The image contains the following problem:

Problem Statement:

• Consider two events $C$ and $D$ such that $P(C) = 0.4$, $P(D) = 0.6$, and $P(C \cap D) = 0.25$.
• a) Draw a Venn diagram showing the probabilities for events $C$ and $D$.

Solution:

To draw a Venn diagram showing the probabilities, we need to understand the relationships between the given probabilities:

1. $P(C) = 0.4$ is the probability of event $C$ occurring.
2. $P(D) = 0.6$ is the probability of event $D$ occurring.
3. $P(C \cap D) = 0.25$ is the probability of both events $C$ and $D$ occurring together (the intersection of $C$ and $D$).

Using this information, we can calculate the probabilities of the other regions:

• Only $C$: $P(C \text{ only}) = P(C) - P(C \cap D) = 0.4 - 0.25 = 0.15$

• Only $D$: $P(D \text{ only}) = P(D) - P(C \cap D) = 0.6 - 0.25 = 0.35$

• Neither $C$ nor $D$: This is calculated as $1 - P(C \cup D)$, where $P(C \cup D)$ is the probability of either $C$ or $D$ or both occurring.

  $P(C \cup D) = P(C) + P(D) - P(C \cap D) = 0.4 + 0.6 - 0.25 = 0.75$

  So, $P(\text{neither } C \text{ nor } D) = 1 - 0.75 = 0.25$.

Thus, the Venn diagram would show:

• The intersection $C \cap D$ with a probability of $0.25$.
• The region corresponding to only $C$ with a probability of $0.15$.
• The region corresponding to only $D$ with a probability of $0.35$.
• The region outside both circles representing neither $C$ nor $D$ with a probability of $0.25$.

Would you like an illustration of this Venn diagram, or do you need further details on any part of this explanation?

Relative Questions:

1. What is the general formula for calculating $P(C \cup D)$?
2. How would you calculate $P(C \cap D)$ if only $P(C)$ and $P(D)$ were known, assuming independence?
3. What is the significance of the union and intersection in probability theory?
4. How would the Venn diagram change if $C$ and $D$ were mutually exclusive?
5. How does the addition rule for probability apply when events are not mutually exclusive?

Tip:

When dealing with probabilities, always ensure that the sum of all disjoint events (including their complements) adds up to 1. This consistency check can help you avoid mistakes.