Conditional Probability
Introduction to Global Health Data Science
Duke University
STA/GLHLTH 198 Fall 2026
2026-09-02
Conditional Probability
Often we wish to know the probabilty an event will occur given that another event has occured.
For example, instead of the marginal probability of contracting COVID (regardless of vaccination status), we may wish to know the probability that someone will contract COVID given that they have been recently vaccinated, or the probability someone will contract COVID given that they have not been recently vaccinated.
These are examples of conditional probability. The conditional probability that an ant is carried back (A) given that they are lightly injured (B) is denoted \(P(A|B)\), which we say as “probability of A given B.”
Simple Example
Suppose we have a small set containing 9 ants with light injuries carried back to the nest, 1 ant with a light injury that is not carried back, and 10 heavily-injured ants that are not carried back.
If we feel this set is a sample representative of a population of interest, we can estimate the probability an ant is carried back as \(\frac{9}{9+1+10}=0.45\).
We may be interested in the conditional probability an ant is carried back given that it is only lightly injured, which we can estimate as \(\frac{9}{9+1}=0.9\) – we just change the denominator to correspond to our smaller population of interest.
Conditional probability and multiplicative rule
More formally, we define conditional probability as \(P(A|B)=\frac{P(A \cap B)}{P(B)}\) (verify this on your own using the simple example on the prior slide).
Manipulating this formula, we get the multiplicative rule of probability: \(P(A \cap B)=P(B)P(A|B)\).
Law of total probability
One more helpful rule is the law of total probability:
\(P(A)= P(A \cap B)+P(A \cap \overline{B})=P(A \mid B)P(B) + P(A \mid \overline{B})P(\overline{B}),\) which translates to the obvious statement that the probability that A occurs is equal to the sum of the probabilities that A occurs with B and that A occurs without B.
Back to ant outcomes
Recall our termite-raiding ant data:
| Carried Back | Carried Away | Not Carried | |
|---|---|---|---|
| Lightly Injured | 45 | 0 | 5 |
| Heavily Injured | 5 | 5 | 20 |
Three Probabilities
Define events A=carried back to the nest and B=lightly injured. Calculate the following probabilities for a randomly-selected ant in this cohort.
Marginal probability of being carried back to the nest, \(P(A)\)
Joint probability of being carried back to the nest and light injury, \(P(A \cap B)\)
Conditional probability of being carried back to the nest given a light injury, \(P(A \mid B)\)
Independence
Events \(A\) and \(B\) are independent when \(P(A \mid B)=P(A)\) or \(P(B \mid A)=P(B)\).
In other words, knowing that one event has occurred does not lead to any change in the probability we assign to another event.
Checking Independence
We can use the multiplicative rule to check if two events are independent.
If events A and B are independent, then \[P(A \cap B)=P(A)\times P(B).\]
Check it out: Are light injury and being carried back to the nest independent in our cohort?
Food for thought: Is this result surprising to you? Why or why not?
Independent vs Disjoint Events
For independent events \(A\) and \(B\), \(P(A \mid B)=P(A)\) and \(P(B \mid A)=P(B)\), so knowing one event occurred tells us nothing about the chances the other event will occur.
For two disjoint or mutually exclusive events, knowing that one event has occurred tells us that the other event definitely has not occurred, e.g. \(P(A \cap B)=0\).
Disjoint events are therefore not independent!
Relative Risk
Relative risk is a useful concept for comparing the probabilities of an outcome event across groups. Sometimes we think of these groups as exposed and unexposed, but we can think of any two groups.
The relative risk or RR is the probability a member of the exposed group will experience the outcome relative to the probability a member of the unexposed group will experience the outcome.
\[RR = \frac{P(\text{outcome}|\text{exposed})}{P(\text{outcome}|\text{unexposed})}\] A relative risk of 1.0 indicates that the probabilities of the outcomes in the two groups are identical.
Note that the value of the relative risk is independent of the size of the probabilities – it is important to bear in mind that a RR=6 may not be that dramatic if the probability of an event is near zero in both groups.
Termite-Raiding Ants: Relative Risk
Recall our termite-raiding ant data:
| Carried Back | Carried Away | Not Carried | |
|---|---|---|---|
| Lightly Injured | 45 | 0 | 5 |
| Heavily Injured | 5 | 5 | 20 |
What is the relative risk of not being carried back among those heavily injured, relative to that among those who are lightly injured?
\[RR = \frac{P(\text{outcome}|\text{exposed})}{P(\text{outcome}|\text{unexposed})}\]
\[RR = \frac{P(\text{not carried back}|\text{heavily injured})}{P(\text{not carried back}|\text{lightly injured})}\]
\[RR = \frac{\frac{5+20}{5+5+20}}{\frac{5}{45+5}}=\frac{\frac{25}{30}}{\frac{5}{50}}=\frac{25}{30}\times \frac{50}{5}=\frac{25}{3}=8\frac{1}{3}\]
What would you say about the relationship between injury level and the risk of not being carried back?
Example: Breast Cancer Screening
Mammogram result
Let \(A\) be the event that a woman has breast cancer (e.g., prevalence in population for a certain age group). Say \(P(A)=0.01\) for a 40-year-old woman.
Let \(B\) be the event that a screening mammogram is positive.
Once a person has a positive mammogram, our mental estimate of the probability she has breast cancer, now \(P(A|B)\), has increased. How much should it increase? Are we certain she has cancer, e.g. P(cancer \(\mid\) mammo positive)=1, or is there some chance the test is wrong?
Sensitivity and Specificity
A=has cancer B=mammogram positive
A diagnostic test like a mammogram is often characterized by its quality – we want a test to have good sensitivity (picking up cancer when a person really has it) and specificity (ruling out cancer when a person is cancer-free).
Sensitivity is \(P(B|A)\), and specificity is \(P(\overline{B} \mid \overline{A})\).
A typical screening mammogram has sensitivity of 85% and specificity of 90%.
Sensitivity and Specificity
Increasing sensitivity usually comes at the expense of specificity, and increasing specificity usually comes at the expense of sensitivity. Choosing the optimal balance depends on the consequences of false positives versus false negatives.
- Prioritize sensitivity when
- the disease is serious or highly contagious
- early detection improves outcomes
- a safe, cheap confirmatory test is available
- examples: in-hospital COVID screening, blood donor screening
- Prioritize specificity when
- false positives have severe consequences
- examples: Tour de France drug testing, forensic DNA testing
Bayes’ Theorem
Bayes’ theorem gives us a formal way to update our beliefs based on new information. It says \(P(A \mid B)=\frac{P(B \mid A)P(A)}{P(B)}=\frac{P(A \cap B)}{P(B)}\).
In this example, a 40 year old woman with a positive screening mammogram may wish to know her chances of having cancer. Several papers have shown that even doctors tend to strongly overestimate her chances of having cancer.
We’ll consider two ways to solve this problem: one way using Bayes’ formula directly, and another based on a “hypothetical 10000” table, which applies known probabilities to a hypothetical population of 10,000 40 year old women.
Hypothetical Population of 10,000
We are still using Bayes’ theorem, but it is hidden behind the scenes.
Here’s what we know:
- The prevalence of breast cancer among 40-year-old women is 1%: \[P(\mathrm{cancer})=0.01\]
- The sensitivity of a screening mammogram is 85%
- The specificity of a screening mammogram is 90%.
Let’s construct a 2 × 2 table comparing true cancer status and mammogram results in our hypothetical population of 10,000 women.
| Cancer | No Cancer | Total | |
|---|---|---|---|
| Mammo + | |||
| Mammo − | |||
| Total | 10,000 |
Hypothetical Population of 10,000
- The prevalence of breast cancer among 40-year-old women is 1%: \[P(\mathrm{cancer})=0.01\]
- The sensitivity of a screening mammogram is 85%
- The specificity of a screening mammogram is 90%.
Item 1 says the prevalence in this group is 1%, so then we expect to have \(10000\times0.01=100\) cases and \(10000\times0.99=9900\) cancer-free women.
| Cancer | No Cancer | Total | |
|---|---|---|---|
| Mammo + | |||
| Mammo − | |||
| Total | 100 | 9900 | 10,000 |
Hypothetical Population of 10,000
- The prevalence of breast cancer among 40-year-old women is 1%: \[P(\mathrm{cancer})=0.01\]
- The sensitivity of a screening mammogram is 85%
- The specificity of a screening mammogram is 90%.
Item 2 gives the sensitivity, so \(P(\mathrm{mammo +} \mid \mathrm{cancer})=0.85\).
Thus in the group of 100 women with cancer, the mammogram should pick up \(100\times0.85=85\) of them, and miss the remaining \(100-85=15\).
| Cancer | No Cancer | Total | |
|---|---|---|---|
| Mammo + | 85 | ||
| Mammo − | 15 | ||
| Total | 100 | 9900 | 10,000 |
Hypothetical Population of 10,000
- The prevalence of breast cancer among 40-year-old women is 1%: \[P(\mathrm{cancer})=0.01\]
- The sensitivity of a screening mammogram is 85%
- The specificity of a screening mammogram is 90%.
Item 3 gives the specificity, so \(P(\mathrm{mammo -} \mid \mathrm{no cancer})=0.90\).
Thus in the group of 9900 women without cancer, the mammogram should correctly identify \(9900*0.90=8910\) of them as being cancer-free, and it will mistakenly identify \(9900-8910=990\) as having cancer (false positives).
| Cancer | No Cancer | Total | |
|---|---|---|---|
| Mammo + | 85 | 990 | |
| Mammo − | 15 | 8910 | |
| Total | 100 | 9900 | 10,000 |
Hypothetical Population of 10,000
- The prevalence of breast cancer among 40-year-old women is 1%: \[P(\mathrm{cancer})=0.01\]
- The sensitivity of a screening mammogram is 85%
- The specificity of a screening mammogram is 90%.
Now we complete the table by filling in the row totals.
| Cancer | No Cancer | Total | |
|---|---|---|---|
| Mammo + | 85 | 990 | 1075 |
| Mammo − | 15 | 8910 | 8925 |
| Total | 100 | 9900 | 10,000 |
At this point, it’s easy to calculate the conditional probability of cancer given a positive mammogram as \(\frac{85}{1075}=0.079\).
Bayes’ Theorem in Action
Alternatively, we could just use Bayes’ Theorem directly.
Baseline probability of cancer \(P(A)=0.01\) (prevalence)
She wants to know \(P(A \mid B)\), or her chances of having cancer given that her mammogram is positive (also called positive predictive value).
Bayes’ Theorem: \(P(A \mid B)=\frac{P(B \mid A)P(A)}{P(B)}\).
Sensitivity is \(P(B \mid A)=0.85\).
How do we get \(P(B)\)?
Bayes’ Theorem in Action
How do we get \(P(B)\)? We can get this using the law of total probability: \(P(B)=P(B \mid A)P(A) + P(B \mid \overline{A})P(\overline{A})\).
We can get \(P(B \mid \overline{A})\) using the specificity \(P(\overline{B} \mid \overline{A})=0.90\) and the fact that \(P(\overline{B} \mid \overline{A})+P(B \mid \overline{A})=1\). So \(P(B \mid \overline{A})=1-0.9=0.1\).
Then \(P(B)=P(B \mid A)P(A) + P(B \mid \overline{A})P(\overline{A})=0.85\times0.01+0.1\times0.99=0.1075\)
Then \(P(A \mid B)=\frac{P(B \mid A)P(A)}{P(B)}=\frac{0.85*0.01}{0.1075}=0.079\).
Bayes’ Theorem and Baseline Prevalence
Here, we can think of the 1% prevalence of breast cancer among 40 year old women as our prior probability a woman has cancer, and 7.9% as the posterior probability she has cancer after we see the data that her screening mammogram is positive.
Bayesian statistics is a framework for updating our beliefs about an event or hypothesis as new evidence emerges.
In this example, cancer would be confirmed or ruled out by subsequent medical care, such as a diagnostic mammogram, ultrasound, and/or biopsy.
You Try It!
Most people who have a negative test result (e.g., mammogram looks good or COVID test negative) don’t worry any longer about whether they really do have disease. Are they right not to worry? Suppose our 40 year old woman with a baseline 1% breast cancer risk instead had a negative (all clear) mammogram. What is the updated probability she has breast cancer given this test result?
You Try It!
The prevalence of cancer is strongly dependent on age. Calculate the conditional probability of cancer given a positive mammogram for women in the following age groups.
- Women in their 20’s, for whom cancer prevalence is roughly \(\frac{1}{10000}\)
- Women in their 50’s, for whom cancer prevalence is roughly \(\frac{1}{50}\)
- Women in their 60’s, for whom cancer prevalence is roughly \(\frac{1}{30}\)
What is the relative risk of breast cancer comparing women in their 60’s to women in their 40’s? What about comparing women in their 40’s to women in their 20’s?
What other factors might influence this conditional probability (that is, are we happy with our assumptions here)?
More practice: Vaccines and COVID
A study examined 85,725 emergency department and urgent care encounters involving immunocompetent adults with COVID-19-like symptoms.
9659 had received the 2025-2026 COVID-19 vaccine
3941 tested positive for SARS-CoV-2
206 were both vaccinated and SARS-CoV-2 positive
Are having received the vaccine and SARS-CoV-2 positivity mutually exclusive? Explain why or why not and use data to support your answer.
What is the relative risk of SARS-CoV-2 positivity comparing those with and without the 2025-2026 vaccine?
Are vaccination status and SARS-CoV-2 positivity independent events in this sample? Use additional data to support your answer.
Do you think these results are generalizable to a broader population? Why or why not?
Even more practice
Pagano, Gauvreau, and Mattie
Section 5.6, problems 3, 4, 6, 8, 13
Section 6.8, problems 1, 6, 7 (a and b only)