A probability is the chance that a defined event will occur. Given two events A and B, we often want to determine the probability that one or both of them will occur. It is quantified as a positive number between 0 (the event is impossible) and 1 (the event is certain). Therefore, the higher the probability of a given event, the more likely it is to occur. If A is a definite event, the probability that A occurs is expressed as P(A). In this article we will provide you the information about Elementary Probability Theory.
Probability can be expressed in various ways. A frequentist approach consists in observing a number of specific events out of a total number of events. Thus, we could say that the probability of being born in a period of one year is 0.52, because of a large number of single births we observe that 52% are boys. A model-based approach is one in which a model or mechanism determines the event. In case there are 6 possibilities of a given event, the probability of “1” is 1/6, each one of them equally probable and they all add up to one. An opinion-based approach is one in which we use our past experience to predict a future event, so we might predict the probability that our favorite soccer team will win the next game, or that it will rain tomorrow.
addition rule
The addition rule is used to determine the probability that at least one of two (or more) events will occur. In general, the probability of an event A or B is given by:
P(A or B) = P(A) + P(B) – P(A and B)
If A and B are mutually exclusive, it means that they cannot occur together, that is, P(A and B)=0. Therefore, for mutually exclusive events, the probability of A or B occurring is given by:
P(A or B) = P(A) + P(B)
Example: If event A is that a person is of blood type O and event B is that they are of blood type B, these events are mutually exclusive, since a person can only be one or the other. Therefore, the probability that a given person is from group O or B is P(A)+P(B).
multiplication rule
The multiplication rule gives the probability that two (or more) events will occur together. In general, the probability that the two events A and B occur is given by:
P(A and B) = P(A) x P(B|A) = P(B) x P(A|B)
The notation P(B|A) is the probability that event B occurs given that event A has occurred, where the symbol ‘|’ it reads ‘given’. This is an example of conditional probability, the condition being that event A has occurred.
Example: If event A is that a person gets neuropathy and event B is that they have diabetes, then P(A|B) is the probability of getting neuropathy given that they have diabetes.
You are reading the Elementary Probability Theory.
If A and B are independent events, the probability of event B is not affected by the probability of event A (and vice versa). In other words, P(B|A) = P(B). Therefore, for independent events, the probability of both event A and event B occurring is given by:
P(A and B) = P(A) x P(B)
Example: If event A is that a person has blood type O and event B is that they are diabetic, then the probability that someone has blood type O and is diabetic is P(A)xP(B), assuming that getting Diabetes is not related to a person’s blood group.
Note that if A and B are mutually exclusive, then P(A|B)=0
Bayes theorem
From the multiplication rule above, we see that:
P(A) x P(B|A) = P(B) x P(A|B)
This leads us to what is known as Bayes’ theorem:
P(B|A)=P(A|B)P(B)P(A)
Thus, the probability of B given A is the probability of A given B multiplied by the probability of B divided by the probability of A. This formula is not suitable if P(A)=0, that is, if A is an event that can’t happen.
Sensitivity and specificity
Many diagnostic test results are given as a continuous variable (that is, they can take any value within a given range), such as diastolic blood pressure or hemoglobin level. However, to facilitate discussion, we will first assume that they have been divided into positive or negative results. For example, a positive diagnostic result for hypertension is a diastolic blood pressure greater than 90 mmHg, while a hemoglobin level of less than 12 g/dl is required for anemia.
For each diagnostic procedure (which may involve laboratory testing of a sample taken) there are a number of fundamental questions that need to be asked. First, if the disease is present, what is the probability that the test result will be positive? This brings us to the notion of test sensitivity.
Second, if the disease is absent, what is the probability that the test result will be negative? This question refers to the specificity of the test. These questions can only be answered if the “true” diagnosis is known. In the case of organic disease, this can be determined by biopsy or, for example, an expensive and risky procedure such as angiography for heart disease. In other situations it may be due to the opinion of an “expert”. These tests constitute the so-called “gold standard”.
Example
Let us consider the results of an N-terminal pro-brain natriuretic peptide (NT-proBNP) test for the diagnosis of heart failure in a general population survey in patients older than 45 years and in patients with an existing diagnosis of heart failure. Heart failure was identified when NT-proBNP > 36 pmol/l.
We denote a positive test result by T+ and a positive diagnosis of heart failure (the disease) by D+. The prevalence of heart failure in these subjects is 103/410=0.251, that is, approximately 25%. Therefore, the probability that a randomly selected subject from the combined group has the disease is estimated to be 0.251. We can write it as P(D+)=0.251.
The sensitivity of a test is the proportion of people with the disease who also have a positive test result. Thus, the sensitivity is given by e/(e+f)=35/103=0.340 or 34%. Sensitivity is the probability that the test result is positive (event T+) if disease is present (event D+) and can be written as P(T+|D+), where “|” means “given”.
The specificity of the test is the proportion of people without disease who give a negative result. Thus, the specificity is h/(g+h)=300/307=0.977 or 98%. Now, the specificity is the probability of a negative test result (event T-) given that the disease is absent (event D-) and can be written as P(T-|D-).
Since the sensitivity is conditional on the presence of the disease and the specificity on its absence, in theory they are not affected by the prevalence of the disease. For example, if we were to double the number of subjects with true heart failure from 103 to 206, so that the prevalence was now 103/(410+103)=20%, then we might expect twice as many subjects to test positive in the proof. Thus, 2×35=70 would have a positive result. In this case, the sensitivity would be 70/206=0.34, which does not vary from the previous value. A similar result is obtained for specificity.
Sensitivity and specificity are useful statistics because they will give consistent results for the diagnostic test in a variety of patient groups with different disease prevalences. This is an important point. Sensitivity and specificity are characteristics of the test, not of the population to which the test is applied. In practice, however, if the disease is very rare, the precision with which sensitivity can be estimated may be limited. This is because the number of subjects with the disease may be small and in this case the proportion correctly diagnosed will have considerable uncertainty.
False Negatives and False Positives
Two other terms in common use are the false negative rate (or probability of a false negative) which is given by f/(e+f)=1-sensitivity and the false positive rate (or probability of a false positive) og /(g+h)=1-specificity.
It is important, for consistency, to always put the true diagnosis at the top and the test result on the side. Since sensitivity=1-P(false negative) and specificity=1-P(false positive), a possibly useful mnemonic to remember this is that “sensitivity” and “negative” have “n” and “specificity” and ” positive” have “p”.
probability ratio
In terms of probabilities, we can summarize Bayes’ theorem using what is known as the positive likelihood ratio (LR+), defined as:
LR+=P(T+|D+)P(T+|D-)=Sensitivity1-Specificity
Therefore, the odds ratio of a positive test is the probability of a positive result when a subject has the disease, relative to the probability of a positive test given that the subject does not have the disease.
It can be shown that Bayes’s theorem is summarized as:
Probability of disease after test = Probability of disease before test x odds ratio.
In this case, the odds ratio is 0.80/(1-0.74)=3.08, so the odds of disease after the test are 3.08×2.33=7.2. This can be verified from the post-test probability of 0.88 calculated above, so that the post-test probabilities are 0.88/(1-0.88)=7.3. (This differs from 7.2 due to rounding errors in the calculation.)
Example 1
This example illustrates Bayes’s theorem in practice by calculating the positive predictive value.
We use the formula:
P(D+|T+)=P(T+|D+)P(D+)P(T+)
With the example data: P(T+)=930/1465=0.63, P(D+)=0.70 and P(T+|D+)=0.80, therefore
Positive predictive value = (0.8 x 0.7)/0.63 = 0.89
…. which is what we have calculated above.
Example 2
The prevalence of a disease is 1 in 1000 and there is a test that can detect it with a sensitivity of 100% and a specificity of 95%. What is the probability that a person has the disease, given a positive test result?
Many people might unthinkingly guess that the answer is 0.95, given the specificity.
However, using Bayes’s theorem
P(D+|T+)=Sensitivity×PrevalenceProbability of positive result
To calculate the probability of a positive result, consider 1,000 people, one of whom has the disease. The test will certainly detect this person. However, it will also give a positive result in 5% of the 999 people without the disease. Thus, the total number of positives is 1+(0.05×999)=50.95 and the probability of a positive result is 50.95/1000=0.05095.
Therefore:
P(D+|T+)=1×0.0010.05095=0.02
The usefulness of a test will depend on the prevalence of the disease in the population to which it has been applied. In general, a useful test is one that significantly modifies the pretest probability. If the disease is very rare or very common, the odds of disease from a negative or positive test are relatively close, so the test is of questionable value.
Independence and mutually exclusive events
In the example, if the results of the exercise tolerance test were not related to whether or not a patient had coronary artery disease, that is, they were independent, we might expect:
P(D+ and T+ )= P(T+) x P(D+)
If we estimate P(D+ and T+) as 815/1465=0.56, P(D+)=1023/1465=0.70 and P(T+)=930/1465=0.63, then the difference:
P(D+ and T+) – P(D+)P(T+)=0.56-(0.70×0.63)=0.12
…. is a rough measure of the independence of these events. In this case, the size of the difference suggests that they are not independent. The question of deciding whether or not the events are independent is clearly important and belongs to statistical inference.
In general, a researcher is not faced with a simple question Does the patient have heart disease?, but with a whole series of different diagnoses. Typically, these diagnoses can be considered mutually exclusive. That is, if the patient has a disease, she does not have any of the alternative differential diagnoses. However, especially in older people, a patient may have several diseases that give similar symptoms.
Students sometimes confuse independent events with mutually exclusive ones, but it follows from the above that mutually exclusive events cannot be independent. The concepts of independence and mutually exclusive events are used to generalize Bayes’ theorem and lead to decision analysis.
We hope that you have understood the concept of Elementary Probability Theory.
