Probability
I believe that we do not know anything for certain, but everything probably.
- Christiaan Huygens, Oeuvres Completes

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Conditional Probabilities

During the Coronavirus outbreak, there was a great deal of discussion of mortality rates. One article reporting on the fatality rate included these lines:
  1. "The fatality rate for people infected with the novel coronavirus is estimated to be less than 1%" and
  2. "Among those whose infections cause them to become sickened by the disease known as COVID-19, the fatality rate is 1.38%."

What is the difference between the two statements? In the first, the fatality rate is reported for all people infected with the virus whereas in the second, that population is restricted to those who not only contract the virus but are sickened by it. Restricting the sample space to a particular event is called 'conditioning'.

Statement 2 above could be restated as "the probability that a person infected with the virus dies, conditional on them being sickened, is 0.0138." This is called a conditional probability.

The conditional probability of event A given B is the probability that event A occurs given that event B occurs. We write P(A|B) and say, the probability of "A given B" or "A conditional on B".


The conditional probability of event A given B is the probability that event A occurs given that event B occurs.


NOTATION: P(A|B) indicates the probability of event A given event B.


Other conditional probabilities related to the virus abound. In that same article it was reported that

Conditional probabilities are common in other areas as well. A political scientist might consider the probability that a person votes given that he is a republican. A computer tech might wonder about the probability that malware is detected given that a certain software is installed on a computer.

When thinking about a conditional probability, the sample space is restricted to a smaller set possessed of a specified characteristic.


Example: Compare the following questions. When rolling a standard die
To answer the first question,divide the number of 6's, 1, by the total number of possible outcomes, 6. The probability is 1/6.

To answer the second question, restrict possible outcomes to even numbers. Divide the number of 6's, 1, by the total number even outcomes, 3. The probability is 1/3.


Alt text: Two Venn diagrams compare ( P(A) ) and ( P(A|B) ). On the left, a rectangle labeled S contains two overlapping circles labeled A and B. The shaded region includes all of A, including the intersection ( A \cap B ), representing ( P(A) ). On the right, only circle B is shown, with the shaded region limited to the overlap ( A \cap B ), representing ( P(A|B) ), or the probability of A given B. The overlap is highlighted in green.

The figure above suggests the formula for finding P(A|B): divide the probability of the intersection of events A and by by the probability of event B.


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



A probability diagram illustrating the difference between 𝑃(𝐴)P(A) and 𝑃(𝐴∣𝐵)P(A∣B). At the top left, a rectangle labeled S contains two overlapping circles labeled A and B. The entire area of A is shaded, representing ( P(A) = \frac{P(A)}{P(S)} ). To the right, only the overlap between A and B, labeled ( A \cap B ), is shaded inside circle B. This represents ( P(A|B) = \frac{P(A \cap B)}{P(B)} ). In the lower part of the image, the same Venn diagram includes numerical values. The area inside A only is labeled 0.2, the overlap ( A \cap B ) is labeled 0.1, the area inside B only is labeled 0.5, and the area outside both circles is labeled 0.2. The equation below shows ( P(A) = \frac{0.3}{1} = 0.3 ), meaning that the total probability of A is 0.3. On the right, only circle B is shown with the overlap labeled 0.1 and the rest of B labeled 0.5. The equation ( P(A|B) = \frac{0.1}{0.1 + 0.5} = 0.167 ) shows the probability of A given B. An orange arrow points from the full Venn diagram to the conditional diagram in both the top and bottom sections, visually emphasizing the difference between total probability and conditional probability.

Example: The internet and social media have growing effects on the way people interact including in their search for romantic partners. According to research from the Pew Research Center, about 10% of American adults have never been married and have used an online dating app. If 20% of American adults have never married, what is the probability that an American adult has used an online dating app given that they have never married?

Let A denote the event that an American adult has used an online dating app and let B denote the event that they have never married. $P(B)=0.2$ and $P(A\cap B) = 0.1$.

The probability that an American adult has used an online dating app given that they have never married is $P(A|B)$.

$$\small{\begin{array}{rcl} P(A|B) & = & \frac{P(A\cap B)}{P(B)} \\ & = & \frac{0.1}{0.2} \\ & = & 0.5 \end{array}}$$
The probability that an American adult has used an online dating app given that they have never married is 0.5.

Let A be the event that a participant is overweight and let B be the event that they drink sugary beverages.

The probability that a person was overweight (event A) given that they drank sugary beveragesis P(A|B).

$$\small{\begin{array}{rcl} P(A|B) & = & \frac{P(A\cap B)}{P(B)} \\ & = & \frac{0.297}{0.4}\\ & = & 0.743\end{array}}$$
The percentage of overweight participants who drank sugary beverages is analogous to the probability that a participant drank sugary beverages given that they were overweight, P(B|A).

$$\small{\begin{array}{rcl} P(B|A) & = & \frac{P(A\cap B)}{P(A)} \\ & = & \frac{0.297}{0.544} \\ & = & 0.546\end{array}}$$
54.6% of overweight participants drank sugary beverages.