https://file.aiquickdraw.com/mathbot/file/1728762808825uskz2ufy.jpg

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?

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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!

If P(A) = 0.2 and P(B) = 0.4 and P(A ∪ B) = 0.5, find: a) P(A ∩ B) b) P(A' ∪ B)

Solve a probability problem where P(A) = 0.2, P(B) = 0.4, and P(A ∪ B) = 0.5. Learn how to find P(A ∩ B) and P(A' ∪ B) using key probability formulas.

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