac{62+57-32}{126}=\frac{87}{126}}$

We can rewrite the above fraction as:

$P\{$senior or female$\displaystyle{\}=\frac{62}{126} + \frac{57}{126} -\frac{32}{126}}$

Which is equal to

$P\{$senior or female$\}=P\{$female$\}+P\{$senior$\}-P\{$senior and female$\}$

And this is what we call the addition rule.

#### Generally

Lets $A$ and $B$ be two events.
Remember,

$\bullet$ $A$ or $B= A \cup B$.
$\bullet$ $A$ and $B= A \cap B$

We want to know the probability of $A \cup B$. Firstly, lets add up their separate probabilities, $P(A) + P(B)$. Consequently, we counted the intersection twice. Since that’s not what we wanted, we have to subtract $A\cap B$. Addition rule is: $$P(A \cup B)=P(A)+P(B)-P(A\cap B)$$

#### Three sets Following the same way of thinking, we can make the addition rule for $3$ events. Firstly, we sum up probabilities of all events. Further, we’ll subtract the intersections of every pair of events. However, by doing that we would be left without the intersection of all three event so we add one back. In conclusion, the addition rule for three events is $$P(A \cap B)=P(A)+P(B)+P(C)-P(A\cup B) – P(A\cup C)-P(B\cup C)+P(A\cap B \cap C)$$

#### Example

In a group of $100$ students, $45$ of them play football, $53$ basketball and $55$ of them play tennis. Further, $28$ of them play football and basketball, $32$ play football and tennis, $35$ of them play basketball and tennis. $20$ of them play all three sports. What’s the probability that random picked student doesn’t play any sport?

Solution

Let’s say:  $F=\{$students who play football$\}$; $B=\{$students who play basketball$\}$ and $T=\{$students who play tennis$\}$. The date we were given is:

• $F= 45, B=53, T=55$
• $F \cap B=28, F \cap T=32, B\cap T =35$
• $F \cup B \cup C=100, F\cap B \cap T=20$

Additionally, lets fill in Venn diagram. We start from the intersection of three events, then we fill in the intersections of two events and so on.

We know that $F\cap B \cap T=20$ and $F\cap B=28$. Therefore, $(F\cap B)\setminus T=28 – 20=8$.
Similarly, $(F\cap T)\setminus B=32 – 20=12$ and $(B\cap T)\setminus F=35 – 20=15$. Now that we have all the intersections, let’s find the number of students who only play one sport: football, basketball or tennis. Lets start with football. The number of people who play football is $45$. However, that’s not the group of people who play only football. We have to subtract the players who play others sports besides football. Specifically, $$F- [(F \cap B)\setminus T] – [(F \cap T)\setminus B] – (F \cap B \cap T)=45-8-12-20=5$$ The number of people who only play football is $5$. Similarly,

Basketball only : $B- [(B \cap F)\setminus T] – [(B \cap T)\setminus F] – (F \cap B \cap T)=53-8-15-20=10$

Tennis only : $T- [(T \cap F)\setminus B] – [(T \cap B)\setminus F] – (F \cap B \cap T)=55-12-15-20=8$

Consequently, our Venn diagram looks like this. We can see we have one more unknown data. That’s precisely the part of the question we need to solve. To find the wanted probability we need the number or players who don’t play any sport.

If a student doesn’t play any given sport, we write it as $F^{c} \cap B^{c} \cap T^{c}$. How do we do that? We know we have $100$ students in total, so we need to subtract all the football, basketball and tennis players; all the football and basketball players.. and so on. The reason we did the Venn diagram is that we don’t count some students twice. Thanks to the diagram, we now have a nice representation of each player group. $$F^{c} \cap B^{c} \cap T^{c} = 100- (F \cap B \cap T) -F- [(F \cap B)\setminus T]-…=$$ $$=100-20-5-8-12-8-15-10=22$$

Consequently, Finally, probability of randomly selected player not playing any sports is :

$$P(F^{c} \cap B^{c} \cap T^{c})=\displaystyle{\frac{22}{100}=22\%}$$