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.