The conditional probability It is the possibility of the occurrence of a certain event, given that another occurs as a condition. This additional information may (or may not) modify the perception that something will happen..
For example, we can ask ourselves: "What is the probability that it will rain today, given that it has not rained for two days?" The event for which we want to know the probability is that it rains today, and the additional information that would condition the answer is that "it has not rained for two days".
Be a probabilistic space composed of Ω (sample space), ℬ (the random events) and P (the probability of each event), plus the events A and B that belong to ℬ.
The conditional probability that A occurs, given that B occurred, denoted as P (A│B), is defined as follows:
P (A│B) = P (A∩B) / P (B) = P (A and B) / P (B)
Where: P (A) is the probability of occurrence of A, P (B) is the probability of event B and is different from 0, and P (A∩B) is the probability of the intersection between A and B, that is, , the probability that both events occur (joint probability).
This is an expression for Bayes' theorem applied to two events, proposed in 1763 by the English theologian and mathematician Thomas Bayes.
Article index
-Every conditional probability is between 0 and 1:
0 ≤ P (A│B) ≤ 1
-The probability that event A occurs, given that such event occurs, is obviously 1:
P (A│A) = P (A∩A) / P (A) = P (A) / P (A) = 1
-If two events are exclusive, that is, events that cannot happen simultaneously, then the conditional probability that one of them happens is 0, since the intersection is null:
P (A│B) = P (A∩B) / P (B) = 0 / P (B) = 0
-If B is a subset of A, then the conditional probability is also 1:
P (B│A) = P (A∩B) / P (A) = 1
Important
P (A│B) is generally not equal to P (B│A), therefore we must be careful not to interchange the events when finding the conditional probability.
Many times you want to find the joint probability P (A∩B), rather than the conditional probability. Then, by means of the following theorem we have:
P (A∩B) = P (A and B) = P (A│B). P (B)
The theorem can be extended for three events A, B and C:
P (A∩B∩C) = P (A and B and C) = P (A) P (B│A) P (C│A∩B)
And also for various events, such as A1, TOtwo, TO3 and more, it can be expressed as follows:
P (A1∩ Atwo ∩ A3… ∩ An) = P (A1). P (Atwo│A1). P (A3│A1∩ Atwo) ... P (An││A1∩ Atwo∩… An-1)
When it is the case of events that occur in sequence and through different stages, it is convenient to organize the data in a diagram or a table. This makes it easier to visualize the options to reach the requested probability..
Examples of this are tree diagram and the contingency table. From one of them you can build the other.
Let's look at some situations in which the probabilities of one event are altered by the occurrence of another:
Two types of cakes are sold in a sweet shop: strawberry and chocolate. When registering the preferences of 50 clients of both sexes, the following values were determined:
-27 women, of which 11 prefer strawberry cake and 16 chocolate cake.
-23 men: 15 choose chocolate and 8 strawberry.
The probability that a customer chooses a chocolate cake can be determined by applying Laplace's rule, according to which the probability of any event is:
P = number of favorable events / total number of events
In this case, out of 50 customers, a total of 31 prefer chocolate, so the probability would be P = 31/50 = 0.62. That is, 62% of customers prefer chocolate cake.
But would it be different if the client is a woman? This is a case of conditional probability.
Using a contingency table like this, the totals are easily displayed:
Then the favorable cases are observed and Laplace's rule is applied, but first we define the events:
-B is the "female client" event.
-A is the event "prefer chocolate cake" being a woman.
We go to the column labeled "women" and there we see that the total is 27.
Then the favorable case is sought in the "chocolate" row. There are 16 of these events, therefore the probability sought is, directly:
P (A│B) = 16/27 = 0.5924
59.24% of female customers prefer chocolate cake.
This value coincides when we contrast it with the initially given definition of conditional probability:
P (A│B) = P (A∩B) / P (B)
We make sure using Laplace's rule and the values of the table:
P (B) = 27/50
P (A and B) = 16/50
Where P (A and B) is the probability that the customer prefers chocolate and is a woman. Now the values are substituted:
P (A│B) = P (A and B) / P (B) = (16/50) / (27/50) = 16/27 = 0.5924.
And it is proven that the result is the same.
In this example the rule of multiplication applies. Suppose there are pants in three sizes on display in a store: small, medium, and large..
In a lot with a total of 24 pants, of which there are 8 of each size and all are mixed, what would be the probability of extracting two of them and that both were small?
It is clear that the probability of removing a small pants on the first attempt is 8/24 = 1/3. Now, the second extraction is conditioned to the first event, since when removing a pair of pants, there are no longer 24, but 23. And if a small pants are removed, there are 7 instead of 8.
Event A is pulling one small pants, having pulled another one on the first try. And event B is the one with the small pants first time. Therefore:
P (B) = 1/3; P (A│B) = 7/24
Finally, using the multiplication rule:
P (A∩B) = (7/24). (1/3) = 7/72 = 0.097
In a study of punctuality on commercial air flights, the following data is available:
-P (B) = 0.83, is the probability that a plane takes off on time.
-P (A) = 0.81, is the probability of landing on time.
-P (B∩A) = 0.78 is the probability that the flight arrives on time taking off on time.
It is asked to calculate:
a) What is the probability that the plane will land on time given that it took off on time?
b) Is the previous probability the same as the probability that he left on time if he managed to land on time??
c) And finally: what is the probability that it will arrive on time given that not came out on time?
To answer the question, the definition of conditional probability is used:
P (A│B) = P (A∩B) / P (B) = P (A and B) / P (B) = 0.78 /0.83 = 0.9398
In this case the events in the definition are exchanged:
P (B│A) = P (A∩B) / P (A) = P (A and B) / P (A) = 0.78 /0.81 = 0.9630
Note that this probability is slightly different from the previous one, as we previously pointed out.
The probability of not leaving on time is 1 - P (B) = 1 - 0.83 = 0.17, we will call it P (BC), because it is the complementary event to take off on time. The conditional probability sought is:
P (A│BC) = P (A∩BC) / P (BC) = P (A and BC) / P (BC)
On the other hand:
P (A∩BC) = P (landing on time) - P (landing on time and take off on time) = 0.81-0.78 = 0.03
In this case the conditional probability sought is:
P (A│BC) = 0.03 / 0.17 = 0.1765
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