Chapter 3 Joint Probability
3.1 Blood Types
Blood plays a role in many forensic science applications, with the “blood type” determined by a combination of two different blood group systems:
Blood types are inherited and represent contributions from both parents. As of 2019, a total of 41 human blood group systems are recognized by the International Society of Blood Transfusion (ISBT). The two most important blood group systems are ABO and Rh; they determine someone’s blood type (A, B, AB, and O, with +, − or null denoting RhD status) for suitability in blood transfusion.1
Suppose that we have a collection of 1000 people, with each classified based on both ABO (A, B, AB, or O) and Rh (\(+\) or \(-\)). The number of people of each combination of ABO and Rh class (the “blood type”) is shown in the table below.2
| Group | \(+\) | \(-\) |
|---|---|---|
| A | 360 | 60 |
| B | 76 | 14 |
| AB | 25 | 5 |
| O | 390 | 70 |
Suppose that one of these people is selected at random and note that person’s groups. There are various possibilities for outcomes, such as the person selected has ABO group O and Rh group \(+\). The likelihood that this combination occurs is the joint probability of the two classifications, ABO and Rh groupings. There are 390 out of the 1000 people are O and \(+\), so the probability of randomly selecting such a person is
\[\text{P(O and +)} = \frac{390}{1000} = 0.39\]
Here P(O and \(+\)) stands for the probability that the person selected is ABO group O and Rh group \(+\). A sample of other probabilities that come from the table are
\[\begin{array}{rcccl} \text{P(AB and }-) & = & {\displaystyle\frac{5}{1000}} & = & 0.005 \\[5pt] \text{P(B and +)} & = & {\displaystyle\frac{76}{1000}} & = & 0.076 \\[5pt] \text{P(B and }-) & = & {\displaystyle\frac{14}{1000}} & = & 0.014 \\ \end{array}\]
We also can combine table entries to compute other probabilities. For instance, there are a total of \(76 + 14 = 90\) people with blood group B, so the probability of randomly selecting a person with blood group B is
\[ \mbox{P(B)} = \frac{76+14}{1000} = \frac{90}{1000} = 0.09 \]
As another example, the number of people with Rh group \(+\) is \(360 + 76 + 25 + 390 = 851\). Thus the probability of randomly selecting a person with Rh group \(+\) is
\[ \mbox{P(+)} = \frac{360 + 76 + 25 + 390}{1000} = \frac{851}{1000} = 0.851 \]
3.1.1 Sample Questions
1. Find the probability that a randomly selected person has blood groups A and \(-\).
Answer:
2. Find the probability that a randomly selected person has ABO group AB.
Answer:
3. Find the probability that a randomly selected person has ABO group A and ABO group B and Rh group \(+\).
Answer:
3.2 And vs Or
Above we saw that \(\text{P(B and }+) = 0.076\). Suppose we instead would like to know \(\text{P(B or }+)\)? That is, we want to know the probability that a randomly selected person is either ABO group B or Rh group \(+\). Note that this includes those people who are both ABO group B and Rh group \(+\).
From our table we see that the total number of people satisfying one or both conditions is \(360 + 76 + 24 + 390 + 14 = 864\) so that
\[ \underbrace{(14)}_\text{ABO group B} \hspace{-.15in}+ \hspace{.05in}\underbrace{(360 + 25 + 390)}_\text{Rh group +} + \underbrace{(76)}_\text{Both}= 864 \] The corresponding probability is
\[ \text{P(B or +)} = \frac{864}{1000} = 0.864 \]
Other combinations are possible. For instance, the number of people with ABO group A blood or ABO group B blood is \[ \underbrace{(360 + 60)}_\text{ABO group A} + \underbrace{(76 + 14)}_\text{ABO group B} = 510 \]
Therefore we have
\[ \mbox{P(A or B)} = \frac{510}{1000} = 0.51 \]
3.2.1 Sample Questions
1. Find the probability that a randomly selected person has ABO group AB or Rh group \(-\) blood.
Answer:
2. Find the probability that a randomly selected person has ABO group O or Rh group \(+\) blood.
Answer:
3. Find the probability that a randomly selected person has ABO group AB or O.
Answer:
4. Find the probability that a randomly selected person has ABO group A or AB, and also Rh group \(-\).
Answer:
3.3 Probability Tables
Frequently, tables provide probabilities (or percentages) instead of counts. To make the conversion from counts to probabilities, all we do is divide each table entry by the total. Thus for our table of blood type counts above, we divide each entry by 1000 to arrive at Table 3.2.
| Group | \(+\) | \(-\) |
|---|---|---|
| A | 0.360 | 0.060 |
| B | 0.076 | 0.014 |
| AB | 0.025 | 0.005 |
| O | 0.390 | 0.070 |
Table 3.2 entries give the probability of selecting each blood type (determined by the combination of ABO and Rh). For instance,
\[ \mbox{P(B}-) = 0.014 \]
We can add entries in Table 3.2 to compute other probabilities.
For example, suppose we want to compute the probability that a randomly
selected person has ABO group O blood.
Based on the table, this can happen if a person is either O\(+\) or O\(-\).
Because these two groups are disjoint, we can add the probabilities to arrive at
\[ \mbox{P(O}) = \mbox{P(O}+) + \mbox{P(O}-) = 0.390 + 0.070 = 0.460 \]
If instead (for instance) we want to compute P(\(+\)), then we add the entries in the corresponding column, giving us
\[ \mbox{Pr(+}) = 0.360 + 0.076 + 0.025 + 0.390 = 0.851 \]
Note that this matches what we saw earlier.
3.4 Exercises
- \(\text{Put exercises here}\)
Blood type - Wikipedia: https://en.wikipedia.org/wiki/Blood_type↩︎
This table is based on the distribution of blood types in Canada, reported at https://www.blood.ca/en/blood/donating-blood/whats-my-blood-type↩︎