How could we use this **Venn diagram **to find the **conditional probability** that a randomly selected student goes to debating given that they go to chess?

We know that **conditional probability** restricts the pool of students out of which we are choosing using the condition.

In this case, the condition is the student taking chess.

From the **Venn diagram**, we can see that 3 + 2 = 5 students go to chess.

So 5 is the number of **possible outcomes** for our conditional probability.

We want the probability that a student also goes to the debating club.

We can see that there are 2 students in the 'overlap', i.e. the **intersection**.

So 2 students is the number of **desired outcomes** for our conditional probability.

So 2 out of the 5 students who go to chess also go to debating.

That means the probability that a randomly selected student goes to debating given that they go to chess is 2/5!

If we call the sets of students who go to chess and debating C and D, respectively, then we can denote this probability using the symbol for given, i.e. the vertical line $∣:$

**$P(D$**$∣$**$C)$**$= 2/5$

Using the same Venn diagram, how could we find P(C$∣$C $∪ D)?$

Oh wow, C$∣$C $∪ D is$ a lot of symbols, isn't it?!

Let's decipher what it means:

**C** means **goes to chess.**

$∣$means **given.**

$∪union$

So P(**C**$∣C ∪ D) means the probability that a randomly selected student goes to chess$**given**$that they go to chess or debating (or both).$

$That's not so awful, is it?!$

$union$

We can see that there are 3 + 2 + 5 = 10 students in the two circles together.

So, $C ∪ D = 10$

There are 3 + 2 = 5 students who **go**** to chess** (and are still in our union).

So 5 out of the 10 students who $go to chess or debating (or both)$ **go to chess.**

P(**C**$∣C ∪ D)$= 5/10 = 1/2

Ready to put all this into practice?