https://file.aiquickdraw.com/mathbot/file/1727197630238z2xt2w0r.jpg

We are given the following information:

\[
P(A') = 0.3, \quad P(B') = 0.4, \quad P(A \cap B) = 0.25
\]

We are asked to find \( P(A \cup B) \).

### Step 1: Use the Complement Rule
We know that:

\[
P(A') = 1 - P(A) \quad \text{and} \quad P(B') = 1 - P(B)
\]

Thus, we can find \( P(A) \) and \( P(B) \):

\[
P(A) = 1 - 0.3 = 0.7
\]
\[
P(B) = 1 - 0.4 = 0.6
\]

### Step 2: Use the Formula for Union of Two Sets
The formula for the union of two events is:

\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Substituting the values:

\[
P(A \cup B) = 0.7 + 0.6 - 0.25
\]
\[
P(A \cup B) = 1.3 - 0.25 = 1.05
\]

Thus, \( P(A \cup B) = 1.05 \), which indicates the sum exceeds 1. In probability, this typically means the events are not mutually exclusive but might involve an overlap beyond normal probability interpretation, suggesting revisiting the conditions.

Would you like further details on this or have any questions?

### 5 Related Questions:
1. How do we interpret a probability value greater than 1 in certain cases?
2. How would this problem change if \( P(A') \) or \( P(B') \) were different?
3. What is the significance of using the complement rule in probability problems?
4. Can events ever have negative probability values in theory or practice?
5. What are other ways to calculate \( P(A \cup B) \) when additional data is provided?

### Tip:
When calculating probabilities, always ensure that the final probability lies between 0 and 1 to reflect realistic outcomes in a probability space.

If P(A') = 0.3, P(B') = 0.4, and P(A ∩ B) = 0.25, find P(A ∪ B).

Solve a probability problem where P(A') = 0.3, P(B') = 0.4, and P(A ∩ B) = 0.25, to find P(A ∪ B). Learn key concepts in probability and set theory, and understand how to apply the formula for union and complement of events.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation