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The problem provides a Venn diagram that displays the probabilities of three events, \( A \), \( B \), and \( C \), along with several related probability questions. I'll break it down and solve each part:

### The information in the Venn diagram:
- \( P(A \cap B \cap C) = 0.1 \)
- \( P(A \cap B \cap C^c) = 0.08 \) (where \( C^c \) is the complement of \( C \))
- \( P(A \cap B^c \cap C) = 0.05 \) 
- \( P(A^c \cap B \cap C) = 0.12 \)
- \( P(A \cap B^c \cap C^c) = 0.2 \)
- \( P(A^c \cap B \cap C^c) = 0.1 \)
- \( P(A^c \cap B^c \cap C) = 0.15 \)

Now let's solve the questions:

### (a) \( P(A | B) \) 
This is the conditional probability of \( A \) given \( B \), calculated as:

\[
P(A | B) = \frac{P(A \cap B)}{P(B)}
\]

From the diagram:
- \( P(A \cap B) = P(A \cap B \cap C) + P(A \cap B \cap C^c) = 0.1 + 0.08 = 0.18 \)
- \( P(B) = P(A \cap B) + P(A^c \cap B \cap C) + P(A^c \cap B \cap C^c) = 0.18 + 0.12 + 0.1 = 0.4 \)

Thus:

\[
P(A | B) = \frac{0.18}{0.4} = 0.45
\]

### (b) \( P(C | A') \)
This is the conditional probability of \( C \) given \( A' \) (the complement of \( A \)):

\[
P(C | A') = \frac{P(C \cap A')}{P(A')}
\]

From the diagram:
- \( P(C \cap A') = P(A^c \cap B \cap C) + P(A^c \cap B^c \cap C) = 0.12 + 0.15 = 0.27 \)
- \( P(A') = 1 - P(A) = 1 - (0.2 + 0.1 + 0.05 + 0.08 + 0.1) = 1 - 0.53 = 0.47 \)

Thus:

\[
P(C | A') = \frac{0.27}{0.47} \approx 0.574
\]

### (c) \( P(A \cap B | C^c) \)
This is the conditional probability of \( A \cap B \) given \( C^c \) (the complement of \( C \)):

\[
P(A \cap B | C^c) = \frac{P(A \cap B \cap C^c)}{P(C^c)}
\]

From the diagram:
- \( P(A \cap B \cap C^c) = 0.08 \)
- \( P(C^c) = 1 - P(C) = 1 - (0.1 + 0.05 + 0.12 + 0.15) = 1 - 0.42 = 0.58 \)

Thus:

\[
P(A \cap B | C^c) = \frac{0.08}{0.58} \approx 0.138
\]

### (d) \( P(C | A' \cup B') \)
This is the conditional probability of \( C \) given \( A' \cup B' \) (the union of the complements of \( A \) and \( B \)):

\[
P(C | A' \cup B') = \frac{P(C \cap (A' \cup B'))}{P(A' \cup B')}
\]

First, calculate \( P(A' \cup B') \):
- \( A' \cup B' \) covers all regions outside of \( A \cap B \), i.e., it includes \( A^c \cap B \cap C \), \( A^c \cap B \cap C^c \), \( A^c \cap B^c \cap C \), and \( A^c \cap B^c \cap C^c \).

So, we have:
\[
P(A' \cup B') = 0.12 + 0.1 + 0.15 + 0.2 = 0.57
\]

Now, calculate \( P(C \cap (A' \cup B')) \):
- This includes the regions \( A^c \cap B \cap C \) and \( A^c \cap B^c \cap C \), so:
\[
P(C \cap (A' \cup B')) = 0.12 + 0.15 = 0.27
\]

Thus:

\[
P(C | A' \cup B') = \frac{0.27}{0.57} \approx 0.474
\]

---

Let me know if you'd like further details or clarifications on any part.

### Related Questions:
1. How do we find the complement of a set in probability?
2. What are the different ways to compute the union of events in probability?
3. Can you explain the concept of conditional probability with a simple example?
4. How do independent and mutually exclusive events differ in probability?
5. What is the significance of the intersection of sets in probability?

### Tip:
In conditional probability, always ensure the denominator reflects the total probability of the condition being true.

The Venn diagram shows the probabilities of three events A, B, and C. Find: a) P(A | B), b) P(C | A'), c) P(A ∩ B | C^c), d) P(C | A' ∪ B')

Learn how to solve conditional probability problems using Venn diagrams. This example covers calculating P(A | B), P(C | A'), and more for events A, B, and C. Suitable for Grades 10-12 students studying probability and set theory.

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