Learn By Doing
So far we have introduced the addition rule for the special case in which the events being considered are disjoint. The purpose of this activity is to make you aware of the danger in wrongly using the addition rule for disjoint events in cases where the events are actually not disjoint. Consider the blood type example again.
Recall the blood type example:
| Blood type | O | A | B | AB |
| Probability | 0.44 | 0.42 | 0.10 | 0.04 |
with the following additional information:
A person with type A can donate blood to a person with type A or AB.
A person with type B can donate blood to a person with type B or AB.
A person with type AB can donate blood to a person with type AB only.
A person with type O blood can donate to anyone.
Suppose that there are two patients who are each in need of a blood donation. Patient 1 has blood type A and patient 2 has blood type B. Consider the following events:
D1; a randomly chosen person can be a donor for patient 1.
D2; a randomly chosen person can be a donor for patient 2.
We are interested in finding the probability that a randomly chosen person can be a donor for patient 1 or patient 2. In other words, we are interested in finding P(D1 or D2).
- Find P(D1) and P(D2). Write your answers in the text box below.
- Are events D1 and D2 disjoint or overlapping?
- Try to (wrongly) apply the addition rule for disjoint events to P(D1 or D2), and explain why the answer you got proves that the addition rule for disjoint events does not work in cases in which the events are not disjoint.