Conditional Probability

Conditional probability quantifies the likelihood of event E, called the conditioned event, given that event A, known as the conditioning or auxiliary event, has already occurred.

It is also commonly referred to as a posteriori probability.

When the events are dependent, the conditional probability P(E|A) is evaluated using the following formula:

$$ P(E|A) = \frac{P(E \cap A)}{P(A)} $$

Here, E represents the conditioned event, while A represents the conditioning event.

• P(E|A) denotes the conditional probability of E given A.
• P(E∩A) is the probability that both events E and A occur simultaneously, also called the joint probability.
• P(A) is the probability of the conditioning event A.

Conditional probability allows us to revise the probability of event E by incorporating the additional information that event A has already taken place.

Note. The probability of the conditioning event P(A) must be nonzero. Otherwise, the expression would involve division by zero. $$ P(A) \ne 0 $$

The two events may exhibit either positive or negative correlation.

• If P(E|A)>P(E), the events are positively correlated.
• If P(E|A)<P(E), the events are negatively correlated.

If the two events are independent, the conditional probability of one event is not influenced by the occurrence of the other. In this case, the probability of event E remains unchanged.

$$ P(E∣A)=P(E) $$

Examples. Some representative practical examples include the following.


• The probability that it rains (event E), given that it rained yesterday (event A).
• The probability of failing an exam (event E), given that a student has not studied sufficiently (event A).
• The probability of suffering a heart attack (event E), given that the patient is elderly (event A).

A Practical Example

Example 1

What is the probability of drawing an ace from a deck of cards, given that the card drawn is a heart?

In this scenario, the conditioned event E is “drawing an ace”, while the conditioning event A is “drawing a heart”.

In a standard deck of playing cards, there are 52 cards in total. Among these, there are 4 aces and 13 hearts, including the ace of hearts.

The probability of drawing a heart is P(A)=13/52.

$$ p(A) = \frac{13}{52} $$

The probability of drawing an ace is P(E)=4/52, which corresponds to 1 out of 13.

$$ p(E) = \frac{4}{52} = \frac{1}{13} $$

Next, compute the probability $ p(E∩A) $ that both events occur simultaneously.

$$ p(E∩A) = \frac{13}{52} \cdot \frac{1}{13} = \frac{1}{52} $$

Now apply the conditional probability formula:

$$ P(E|A) = \frac{P(E \cap A)}{P(A)} $$

$$ P(E|A) = \frac{ \frac{1}{52} }{ \frac{13}{52} } $$

$$ P(E|A) = \frac{1}{52} \cdot \frac{52}{13} $$

$$ P(E|A) = \frac{1}{13} $$

$$ P(E|A) = 0.0769 $$

Therefore, the probability of drawing an ace, given that the card drawn is a heart, is approximately 0.076 or 7.6%.

This result indicates that, once it is known that the extracted card is a heart, there is about a 7.7% chance that it is also an ace.

Example 2

70% of the population wears pants, while 30% wears skirts.

The probability that a random passerby is wearing pants is P(E)=70%.

$$ P(E) = 70\% $$

Where E is the event that a person wearing pants passes by.

Given that the next passerby is a woman (conditioning event A), we need to calculate the conditional probability P(E|A) of event E given A.

$$ P(E|A) = \frac{P(E \cap A)}{P(A)} $$

The probability of encountering a woman is 60%, as 6 out of 10 people are women: P(A)=0.6.

The probability of encountering a woman wearing pants (E⋂A) is 30%, as 3 out of 10 are women wearing pants: P(E⋂A)=0.3.

So, the conditional probability is:

$$ P(E|A) = \frac{0.3}{0.6} = 0.50 $$

The probability that the next passerby is wearing pants (event E), given that the person is a woman (event A), is 50%.

Example 3

Rolling two dice results in a sample space S consisting of 36 possible outcomes.

$$ (i,j) $$

Where i=1,2,3,4,5,6 represents the outcome of the first die, and j=1,2,3,4,5,6 represents the outcome of the second die.

$$ S = \begin{pmatrix} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{pmatrix} $$

If the first die shows a 4 (conditioning event A), what is the probability that the sum of the dice is 7 (conditioned event E)?

$$ P(E|A)= \frac{P(E \cap A)}{P(A)} $$

The probability P(E ∩ A) that the sum is 7 and the first die is 4 is 1/36, as the only outcome satisfying this condition is (4,3) in a sample space S with 36 total outcomes.

Thus, P(E ∩ A) = 1/36

$$ P(E|A)= \frac{\frac{1}{36}}{P(A)} $$

The probability of the conditioning event, which is rolling a 4 on the first die, is P(A) = 1/6, since the die has six faces.

$$ P(E|A)= \frac{\frac{1}{36}}{\frac{1}{6}} $$

Simplifying the calculation gives:

$$ P(E|A)= \frac{\frac{1}{36}}{\frac{1}{6}} = \frac{1}{36} \cdot \frac{6}{1} = \frac{1}{6} $$

Therefore, the conditional probability that the sum is 7 given that the first die shows a 4 is P(E|A) = 1/6.

And so on.

Proof

Consider two events $ E_1 $ and $ E_2 $ defined on the same sample space $ U $.

$$ E_1 \subset U $$

$$ E_2 \subset U $$

Each event is a nonempty subset of the sample space.

• Event $ E_1 $ contains $ n_1 = |E_1| $ outcomes.
• Event $ E_2 $ contains $ n_2 = |E_2| $ outcomes.

Assume that the intersection of the two subsets is nonempty.

$$ E_1 \cap E_2 \ne \emptyset $$

From the viewpoint of Euler - Venn diagrams, the situation can be represented as follows.

The intersection $ E_1 \cap E_2  $ consists of the favorable outcomes $ k $ of event $ E_1 $ under the condition that event $ E_2 $ has occurred.

The conditional probability of event $ E_1 $ given event $ E_2 $ is defined as the ratio between the number of favorable outcomes $ k = |E_1 \cap E_2| $ and the total number of outcomes in $ E_2 $, namely $ n_2 = |E_2| $.

$$ P(E_1|E_2) = \frac{k}{n_2} $$

Now divide both the numerator and the denominator by the total number of outcomes $ n $ in the sample space $ U $.

$$ P(E_1|E_2) = \frac{ \frac{k}{n} }{ \frac{n_2}{n} } $$

Since $ P(E_2) = \frac{n_2}{n} $ and $ P(E_1 \cap E_2)  = \frac{k}{n} $, it follows that

$$ P(E_1|E_2) = \frac{ \frac{k}{n} }{ \frac{n_2}{n} } = \frac{P(E_1 \cap E_2)}{P(E_2)} $$

This derivation leads to the fundamental formula of conditional probability.

$$ P(E_1|E_2) = \frac{P(E_1 \cap E_2)}{P(E_2)} $$

This relation shows that conditional probability measures the likelihood of event $ E_1 $ by restricting the sample space to the subset $ E_2 $.

And so on.