We are given the following information:

• $P(A) = 0.9$
• $P(B) = 0.9$
• $P(A' \cup B') = 0.2$

We are tasked to find $P(AB' \cup A'B)$, which represents the probability that either $A$ occurs but $B$ does not, or $B$ occurs but $A$ does not. This is equivalent to $P(A \Delta B)$ (the symmetric difference of $A$ and $B$).

Step 1: Relation between the complements and union

We know that: $P(A' \cup B') = 1 - P(A \cap B)$ Thus: $P(A \cap B) = 1 - P(A' \cup B') = 1 - 0.2 = 0.8$

Step 2: Decompose the symmetric difference

The event $AB' \cup A'B$ corresponds to the symmetric difference between $A$ and $B$. The formula for the symmetric difference is: $P(A \Delta B) = P(A) + P(B) - 2P(A \cap B)$ Substitute the known values: $P(A \Delta B) = 0.9 + 0.9 - 2 \times 0.8 = 1.8 - 1.6 = 0.2$

Thus, the probability $P(AB' \cup A'B) = 0.2$.

Final Answer:

$P(AB' \cup A'B) = 0.2$

Do you want more details or have any questions?

Here are 5 related questions to explore further:

1. How do you calculate the union of two independent events?
2. What is the meaning of the symmetric difference in probability theory?
3. How does $P(A \cup B)$ relate to the individual probabilities of $A$ and $B$?
4. What happens if $A$ and $B$ are mutually exclusive events?
5. How does Bayes' theorem relate to conditional probabilities of intersections?

Tip: When solving problems involving multiple events, Venn diagrams can be a helpful visual tool to understand the relationships between intersections and unions.