| Conditional Probabilities |
|---|
Consider a sample space Ω of events. Suppose that two experiments are to be performed that result in two events A and B. We shall study how the probability of one event impacts the probability of another event. We learn in the previous section that for two independent events occurence of one of them does not change the probability of observing another one. However, when two events are dependent, then knowledge of observing one of them effects the probability of another one. If you know that event B has already occured, then the outcome of the experiment is one of those included in B. How should we change the probabilities of the remaining events? To evaluate the probability that A will occur, we must consider the set of those outcomes in B that also result in the occurence of A. We shall call the new probability for an event A the conditional probability of A given B and denote it by Pr[A | B]. So we need a strict definition, which belongs to Andrey Kolmogorov (1903--1987).
Given two events A and B, from the sigma-algebra of a probability space, with the unconditional probability of B (that is, of the event B occurring) being greater than zero – Pr(B) > 0 – the conditional probability of A given B is defined as the quotient of the probability of the joint of events A and B, and the probability of B
Some authors, such as Bruno de Finetti, prefer to introduce conditional probability as an axiom of probability:
\[
\Pr [A | B ] = \mbox{Pr}_B [A ] = \Pr \left( A | B \right) =
\frac{\Pr [A \cap B]}{\Pr[B]} ,
\]
where Pr[A ∩ B] is the probability that both events
A and B occur.
\[
\Pr \left[ A \cap B\right] = \Pr [A | B ] \,\Pr[B] ,
\]
Although mathematically equivalent, this may be preferred philosophically;
under major probability interpretations such as the subjective theory, conditional probability is considered a primitive entity.
The following properties of conditional probability follow immediately from definition:
- Pr[ω | ω] = 1 , Pr[∅ | ω] = 0,
- Pr[A | B] = 1 if B ⊊ A,
- Pr[A1∪A2 | B] = Pr[A1 | B] + Pr[A2 | B] .
\[
\frac{\mbox{Pr}_B \left[ A_1 A_2 \right]}{\mbox{Pr}_B \left[ A_1 \right]\,
\mbox{Pr}_B \left[ A_2 \right]} \ge \frac{\mbox{Pr}_C \left[ A_1 A_2 \right]}{\mbox{Pr}_C \left[ A_1 \right]\,
\mbox{Pr}_C \left[ A_2 \right]} .
\]
The events A1 and A2 are equally
associated conditionally under B and C if equality holds in the
above inequality.
Marginal probability is the probability of a single event without
consideration of any other event. Marginal probability is also called
simple probability.
▣
Example:
Suppose all 250 employees of a company were asked whether they do not mind to
consume genetically modified food (GMO for short of genetically modified organisms) or against them. The following table gives a two-way classification of the responses of these 250 employees.
All 250 employees are divided into two categories regarding their gender and are classified according their opinion. Each box that contains a number is called a cell.
| In favor | Against | Total | |
| Male | 38 | 112 | 150 |
| Female | 10 | 90 | 100 |
| Total | 48 | 202 | 250 |
Suppose that one employee is selected at random from these 250 employees. This employee may be classified either on the basis of gender alone or on the basis of opinion. If only one characteristic is considered at a time, the employee selected can be a male, a female, in favor of GMO, or against. The probability of each of these four characteristics or events is called marginal probability.
\[
\Pr\left[ \mbox{Female} \right] = \frac{\mbox{Number of females}}{\mbox{Total number of employees}} = \frac{100}{250} = \frac{2}{5} = 0.4 .
\]
Similarly, we obtain marginal probabilities:
\begin{align*}
\Pr \left[ \mbox{male} \right] &= 150/250 =0.6 ,
\\
\Pr \left[ \mbox{in favor} \right] &= 48/250 = 0.192 ,
\\
\Pr \left[ \mbox{against} \right] &= 202/250 = 0.808 .
\end{align*}
Now suppose that one employee is selected at random from these 250
employees. Furthemore, assume it is known that this (selected) employee is a
female. In other words, the event that the employee selected is a female has
already occured. Then the conditional probability of event "in favor" is
\[
\Pr\left[ \mbox{In favor given that she is female} \right] =
\frac{\mbox{Number of in favor among females}}{\mbox{Number of females}}
= \frac{10}{100} = 0.1.
\]
So this conditional probability is almost twice less that its marginal
probability of event "in favor."
■
Example:
An experiment consists of rolling a die once. Let X
be the outcome. Let A be the event { X = 5 }, and let
B be the event{ X > 3 }. We assign the uniform
distribution function Pr[ω] = 1/6, for ω = 1, 2, ..., 6. Thus,
Pr[A] = 1/6. Now suppose that the die is rolled and we are told that
the event B has occurred. This leaves only three possible outcomes:
4, 5, and 6. In the absence of any other information, we would still regard
these outcomes to be equally likely, so the probability of A
becomes
\[
\Pr [A | B ] = \frac{\Pr [A \cap B]}{\Pr[B]} = \frac{1/6}{1/2} = \frac{1}{3} ,
\]
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Example:
■
Example:
US life expectancy drops for second year in a row, but women can still expect
to live longer than men: 81.1 years vs. 76.1 year in 2016 (according to
U.S. Department of Health and Human Services).
Suppose one finds that in a population of 100,000 females, 89% can expect to live to age 60, while 57% can expect to live to age 80. Given that a woman is 60, what is the probability that she lives to age 80?
This is an example of a conditional probability. In this case, the original sample space can be thought of as a set of 100,000 females. The events A and B are the subsets of the sample space consisting of all women who live at least 60 years, and at least 80 years, respectively. We consider A to be the new sample space, and note that B is a subset of A. Thus, the size of A is 89,000, and the size of B is 57. So, the probability in question equals 57/89 ≈ 0.640449. Thus, a woman who is 60 has a 64% chance of living to age 80.
Example:
Out of many countries that participate in the World Cup championship since 1930, only countries from Europe and American continent where champions. So the probability to win for a European country and for American country will be (based on data of 21 championships, 1930--2018) are
\[
\Pr \left[ E \right] = \frac{13}{21} \approx 0.619048 \qquad \mbox{and}\qquad
\Pr \left[ A \right] = \frac{8}{21} \approx 0.380952,
\]
respectively. However, only twice a European country won FIFA championship when
competition was held at American continent, and correspondingly American
team became a champion in Europe. Therefore, the conditional probability of
European team to win in soccer championship that is held in Europe is
\[
\Pr \left[ \mbox{win}\,|\,E \right] = \frac{8}{11} \approx 0.727273 ,
\]
Correspondingly, the conditional probability to become a champion for American
team if competition is held in America is
\[
\Pr \left[ \mbox{win}\,|\,A \right] = \frac{6}{8} = 0.75 .
\]
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Example:
Consider families with two children, assuming that all four gender
distributions---BB, BG, GB, and GG---are equally likely. Here notation "BG"
stands for older boy and younger girl in a family. Actually, this problem is
equivalent for tossing a fair coin twice.
Let A be the event that both children in the family are boys. Since A consists of one event, MM, its probability is Pr[A] = 1/4. Let B be the event that a family has at least one boy. Then its priobability becomes Pr[B] = Pr[{BB, BG, GB}] = 3/4. Let C be the event that the older chilld is a boy. Then Pr[C] = Pr[{BB, GB}] = 2/4 = 1/2.
Calculating the conditional probabilities, we get
\begin{align*}
\Pr\left[ A\,|\,B\right] &= \frac{\Pr [A \cap B]}{\Pr [B]} = \frac{1/4}{3/4} =
\frac{1}{3} ,
\\
\Pr\left[ A\,|\,C\right] &= \frac{\Pr [A \cap C]}{\Pr [C]} = \frac{1/4}{2/4} =
\frac{1}{2} .
\end{align*}
Among all two-child families with a male older child, half have a younger boy.
However, among all two-child families with at least one boy, only one-third are
BB families, one-third are GB, and one-third are BG.
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