Let P(E) = 0.25, P(F) = 0.35, and P(F ∩ E) = 0.1. Draw a Venn diagram and find the conditional probabilities.
(a)    P(E | FC )



(b)    P(F | EC )

Let's first interpret the given information:

- \( P(E) = 0.25 \) is the probability of event \( E \).
- \( P(F) = 0.35 \) is the probability of event \( F \).
- \( P(F \cap E) = 0.1 \) is the probability of both events \( F \) and \( E \) occurring simultaneously (the intersection of \( F \) and \( E \)).

We are asked to find the conditional probabilities:
- \( (a) \, P(E | F^C) \), the probability of event \( E \) given that \( F \) does not occur.
- \( (b) \, P(F | E^C) \), the probability of event \( F \) given that \( E \) does not occur.

### Venn Diagram
To visualize this problem, we can draw two intersecting circles representing events \( E \) and \( F \), with an overlap indicating \( P(F \cap E) \). The rest of the diagram will show the areas representing \( F^C \) (complement of \( F \)) and \( E^C \) (complement of \( E \)).

### Step 1: Solve \( P(E | F^C) \)

The formula for conditional probability is:

\[
P(E | F^C) = \frac{P(E \cap F^C)}{P(F^C)}
\]

- We already know that \( P(F) = 0.35 \), so \( P(F^C) = 1 - P(F) = 0.65 \).
- To find \( P(E \cap F^C) \), we use:

\[
P(E) = P(E \cap F) + P(E \cap F^C)
\]

\[
0.25 = 0.1 + P(E \cap F^C) \quad \Rightarrow \quad P(E \cap F^C) = 0.25 - 0.1 = 0.15
\]

Now, calculate \( P(E | F^C) \):

\[
P(E | F^C) = \frac{0.15}{0.65} \approx 0.2308
\]

### Step 2: Solve \( P(F | E^C) \)

Similarly, for \( P(F | E^C) \):

\[
P(F | E^C) = \frac{P(F \cap E^C)}{P(E^C)}
\]

- We know that \( P(E) = 0.25 \), so \( P(E^C) = 1 - P(E) = 0.75 \).
- To find \( P(F \cap E^C) \), we use:

\[
P(F) = P(F \cap E) + P(F \cap E^C)
\]

\[
0.35 = 0.1 + P(F \cap E^C) \quad \Rightarrow \quad P(F \cap E^C) = 0.35 - 0.1 = 0.25
\]

Now, calculate \( P(F | E^C) \):

\[
P(F | E^C) = \frac{0.25}{0.75} = \frac{1}{3} \approx 0.3333
\]

### Final Answers:
(a) \( P(E | F^C) \approx 0.2308 \)

(b) \( P(F | E^C) \approx 0.3333 \)

---

Would you like me to generate a Venn diagram illustrating this problem?

### Related Questions:
1. How does the complement rule apply to conditional probabilities?
2. What is the significance of \( P(E \cap F) \) in conditional probability?
3. How can conditional probabilities change when events are dependent?
4. What is the law of total probability, and how does it apply here?
5. How can conditional probability be useful in real-life applications?

#### Tip: Always check if two events are independent before applying formulas, as it simplifies many problems!

Let P(E) = 0.25, P(F) = 0.35, and P(F ∩ E) = 0.1. Draw a Venn diagram and find the conditional probabilities. (a) P(E | FC) (b) P(F | EC)

This problem explores conditional probability with given probabilities P(E) = 0.25, P(F) = 0.35, and P(F ∩ E) = 0.1. Learn how to calculate P(E | F^C) and P(F | E^C) with step-by-step solutions. Suitable for students in Grades 11-12 or college-level introductory probability courses.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation