We can solve this problem using the formula for the union of two events:

$P(A \cup B) = P(A) + P(B) - P(A \cap B)$

Where:

• $P(A \cup B)$ is the probability of either event A or event B occurring (given as 0.6),
• $P(A)$ is the probability of event A occurring (given as 0.3),
• $P(B)$ is the probability of event B occurring (given as 0.4),
• $P(A \cap B)$ is the probability of both events A and B occurring, which is what we are trying to find.

Plugging the known values into the formula:

$0.6 = 0.3 + 0.4 - P(A \cap B)$

Now, solve for $P(A \cap B)$:

$0.6 = 0.7 - P(A \cap B)$ $P(A \cap B) = 0.7 - 0.6 = 0.1$

Thus, the probability of both events A and B occurring is:

$P(A \cap B) = 0.1$

So, the correct answer is 0.1.

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Would you like a detailed explanation of the steps, or do you have any questions?

Here are 5 related questions:

1. What is the formula for the union of two events?
2. How would the result change if events A and B were mutually exclusive?
3. How do you interpret the probability of the intersection of two events?
4. What is the difference between mutually exclusive and independent events?
5. How do you calculate $P(A \cup B)$ if events A and B are independent?

Tip: For non-mutually exclusive events, remember that the probability of their union is not simply the sum of their individual probabilities. Subtract the intersection to avoid double-counting.