Let's solve the problem shown in the image. We are given the following information:

• $P(A) = 0.2$
• $P(B) = 0.4$
• $P(A \cup B) = 0.5$

We are asked to find:

a) $P(A \cap B)$

b) $P(A' \cup B)$

a) Finding $P(A \cap B)$

We can use the formula for the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Substituting the given values:

$0.5 = 0.2 + 0.4 - P(A \cap B)$

Simplifying:

$0.5 = 0.6 - P(A \cap B)$

Solving for $P(A \cap B)$:

$P(A \cap B) = 0.6 - 0.5 = 0.1$

So, $P(A \cap B) = 0.1$.

b) Finding $P(A' \cup B)$

Recall that $A'$ is the complement of $A$, and the formula for the union of complements is:

$P(A' \cup B) = 1 - P(A \cap B')$

However, it’s easier to use De Morgan's Law:

$P(A' \cup B) = 1 - P(A \cap B)$

Since we already found $P(A \cap B) = 0.1$:

$P(A' \cup B) = 1 - 0.1 = 0.9$

So, $P(A' \cup B) = 0.9$.

Summary:

• a) $P(A \cap B) = 0.1$
• b) $P(A' \cup B) = 0.9$

Would you like any further explanation or have any questions?

Here are 5 related questions to extend the problem:

1. How would you find $P(A \cup B')$ given the same information?
2. What is the probability of $A \cup B \cup C$ if a third event $C$ is introduced?
3. How do you calculate conditional probability $P(A|B)$ using the values provided?
4. Can $P(A \cap B)$ ever exceed $P(A)$ or $P(B)$? Why or why not?
5. What changes in the calculation if $A$ and $B$ are mutually exclusive?

Tip: Always check for independence or mutual exclusivity when dealing with events in probability. It simplifies calculations!