Conditional probability

Motivational example

Sometimes you have some information about a result of an experiment. For example, you threw two dice twice and someone told you that the first dice is greater than $4$. What is the probability that the total of two dice will be greater than $8$?

You can solve this problem with brute-force method. Overall number of results, where the first one is greater than $4$, is $12$: $(5, 1), \dots, (5, 6), (6, 1), \dots, (6,6)$ . And there are $7$ suitable results: $(5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)$. So, the probability, that the total of two dice will be greater than $8$, is $\frac{7}{12}$.

Let's compare this answer with a result in the situation when it's unknown that the first dice is greater than $4$. Firstly, we have $10$ suitable results: $(3, 6), (4, 5), (4, 6)$ and $7$ that we already mentioned. Secondly, the overall number of results is $6 \cdot 6 = 36$. The probability that the sum of points will be greater than $8$ is $\frac{10}{36} = \frac{5}{18}$.

The main conclusion is the following: if it's known that some event has occurred, you should recalculate the probability.

Formal definition

We can generalize the previous problem and consider a whole class of similar problems. It's known that an event $B$ has occurred. How do we find the probability of an event $A$? So we come to the definition of the conditional probability.

Conditional probability of the event $A$, provided that the event $B$ has occurred, is calculated by a formula:

$\mathbb{P}(A | B) = \dfrac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}$, if $\mathbb{P}(B) > 0$

The fact that the event $B$ has occurred means that the space of all possible outcomes is constricted to $B$. In this constricted space, to say that event A will occur means the same as to say that both A and B will occur. The probability of A and B occurring together is the numerator.

Now we will use the following problem to show that the conditional probability has a simple geometric interpretation.

For example, we want to find the probability of rolling a dice and obtaining a value less than $4$ knowing that the value is an even number.

The event $A$ is that the value is less than $4$ and the event $B$ is that the value is an even number. It's easy to see that the answer is $\frac{1}{3}$ on the following picture:

Problem solving using the definition

Let's solve the problem with $2$ dice using the definition of conditional probability. What is the probability that the total of two dice will be greater than $8$, given that the first dice is greater than $4$?

Let's understand what will be events $A$ and $B$ here? The event $A$ is the fact that total of two dice is greater than $8$, the event $B$ is the fact that the first dice is greater than $4$. Now we need to calculate the probability of the events $B$ and $A \cap B$.

It is easy to understand that $\mathbb{P}(B) = \frac{2}{6} = \frac{1}{3}$. $A \cap B$ means that the sum of two dice is greater than $8$ and the first one is greater than $4$. Let's fill the following table, where sum of two dice is written in $(i, j)$ cell, if the first dice has $i$ points and the second has $j$:

We see that there are $7$ suitable results and $36$ results in total, so probability $\mathbb{P}(A \cap B) = \frac{7}{36}$

The final probability is:

$\mathbb{P}(A | B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)} = \frac{\frac{7}{36}}{\frac{1}{3}} = \frac{7}{12}$

Conclusion

Now we have learned what conditional probability is, and have seen how to find it. Later we'll see why this concept is extremely useful for applications – for example, in data analysis.