From the uploaded image, the problem appears to involve probabilities related to sets $A$ and $B$. Here's the interpretation of the problem:

If $P(A) = 0.35$, $P(B) = 0.25$, $P(A \cap B) = 0.25$, and $P(A \cup B) = 0.6$, find:

1. $P(\text{not } A)$
2. $P(\text{not } B)$

Solutions:

1. For $P(\text{not } A)$: Using the complement rule: $P(\text{not } A) = 1 - P(A)$ Substituting $P(A) = 0.35$: $P(\text{not } A) = 1 - 0.35 = 0.65$

2. For $P(\text{not } B)$: Similarly, using the complement rule: $P(\text{not } B) = 1 - P(B)$ Substituting $P(B) = 0.25$: $P(\text{not } B) = 1 - 0.25 = 0.75$

Summary:

• $P(\text{not } A) = 0.65$
• $P(\text{not } B) = 0.75$

Let me know if you need further clarification or additional questions related to this! Here are some related questions for practice:

1. What is the probability that neither $A$ nor $B$ occurs?
2. How is $P(A \cup B)$ calculated using $P(A)$, $P(B)$, and $P(A \cap B)$?
3. What is the probability of $A$ occurring but not $B$?
4. If $P(B | A)$ is needed, how do you calculate it?
5. What would happen to probabilities if $P(A \cap B) = 0$?

Tip: Always check whether events $A$ and $B$ are independent or mutually exclusive when solving probability problems!