Chapter 4 Conditional Probability

4.1 Restricting the Population

Let’s continue with the blood type example from the preceding sections. For convenience the frequency table is included again below.

Table 4.1: Frequencies by ABO and Rh Groups
Group	\(+\)	\(-\)
A	360	60
B	76	14
AB	25	5
O	390	70

Now suppose that we focus only on those in our the population that have Rh group \(-\). The frequency table for this subpopulation is just the \(-\) column from Table 4.1.

Table 4.2: Frequencies by ABO group among those in Rh\(-\) group
Group	\(-\)
A	60
B	14
AB	5
O	70

There is no reason why we cannot compute probabilities based on this one column of data, provided we specify that is what we are doing. This is an example of conditional probability.

4.2 Conditional Probability

Let’s compute a probability based on the data in Table 4.2. For instance, suppose that a person from the Rh \(-\) group is selected at random. The probability that this person has ABO group A is an example of a conditional probability.

There are a total of 149 people in the Rh group \(-\), with 60 of those having ABO group A. Therefore the probability that an Rh group \(-\) person selected at random is ABO group A is

\[ \mbox{P(ABO group A | Rh group }-) = \frac{60}{149} = 0.403 \]

The vertical bar in the notation means “given that” so that \(\mbox{P(ABO group A | Rh group }-)\) is read as

\[ \mbox{The probability of ABO group A, given that the person is Rh group }-. \] Note that we can abbreviate the notation to \(\mbox{P(A | }-)\) for compactness in cases where the brevity is not confusing. Conditional probabilities arise all the time when evaluating forensic evidence. (In fact almost all probabilities encountered are conditional.)

For another example, suppose that we want to determine the probability that someone is Rh \(+\) given that they are ABO group AB. The notation for this is P(\(+\) | AB).

Referring back to Table 1, we see that there are \(25 + 5 = 30\) people who are ABO group AB, and among those there are 25 who are Rh \(+\). Thus we have

\[ \mbox{P(}+\mbox{ | AB)} = \frac{25}{30} = 0.833 \]

4.2.1 Sample Questions

1. Determine \(\mbox{P(O | }+)\)

Answer: From Table 1 we see that there are a total of \(360 + 76 + 25 + 390 = 851\) in the Rh \(+\) group. Among these are 390 that are ABO group O. Therefore \(\mbox{P(O | }+) = \frac{390}{851} = 0.458\).

2. Determine \(\mbox{P(+ | O})\)

Answer: From Table 1 we see that there are a total of \(390 + 70 = 460\) that are ABO group O. Among these are 390 that are Rh \(+\). Thus we have \(\mbox{P(}+\mbox{ | O)} = \frac{390}{460} = 0.848\).

3. Determine the probability that a random person is ABO group A, given that the person is either ABO group A or ABO group B.

Answer: From Table 1 we see that there are a total of \(360 + 60 + 76 + 14 = 510\) that are ABO group A or ABO group B. Among these are \(360 + 60 = 420\) that are ABO group A. Thus we have \(\mbox{P(A | A or B)} = \frac{420}{510} = 0.824\).

4. Determine the probability that a random person is ABO group AB, given that the person is either ABO group AB or Rh \(-\).

Answer: From Table 1 we see that there are a total of \(60 + 14 + 5 + 70 + 25 = 174\) that are ABO group AB or Rh \(-\). Among these are \(25 + 5 = 30\) that are ABO group AB. Thus we have P(AB | AB or \(-\)) = \(\frac{30}{174}\) = 0.172.

4.3 Exercises

\(\text{Put exercises here}\)